Class 7

For better exam readiness, students can include Class 7 Science MCQ and Curiosity Chapter 5 Changes Around Us: Physical and Chemical MCQ in their daily study routine.

Class 7 Science Chapter 5 Changes Around Us: Physical and Chemical MCQ

Changes Around Us: Physical and Chemical Class 7

MCQ On Changes Around Us: Physical and Chemical Class 7

Question 1.
Which of the following is a physical change?
(a) Rusting of iron
(b) Combustion of magnesium ribbon
(c) Burning of candle
(d) Melting of wax
Answer:
(d) Melting of wax is a physical change, as no new substance is formed.

Question 2.
A cube of ice, 2 cm in length, is left out in a bowl, and it turns to water. The bowl is then kept in the freezer. The ice formed has the diameter of the bowl. What kind of change occurs in this case?
(a) A physical change with a change in shape and state.
(b) A chemical change with a change in state.
(c) A physical change where solid ice directly changes from a cube to a thick circle.
(d) A chemical change with the formation of a new substance.
Answer:
(a) A physical change with a change in shape and state.

Question 3.
What type of changes occur in the wax and the wick of a burning candle?
(a) Physical change in the wick; chemical change in the wax
(b) Chemical change in the wick, physical change in the wax
(c) Chemical change in both the wax and wick; physical change in some wax
(d) Physical change in both the wick and the wax; chemical change in the wick
Answer:
(c) Burning of wax near the wick involves a chemical change because combustion of both wick and wax vapour takes place, while melting, solidification and evaporation of wax involve physical changes.

Question 4.
In an experiment, Ravina, a class 7 science teacher, demonstrated the burning of magnesium ribbon. During the experiment, a white-coloured substance, X, was formed, accompanied by the evolution of heat and light. Which of the following statements is correct based on this experiment?
(a) The substance X is magnesium carbonate, and there occurs only a change in the state of magnesium.
(b) The substance X is magnesium hydroxide, and there occurs no change in the chemical composition of magnesium.
(c) The substance X is magnesium oxide, and the reaction involved is combustion.
(d) The substance X is magnesium carbonate, and a chemical reaction is involved in this process.
Answer:
(c) When a magnesium ribbon is burnt in air, the following chemical reaction takes place:

Hence, the white-coloured substance X formed on burning magnesium ribbon is magnesium oxide. This reaction is classified as a combustion reaction.

Question 5.
Nitin, Rashmi, Sunil and Rohit were discussing the topic ‘Physical and Chemical changes around Us’. During their discussion, they made the following statements:
Nitin: No new substance is formed during the rusting of iron.
Rashmi: Wood and kerosene are combustible substances.
Sunil: Burning of a candle involves only a physical change.
Rohit: Curdling of milk and ripening of fruits are chemical changes.
(a) Nitin and Sunil are correct
(b) Nitin and Rashmi are correct
(c) Rohit and Sunil are correct
(d) Rashmi and Rohit are correct
Answer:
(d) A new substance called iron oxide is formed during the rusting of iron. Burning of a candle involves both physical and chemical changes.

Question 6.
Which of the following statements is incorrect for a chemical reaction?
(a) Heat may be given out but never absorbed.
(b) Sound may be produced.
(c) A colour change may take place.
(d) A gas may be evolved.
Answer:
(a) Heat may be given out but never absorbed.

Question 7.
In a bowl a few fresh fruits were placed. The next day it was observed that the fresh fruits had ripened. What kind of change occurred within the fresh fruits?
(a) Chemical change as new substances are formed during ripening.
(b) Physical change as the colour of fruit changes.
(c) Chemical change as the shape of the fruit changes.
(d) Physical change as the state of the fruit alone changes.
Answer:
(a) Ripening of fruits is an irreversible chemical change.

Question 8.
Read the following passage carefully and fill in the blanks by selecting the correct option.
A black-coloured rock called basalt, containing (i) chemically changes to form a red coloured layer of (ii), in a process called (iii). (i), (ii) and (iii) are respectively,
(a) Magnesium, magnesium oxide and erosion
(b) Iron, iron oxide and weathering
(c) Magnesium, magnesium oxide and combustion
(d) Iron, iron oxide and erosion
Answer:
(b) Iron, iron oxide and weathering

Question 9.
A student collects wax from a burning candle. He melts it and then adds a new wick to it and leaves it to cool. Can a new and functional candle be created by this process? Why or why not?
(a) Yes; it is a reversible physical change in state of wax.
(b) No; it is a physical change that cannot be reversed.
(c) No; it is an irreversible chemical change with the formation of a new substance.
(d) Yes; it a chemical change causing the wax to change from one state to another.
Answer:
(a) Yes, a new functional candle can be made by melting the wax of an old candle. This is because the melting of wax is a reversible physical change.

Question 10.
A woman mixes flour, milk, eggs and water to create a batter, for baking a cake. The steps in the process are:
Step 1: Mix flour, water, eggs, sugar and milk in a bowl.
Step 2: Place the batter in a baking tray and bake a cake in the oven.
Which types of changes do each of these steps represent?
(a) 1: Physical change, 2: Chemical change
(b) 1: Physical change, 2: Physical change
(c) 1: Chemical change, 2: Chemical change
(d) 1: Chemical change, 2: Physical change
Answer:
(a) Since no new substance is formed in step 1, it is a physical change. Step 2 involves baking the cake, hence it is a chemical change as a new substance, i.e., cake is formed, that has different properties from those of the original batter.

Question 11.
Given below are the pictures of certain processes/ events that you may observe in your daily lives:

Based on your observation, select the correct statements from the following:
(a) Processes A and E are reversible, while B, C and D are irreversible.
(b) Only process A is reversible, while the rest are irreversible.
(c) Only process C is reversible, while the rest are irreversible.
(d) Processes A and D are reversible, while B, C and E are irreversible.
Answer:
(b) Only process A is reversible, while the rest are irreversible.

Question 12.
Pooja prepared a birthday card for her mother and the process involved the following steps:
(i) Folding of the paper
(ii) Cutting of the paper
(iii) Pasting stickers on the paper
Which of the following options correctly identifies the types of changes taking place?

	(i)	(ii)	(iii)
(a)	Reversible physical	Irreversible chemical	Irreversible physical
(b)	Irreversible chemical	Reversible physical	Reversible physical
(c)	Reversible physical	Irreversible physical	Reversible physical
(d)	Reversible physical	Reversible physical	Irreversible physical

Answer:
(c)

Question 13.
Which of the following is/are true when milk changes into curd?
(i) Its state is changed from liquid to semi-solid.
(ii) It changes colour.
(iii) It changes taste.
(iv) The change cannot be reversed.
Choose the correct option from below.
(a) (i) and (ii) are correct.
(b) (ii) and (iii) are correct.
(c) (i), (iii) and (iv) are correct.
(d) (i) and (iv) are correct
Answer:
(c) (i), (iii) and (iv) are correct.

Question 14.
The given experimental setup shows that lime water turns milky in the presence of a gas X. Which of the following statements are incorrect regarding this experiment?

(i) The lime water turns milky due to the formation of calcium hydrogen carbonate.
(ii) Y must be an acid like acetic acid.
(iii) The process involves an irreversible physical change, as there occurs only a change in the colour of the original mixture.
(iv) The gas X is hydrogen, which turns lime water milky.
(a) (i) and (ii) only
(b) (ii), (iii), and (iv)
(c) (iii) and (iv) only
(d) (i), (iii), and (iv)
Answer:
(d) The lime water turns milky due to the formation of calcium carbonate, which is insoluble in water. The process involves a chemical change as a new substance (calcium carbonate) is formed. The gas X is carbon dioxide, which turns lime water milky.

Question 15.
Which of the following statements are correct about combustion?
(i) In this reaction, a substance reacts with carbon dioxide and produces heat and light.
(ii) In this reaction, a substance reacts with oxygen and produces heat and/or light.
(iii) Combustion can only take place if the heat supplied is sufficient to reach the ignition temperature of the combustible substance.
(iv) All substances undergo combustion at all temperatures.
(a) (ii) and (iv)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(c) In a combustion reaction, a given substance reacts with oxygen present in the air and produces heat and light. Only combustible substances undergo combustion, and that too at a certain minimum temperature. A given combustible substance cannot burn if the heat supplied is not sufficient enough to reach its ignition temperature.

Question 16.
Divya, a student of class 7, placed two identical candles on two separate Petri dishes and lit them. She covered one of these with a glass tumbler as shown in the given experimental setup:

Based on the experiment, select the correct statements from the following:
(i) The candle in case (I) will get extinguished earlier in comparison to the candle in case (II).
(ii) The candle in case (II) will get extinguished earlier in comparison to the candle in case (I).
(iii) The component of air that supports combustion is oxygen.
(iv) The component of air that supports combustion is carbon dioxide.
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(d) The candle in case (I) has a continuous supply of air (oxygen), while the supply of air gets discontinued in case (II) as the candle gets covered with a glass tumbler. Hence, the candle in case (II) will get extinguished earlier in comparison to the candle in case (I). Also, the component of air that supports combustion is oxygen.

Question 17.
In a class discussion on physical and chemical changes, the teacher asked the students to determine whether the burning of a candle is a physical or a chemical change.
The students had the following discussion:
Student X: It involves only a physical change as only melting and solidification of the wax occur in this process.
Student Y: It involves only a chemical change as the wax burns to produce heat.
Student Z: It involves both physical and chemical changes as the wax melts, evaporates, and solidifies (physical changes), and the vapour of wax burns to produce a flame (chemical change).
The correct statement is provided by which of the above students?
(a) Students X and Y
(b) Student Y only
(c) Students Y and Z
(d) Student Z only
Answer:
(d) On burning a candle, the wax of the candle melts and the melted wax is carried up the wick, which then evaporates due to the heat of the flame. The wax vapour burns to produce a flame. Here, the melting of wax, its solidification, and evaporation are classified as physical changes. The burning of wax vapour in the flame represents a chemical change. Thus, burning a candle involves both physical and chemical changes.

Question 18.
In which of the following processes is a chemical change taking place?
(i) Heating water in a pan on a Bunsen burner
(ii) Folding a paper to make a boat
(iii) Cooking food
(iv) Folding a bed sheet
(a) (i) and (ii) only
(b) (iii) only
(c) (iii) and (iv) only
(d) (i), (ii) and (iii)
Answer:
(c) (iii) and (iv) only

Changes Around Us: Physical and Chemical Class 7 Assertion and Reason Questions

The following questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

1. (A): Melting of wax is an irreversible physical change.
(R): Only the state of matter changes during this process.
Answer:
(d) Melting of wax is a reversible physical change.

2. (A): After dissolving salt in water, it cannot be recovered by any physical method.
(R): Dissolving salt in water is a physical change.
Answer:
(d) After dissolving salt in water, it can be recovered by evaporation, which is a physical method.

3. (A): A physical change is always reversible.
(R): A physical change is accompanied by a change in physical properties of the original substance.
Answer:
(d) A physical change may be reversible as well as irreversible. For example, breaking a glass tumbler into smaller pieces is a physical, irreversible change.

4. (A): If a burning candle is covered with a glass tumbler, it stops burning after some time.
(R): Oxygen is required for combustion.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

5. (A): Ripening of fruits is a chemical change.
(R): During ripening, a new substance is formed.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

6. (A): Turning of lime water milky is a chemical change.
(R): Lime water turns milky in the presence of carbon dioxide due to the formation of a new substance, i.e., calcium carbonate.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Changes Around Us: Physical and Chemical Class 7 Fill in the Blanks

1. Souring of milk is an example of a ………… change.
Answer: chemical

2. Iron benches kept in lawns and gardens acquire …………. This process involves a ………..change because a new substance is formed.
Answer: rust, chemical

3. Grinding of wheat grain changes its size. It is a…………(physical/chemical) change.
Answer: physical

4. The process that results in the formation of new substances during a …………change is known as a …………reaction.
Answer: chemical, chemical

5. Cooking of food is a …………(desirable/undesirable) change.
Answer: desirable

6. ………… gas when passed through lime water turns it milky due to the formation of …………………
Answer: Carbon dioxide, calcium carbonate

Changes Around Us: Physical and Chemical Class 7 True or False

1. When a candle burns, both physical and chemical changes take place.
Answer: True

2. The component of air that supports combustion is carbon dioxide.
Answer: False: The component of air that supports combustion is oxygen.

3. Stretching of a rubber band is not a physical change.
Answer: False: Stretching of a rubber band is a physical change.

4. A physical change does not involve the formation of a new substance.
Answer: True

5. Rusting of iron is an undesirable chemical change.
Answer: True

6. The movement of rock pebbles, soil, and sediments (after breaking down) from one place to another by wind or flowing water to form sand is called erosion.
Answer: True

Changes Around Us: Physical and Chemical Class 7 Match the Following

Question 1.

Answer:

	Column I		Column II
(i)	Chopping of vegetables	C.	Irreversible physical change
(ii)	Rusting of iron	B.	Chemical change
(iii)	Melting of ice cubes	A.	Reversible physical change
(iv)	Burning of a candle	D.	Both physical and chemical changes

Question 2.

Answer:

	Column I		Column II
(i)	Folding a sheet of paper	D.	Reversible physical change
(ii)	Magnesium + Oxygen	A.	Combustion
(iii)	Iron + Oxygen + Water	B.	Rusting
(iv)	Cutting a sheet of paper	C.	Irreversible physical change

Students can use Class 7 Math Solution Odia Medium and Class 7 Maths Chapter 8 Question Answer Odia Medium ଭଗ୍ନସଂଖ୍ୟାର ବ୍ୟବହାର to check their answers after solving exercises.

7th Class Maths Chapter 8 Question Answer Odia Medium

Class 7 Maths Chapter 8 Odia Medium

8.1 ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣନ :

Page No. (176 to 189)

Question 1.
ଅମ୍ଳାନ 1 ଘଣ୍ଟାରେ 3 କିଲୋମିଟର ବାଟ ଚାଲିପାରେ । ତେବେ ସେ 5 ଘଣ୍ଟାରେ କେତେ ଦୂର ବାଟ ଚାଲିପାରିବ ?
Solution:
ଅମ୍ଳାନ 1 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରୁଥିବା ଦୂରତା = 3 କି.ମି. ।
5 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରୁଥିବା ଦୂରତା = (5 × 3) କି.ମି. = (3 + 3 + 3 + 3 + 3) କି.ମି. = 15 କି.ମି. ।
∴ ଅମ୍ଳାନ 5 ଘଣ୍ଟାରେ 15 କି.ମି. ଚାଲିପାରିବ ।
କୌଣସି ସଂଖ୍ୟାର କ୍ରମିକ ଯୋଗକୁ ଆମେ ଗୁଣନ କହିଥାଉ ।

Question 2.
ଅମ୍ଳାନର ପୋଷା ବିରାଡ଼ି 1 ଘଣ୍ଟାରେ ମାତ୍ର ମୁଁ \(\frac{1}{4}\) କି.ମି. ଚାଲିପାରେ । ସେହି ବେଗରେ ସେ 3 ଘଣ୍ଟାରେ କେତେ ଦୂର ଚାଲିପାରିବ ?
Solution:

1 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା = \(\frac{1}{4}\) କି.ମି. ।
3 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା = 3 × \(\frac{1}{4}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\) କି.ମି. ।
(ଏଠାରେ 3ଟି \(\frac{1}{4}\) ଯୋଗଫଳ = 3 × \(\frac{1}{4}\))
∴ ବିରାଡ଼ିଟି 3 ଘଣ୍ଟାରେ \(\frac{3}{4}\) କି.ମି. ଦୂରତା ଅତିକ୍ରମ କରିପାରିବ ।

Question 3.
ଅମ୍ଳାନ 1 ଘଣ୍ଟାରେ 3 କି.ମି. ବାଟ ଚାଲିଲେ \(\frac{1}{5}\) ଘଣ୍ଟାରେ କେତେ ଦୂର ଚାଲିପାରିବ ?
Solution:

1 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରାଯାଇଥିବା ଦୂରତା = 3 କି.ମି. ।
\(\frac{1}{5}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରାଯାଇଥିବା ଦୂରତା = 3 × \(\frac{1}{5}\) କି.ମି. = \(\frac{3}{5}\) କି.ମି. ।

Question 4.
\(\frac{2}{5}\) ଘଣ୍ଟାରେ ଅମ୍ଳାନ କେତେ ଦୂର ଚାଲିପାରିବ ?
Solution:

\(\frac{2}{5}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିବ = \(\frac{2}{5}\) × 3 କି.ମି.= \(\frac{6}{5}\) ଓ କି.ମି. ।
(1) ଆମେ ପ୍ରଥମେ \(\frac{1}{5}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା ନିର୍ଣ୍ଣୟ କରିବା ।
(2) ଯେହେତୁ \(\frac{2}{5}\) ହେଉଛି \(\frac{1}{5}\) ର 2 ଗୁଣ ।

ଏଣୁ ଆମେ ଏହି ଦୂରତାକୁ 2 ଦ୍ଵାରା ଗୁଣନ କଲେ
\(\frac{2}{5}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ମୋଟ ଦୂରତା ପାଇପା `ବା ।
1 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା = 3 କି.ମି. ।
(1) \(\frac{1}{5}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥୁବା ଦୂରତା = 3 କି.ମି.କୁ ସମାନ 5 ଭାଗ କଲେ ପାଇବା \(\frac{1}{5}\) କି.ମି. ।

(2) ଏହି ଦୂରତାକୁ 2 ଦ୍ଵାରା ଗୁଣନ କଲେ ଆମେ ପାଇବା 2 x \(\frac{3}{5}\) = \(\frac{6}{5}\) କି.ମି. ।
ଏଥୁରୁ ଆମେ ଦେଖୁଲୁ ଯେ \(\frac{2}{5}\) × 3 = \(\frac{6}{5}\)
ଆମେ ପ୍ରଥମେ ଗୁଣକ 3 କୁ \(\frac{2}{5}\) ର ହର 5 ଦ୍ଵାରା ଭାଗ କରି \(\frac{3}{5}\) ପାଇଲୁ ଓ ସେହି ଫଳାଫଳକୁ ଗୁଣ୍ୟର ଲବ ଦ୍ଵାରା ଅର୍ଥାତ୍ 2 ଦ୍ଵାରା ଗୁଣନ କରି \(\frac{6}{5}\) ପାଇଲୁ ।

Question 5.
ଉଦାହରଣ – 1 : ଜଣେ ମହିଳା ଚାଷୀଙ୍କର 5 ଜଣ ନାତିନାତୁଣୀ ଥିଲେ । ସେ ତାଙ୍କର ପ୍ରତ୍ୟେକ ନାତି ନାତୁଣୀଙ୍କୁ \(\frac{2}{3}\) ଏକର ଲେଖାଏଁ ଜମି ଭାଗକରି ଦେଲେ ସେ ସମସ୍ତ ନାତିନାତୁଣୀଙ୍କୁ କେତେ ଜମି ଦେଇଥିଲେ ?
Solution:
ସମସ୍ତ ନାତୁଣୀ ମୋଟ ଜମି ପାଇବେ = 5 × \(\frac{2}{3}\) = \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) = \(\frac{2}{3}\) + \(\frac{2}{3}\) = \(\frac{10}{3}\)

Question 6.
ଉଦାହରଣ – 2 : 1 ଘଣ୍ଟାରେ ଇଣ୍ଟରନେଟ୍ ପାଇଁ ଖର୍ଚ୍ଚ 8 ଟଙ୍କା ହେଲେ, 1\(\frac{1}{4}\) ଘଣ୍ଟାରେ ଇଣ୍ଟରନେଟ୍ ମୂଲ୍ୟ କେତେ ହେବ ?
Solution:
1 ଘଣ୍ଟାର ଇଣ୍ଟରନେଟ୍ ପାଇଁ ଖର୍ଚ୍ଚ = 8 ଟଙ୍କା । 1\(\frac{1}{4}\) ଘଣ୍ଟା = \(\frac{5}{4}\) ଘଣ୍ଟା ।
\(\frac{5}{4}\)ଘଣ୍ଟା ପାଇଁ ଇଣ୍ଟରନେଟ୍ ଖର୍ଚ୍ଚ = \(\frac{5}{4}\) × 8 ଟଙ୍କା = 5 × \(\frac{8}{4}\) ଟଙ୍କା = 10 ଟଙ୍କା ।
∴ 1\(\frac{1}{4}\) ଘଣ୍ଟାର ଇଣ୍ଟରନେଟ୍ ପାଇଁ ଖର୍ଚ୍ଚ 10 ଟଙ୍କା ହେବ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ରମେଶ ପ୍ରତ୍ୟେକ ଦିନ \(\frac{1}{2}\) ଗ୍ଲାସ୍ କ୍ଷୀର ପିଏ । ସେ ଏକ ସପ୍ତାହରେ କେତେ ଗ୍ଲାସ୍ କ୍ଷୀର ପିଇବ ? ସେ ଜାନୁଆରୀ ମାସରେ କେତେ କ୍ଷୀର ପିଇବ ?
Solution:
1 ସପ୍ତାହ = 7 ଦିନ । ରମେଶ ଗୋଟିଏ ଦିନରେ କ୍ଷୀର ପିଏ = \(\frac{1}{2}\) ଗ୍ଲାସ୍ ।
ରମେଶ 7 ଦିନରେ କ୍ଷୀର ପିଏ 7 × \(\frac{1}{2}\) ଗ୍ଲାସ୍ = \(\frac{7}{2}\) ଗ୍ଲାସ୍ = 3\(\frac{1}{2}\) ଗ୍ଲାସ୍ ।
∴ ରମେଶ ଏକ ସପ୍ତାହରେ 3\(\frac{1}{2}\) ଗ୍ଲାସ୍ କ୍ଷୀର ପିଏ ।
ଜାନୁୟାରୀ ମାସ = 31 ଦିନ ।
ରମେଶ ଗୋଟିଏ ଦିନରେ କ୍ଷୀର ପିଏ = \(\frac{1}{2}\) ଗ୍ଲାସ୍
ରମେଶ 31 ଦିନରେ କ୍ଷୀର ପିଏ = \(\frac{1}{2}\) × 31 ଗ୍ଲାସ୍ = \(\frac{31}{2}\) କ୍ଲାସ୍ = 15\(\frac{1}{2}\) ଗ୍ଲାସ୍ ।
∴ ରମେଶ ଜାନୁୟାରୀ ମାସରେ 15\(\frac{1}{2}\) ଗ୍ଲାସ୍ କ୍ଷୀର ପିଏ ।

Question 2.
ଏକ ଶ୍ରମିକ ଦଳ 8 ଦିନରେ 1 କିଲୋମିଟର କେନାଲ ତିଆରି କରନ୍ତି । ଏକ ଦିନରେ ସେହି ଶ୍ରମିକ ଦଳଟି କେତେ କିଲୋମିଟର କେନାଲ ତିଆରି କରିପାରିବେ ? ଯଦି ସେମାନେ ଗୋଟିଏ ସପ୍ତାହରେ 5 ଦିନ କାମ କରନ୍ତି, ତେବେ ସେହି ସପ୍ତାହରେ କେତେ କି.ମି. କେନାଲ ତିଆରି କରିପାରିବେ ?
Solution:
ଏକ ଶ୍ରମିକ ଦଳ 8 ଦିନରେ 1 କିଲୋମିଟର କେନାଲ ତିଆରି କରନ୍ତି ।
1 ଦିନରେ କେନାଲ ତିଆରି କରିବେ = \(\frac{1}{8}\) କି.ମି. ।
5 ଦିନରେ କେନାଲ ତିଆରି କରିବେ = \(\frac{1}{8}\) × 5 କି.ମି. = \(\frac{5}{8}\) କି.ମି. ।
∴ ତେବେ ସେହି ସପ୍ତାହରେ \(\frac{5}{8}\) କି.ମି. । କେନାଲ ତିଆରି କରିପାରିବେ ।

Question 3.
ମଞ୍ଜୁ ଓ ତା’ର ଦୁଇଜଣ ପଡ଼ୋଶୀ ପ୍ରତି ସପ୍ତାହରେ 5 ଲିଟର ତେଲ କିଣି, ଏହାକୁ 3 ପରିବାର ମଧ୍ୟରେ ସମାନ ଭାବରେ ବାଣ୍ଟି ନିଅନ୍ତି । ପ୍ରତ୍ୟେକ ପରିବାର ସପ୍ତାହରେ କେତେ ତେଲ ପାଆନ୍ତି ? ପ୍ରତ୍ୟେକ ପରିବାର ଚାରି ସପ୍ତାହରେ କେତେ ତେଲ ପାଆନ୍ତି ।
Solution:
ପ୍ରତି ସପ୍ତାହରେ 5 ଲିଟର ତେଲ କିଣି, ଏହାକୁ 3 ପରିବାର ମଧ୍ୟରେ ସମାନ ଭାବରେ ବାଣ୍ଟି ଦେଲେ ପ୍ରତ୍ୟେକ ପରିବାର ପାଇବେ = 6\(\frac{5}{3}\) ଲିଟର ।
ପ୍ରତ୍ୟେକ ପରିବାର ଚାରି ସପ୍ତାହରେ ପାଇବେ = \(\frac{5}{3}\) × 4 ଲିଟର = \(\frac{20}{3}\) ଲିଟର = 6\(\frac{2}{3}\) ଲିଟର ।
∴ ପ୍ରତ୍ୟେକ ପରିବାର ଚାରିସପ୍ତାହରେ 6\(\frac{2}{3}\) ଲିଟର ତେଲ ପାଆନ୍ତି ।

Question 4.
ସୋମବାର ରାତି 10ଟାରେ ସୋଫିଆ ଚନ୍ଦ୍ର ଦେଖ‌ିଲେ । ତାଙ୍କ ମାଆ ଜଣେ ବୈଜ୍ଞାନିକ । ସେ ତାଙ୍କୁ କହିଲେ ଯେ ପ୍ରତ୍ୟେକ ଦିନ ଚନ୍ଦ୍ର ପୂର୍ବଦିନ ଅପେକ୍ଷା \(\frac{5}{6}\) ଘଣ୍ଟା ବିଳମ୍ବରେ ଦେଖାଦିଏ । ତେବେ ସେହି ସପ୍ତାହର ଗୁରୁବାର ରାତି 10ଟା ପରେ କେଉଁ ସମୟରେ ଚନ୍ଦ୍ର ଦେଖାଦେବ ?
Solution:
ସୋମବାର ରାତି 10 ଟାରେ ସୋଫିଆ ଚନ୍ଦ୍ର ଦେଖ‌ିଲେ ।
ପ୍ରତ୍ୟେକ ଦିନ ଚନ୍ଦ୍ର ପୂର୍ବଦିନ ଅପେକ୍ଷା \(\frac{5}{6}\) ଘଣ୍ଟା ବିଳମ୍ବରେ ଦେଖାଯାଏ ।
ସୋମବାର ଦେଖାଦିଏ = 10 ଘଣ୍ଟାରେ ।
ମଙ୍ଗଳବାର ଦେଖାଯିବ = (10 + \(\frac{5}{6}\)) ଘଣ୍ଟାରେ ।
ବୁଧବାର ଦେଖାଯିବ = (10 + \(\frac{5}{6}\) + \(\frac{5}{6}\) ଘଣ୍ଟାରେ ।
ଗୁରୁବାର ଦେଖାଯିବ = (10 + \(\frac{5}{6}\) + \(\frac{5}{6}\) + \(\frac{5}{6}\)) ଘଣ୍ଟାରେ ।
ସରଳ କଲେ (10 + \(\frac{5}{6}\) + \(\frac{5}{6}\) + \(\frac{5}{6}\)) = (10 + \(\frac{15}{6}\)) = 12 : 30 AM (ରାତି 12 ଟା 30 ମିନିଟ୍‌ରେ)
∴ ସେହି ସପ୍ତାହର ଗୁରୁବାର ରାତି 10ଟା ପରେ 12 ଟା 30 ମିନିଟ୍ (ଅର୍ଥାତ୍ 2 ଘଣ୍ଟା 30 ମିନିଟ୍ ବିଳମ୍ବରେ) ଚନ୍ଦ୍ର ଦେଖାଦେବ ।

Question 5.
ନିମ୍ନସ୍ଥ ପ୍ରଶ୍ନଗୁଡ଼ିକୁ ପ୍ରଥମେ ଗୁଣନ କର । ତା’ପରେ ଗୁଣଫଳକୁ ମିଶ୍ର ସଂଖ୍ୟାରେ ପ୍ରକାଶ କର ।
(a) 7 × \(\frac{3}{5}\)
(b) 4 × \(\frac{1}{3}\)
(c) \(\frac{9}{7}\) × 6
(d) \(\frac{13}{11}\) × 6
Solution:
(a) 7 × \(\frac{3}{5}\) = \(\frac{21}{5}\) = 4\(\frac{1}{5}\)

(b) 4 × \(\frac{1}{3}\) = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)

(c) \(\frac{9}{7}\) × 6 = \(\frac{54}{7}\) = 7\(\frac{5}{7}\)

(d) \(\frac{13}{11}\) × 6 = \(\frac{78}{11}\) = 7\(\frac{1}{11}\)

ଦୁଇଟି ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣନ :

Question 1.
ଆମେ ଜାଣିଛୁ ଯେ, ଅମ୍ଳାନର ପୋଷା ବିରାଡ଼ି 1 ଘଣ୍ଟାରେ ମାତ୍ର \(\frac{1}{4}\) କିଲୋମିଟର ଚାଲିପାରେ । \(\frac{1}{2}\) ବା ଅଧା ଘଣ୍ଟାରେ ସେ କେତେ ଦୂର ଚାଲିପାରିବ ?

Solution:
ଗୁଣନ ପ୍ରକ୍ରିୟା ଅନୁସରଣ କରି ପାଇବା
\(\frac{1}{2}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା = \(\frac{1}{2}\) × \(\frac{1}{4}\)
1 ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା = \(\frac{1}{4}\) କି.ମି. ।
ତେଣୁ \(\frac{1}{4}\) ଧିକୁ ସମାନ 2 ଭାଗରେ ବିଭକ୍ତ କଲେ ଯେଉଁ ଦୂରତା ପାଇବା ତାହା ହେଉଛି \(\frac{1}{2}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା ।

Question 2.
ଏବେ ଆମେ ଏହି \(\frac{1}{4}\) କୁ ସମାନ 2 ଭାଗରେ ବିଭକ୍ତ କଲେ କ’ଣ ପାଇବୁ ?

Solution:

ରେଖାଙ୍କିତ ହୋଇଥ‌ିବା ଅଂଶଟି ପୂରା ଚିତ୍ରର କେତେ ଅଂଶ : ଯେହେତୁ ପୂରା ଚିତ୍ରକୁ 8 ସମାନ ଭାବରେ ବିଭକ୍ତ କରାଯାଇଛି ଏବଂ ଗୋଟିଏ ଅଂଶ ମାତ୍ର ଚିତ୍ରିତ ହୋଇଛି ।
ଏଣୁ ଆମେ କହିପାରିବା ଚିତ୍ରିତ ଅଂଶଟି ପୂରା ଚିତ୍ରର \(\frac{1}{8}\) ଅଂଶ ।
ତେଣୁ ବିରାଡ଼ିଟି \(\frac{1}{2}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ
କରିଥୁବା ଦୂରତା = \(\frac{1}{8}\) କି.ମି. ।
ଏବେ ଆମେ କହିପାରିବା = \(\frac{1}{2}\) × \(\frac{1}{4}\) = \(\frac{1}{8}\) କି.ମି. ।

Question 3.
ଯଦି ବିରାଡ଼ିଟି ଦ୍ରୁତଗତିରେ ଚାଲେ ଓ 1 ଘଣ୍ଟାରେ \(\frac{2}{5}\) କିଲୋମିଟର ଦୂରତା ଅତିକ୍ରମ କରେ, ତେବେ \(\frac{3}{4}\) ଘଣ୍ଟାରେ ସେ କେତେ କିଲୋମିଟର ଦୂରତା ଅତିକ୍ରମ କରିପାରିବ ?
Solution:
ପୂର୍ବ ନିୟମ ଅନୁଯାୟୀ ଦୂରତା ଅତିକ୍ରମ କରିବ = \(\frac{3}{4}\) × \(\frac{2}{5}\)
(i) ପ୍ରଥମେ ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା ନିର୍ଣ୍ଣୟ କରିବା ।
(ii) ଉତ୍ତର ସହିତ 3 ଗୁଣନ କଲେ ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା ନିର୍ଣ୍ଣୟ କରିପାରିବା ।
ଏହାକୁ ବର୍ଗଚିତ୍ରରେ ଦର୍ଶାଇବା –
(i) \(\frac{1}{4}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା :
ବର୍ଗଚିତ୍ରର \(\frac{2}{5}\) ଅଂଶକୁ 4 ସମାନ ଭାଗକରି ଆମେ ଉତ୍ତର କିଲୋମିଟରରେ ପାଇବା । ଦିଆଯାଇଥିବା ବର୍ଗଚିତ୍ରର କଳା ଚିହ୍ନିତ ଅଂଶଟି \(\frac{2}{5}\) ଅଂଶ ।
ଏହି \(\frac{2}{5}\) ଅଂଶକୁ ସମାନ 4 ଭାଗ କରାଯାଇଛି ।
ଏଠାରେ ବର୍ଗଚିତ୍ରକୁ 5ଟି ଧାଡ଼ି ଓ 4ଟି ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରାଯାଇଛି । ଏଣୁ ବର୍ଗଚିତ୍ରଟି 5 × 4 = 20 ଟି ସମାନ ଅଂଶରେ ବିଭକ୍ତ ହୋଇଛି ।
ଚିତ୍ରରେ ରେଖାଙ୍କିତ ଅଂଶ ବର୍ଗଚିତ୍ରର \(\frac{2}{5}\) ଅଂଶର 4 ସମାନ ଭାଗରୁ 2 ଭାଗ । ଏଣୁ \(\frac{1}{4}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରାଯାଇଥିବା ଦୂରତା \(\frac{2}{20}\)।
ଏବେ ଆମେ \(\frac{2}{20}\) କୁ 3 ସହିତ ଗୁଣନ କରିବାକୁ ପଡ଼ିବ ।
∴ \(\frac{3}{4}\) ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରୁଥିବା ଦୂରତା = 3 × \(\frac{2}{20}\) = \(\frac{6}{20}\)
ତେଣୁ \(\frac{3}{4}\) x \(\frac{2}{5}\) = \(\frac{6}{20}\) = \(\frac{3}{10}\) କି.ମି. ।(ଏଠାରେ \(\frac{3}{4}\) ଟି \(\frac{2}{5}\) ର
ଯୋଗଫଳ = \(\frac{3}{10}\) ବୋଲି କହିପାରିବା କି ? ନାଁ, କାରଣ \(\frac{3}{4}\) ଟିର କୌଣସି ଅର୍ଥ ନାହିଁ । 2ଟି, 3ଟି କହିପାରିବା ।
ଗଣିବା ପାଇଁ କେବଳ ଗଣନ ସଂଖ୍ୟା ବ୍ୟବହାର କରାଯାଏ କିନ୍ତୁ ଭଗ୍ନସଂଖ୍ୟା ନୁହେଁ ।)


ଆଲୋଚନା କରିବା :

ଏହି ଭଗ୍ନସଂଖ୍ୟାକୁ ଅନ୍ୟ ଏକ ଭଗ୍ନସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନ କରିବା ସମୟରେ ଆମେ ଏକ ଭଗ୍ନସଂଖ୍ୟାକୁ ଏକ ପୂର୍ଣ୍ଣସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନ କରିବା ପଦ୍ଧତି ଅନୁସରଣ କରିଥାଉ ।
ଆମେ \(\frac{3}{4}\) କୁ \(\frac{2}{5}\) ଦ୍ଵାରା ଏପରି ଗୁଣନ କଲୁ

Question 4.
ଏହି ପଦ୍ଧତି ଅନୁଯାୟୀ ଗୁଣନ କର : \(\frac{5}{4}\) × \(\frac{3}{2}\)
Solution:

ପ୍ରଥମେ ଚିତ୍ର ମାଧ୍ୟମରେ \(\frac{3}{2}\) କୁ ଦର୍ଶାଇବା ।
ଯେହେତୁ \(\frac{3}{2}\) ହେଉଛି 1 + \(\frac{1}{2}\) ବା ଏକ ପୂର୍ବବର୍ଗ ଚିତ୍ର ଓ ଏକ ଅଧା ବର୍ଗଚିତ୍ର ରୂପେ ଦେଖାଇ ପାରିବା ।
ଗୁଣନ ପ୍ରକ୍ରିୟାର ସୋପାନକୁ ଅନୁସରଣ କରି
ଆମେ ପ୍ରଥମେ \(\frac{3}{2}\) କୁ 4 ସମାନ ଭାଗରେ ବିଭକ୍ତ କରିବା ।
ଚିତ୍ରରେ ଦେଖାଯାଉଥ‌ିବା ପରି କରାଯିବ ।
ଯେପରିକି କଳା ଚିତ୍ରିତ \(\frac{3}{2}\) ଅଂଶ କୁ 4 ସମାନ ଭାଗକରି ଦର୍ଶାଉଛି ।
ଏହି ହଳଦିଆ ରଙ୍ଗଦ୍ଵାରା ରଙ୍ଗୀନ ଅଂଶ \(\frac{3}{2}\) ଚିତ୍ରର କେତେ ଅଂଶ ?
ଆମେ ଦେଖୁବା ଯେ ପୂରା ଚିତ୍ରଟିକୁ 2ଟି ଧାଡ଼ି ଓ 4ଟି ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ଫଳରେ ବର୍ଗଚିତ୍ରଟି 8ଟି ସମାନ ଭାଗରେ ବିଭକ୍ତ ହୋଇଛି ।
ମୋଟ ରଙ୍ଗୀନ ହୋଇଥିବା ଅଂଶ ସଂଖ୍ୟା = 3
ତେଣୁ ହଳଦିଆ ରଙ୍ଗୀନ ହୋଇଥିବା ଅଂଶ = \(\frac{3}{8}\)
ବର୍ତ୍ତମାନ ପରବର୍ତ୍ତୀ ସୋପାନ ହେଉଛି ଏହି ଫଳାଫଳକୁ 5 ଦ୍ଵାରା ଗୁଣନ କରିବା ।
ଏହା \(\frac{5}{4}\) ଓ \(\frac{3}{2}\) ର ଗୁଣଫଳକୁ ସୂଚାଇବ । \(\frac{5}{4}\) × \(\frac{3}{2}\) = 5 × \(\frac{3}{8}\) = \(\frac{15}{8}\)

Question 5.
ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ଏବଂ ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣନ ମଧ୍ୟରେ ସଂଯୋଗ ସ୍ଥାପନ : ଚିତ୍ରରେ ରେଖାଙ୍କିତ ଆୟତାକାର କ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥ କେତେ ?
Solution:

ଯେହେତୁ ଆମେ ଏକକ ବର୍ଗ (ପ୍ରତ୍ୟେକ ପାର୍ଶ୍ଵ 1 ଏକକ) ନେଇ ଆ ରମ୍ଭ କରିଥିଲୁ ।
ଏଣୁକରି ରେଖାଙ୍କିତ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ = \(\frac{1}{2}\) ଏକକ
ଓ ପ୍ରସ୍ଥ = \(\frac{1}{4}\) ଏକକ ହେବ ।
ଆମେ ଦେଖୁଛୁ ଯେ ଏହି ବର୍ଗଚିତ୍ରରେ ଏହିପରି 8ଟି ଆୟତକ୍ଷେତ୍ର ସୃଷ୍ଟି ହେଉଛି ।
ଏଣୁ ରେଖାଙ୍କିତ କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{8}\) ବର୍ଗ ଏକକ ହେବ ।

Question 6.
ତୁମେ ରେଖାଙ୍କିତ କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ଏବଂ ତା’ର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥର ଗୁଣଫଳ ମଧ୍ଯରେ କୌଣସି ସମ୍ପର୍କ ଦେଖୁଛ କି ?
Solution:
ଭଗ୍ନସଂଖ୍ୟା ବାହୁ ବିଶିଷ୍ଟ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ଏହାର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥର ଗୁଣଫଳ ସହିତ ସମାନ ।
ସାଧାରଣତଃ ଯଦି ଆମେ ଦୁଇଟି ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କରିବାକୁ ଚାହୁଁ ତେବେ ଭଗ୍ନସଂଖ୍ୟା ଦୁଇଟିକୁ ନେଇ ଗଠିତ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥ ମାଧ୍ୟମରେ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ନିର୍ଣ୍ଣୟ କରିପାରିବା ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନସ୍ଥ ଗୁଣନ କ୍ରିୟାର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର । ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ର ନେଇ ଭଗ୍ନସଂଖ୍ୟାର ଚିତ୍ରିତ ଉପସ୍ଥାପନା କର ।
(a) \(\frac{1}{3}\) × \(\frac{1}{5}\)
(b) \(\frac{1}{4}\) × \(\frac{1}{3}\)
(c) \(\frac{1}{5}\) × \(\frac{1}{2}\)
(d) \(\frac{1}{6}\) × \(\frac{1}{5}\)
Solution:
(a) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 5ଟି ଧାଡ଼ି ଓ 3ଟି ସ୍ତମ୍ଭ, 5 × 3 = 15 ଟି ଛୋଟ କ୍ଷେତ୍ରରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ଏଠାରେ \(\frac{1}{3}\) × \(\frac{1}{5}\) = \(\frac{1}{15}\)

(b) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 3ଟି ଧାଡ଼ି ଓ 4ଟି ସ୍ତମ୍ଭ, 3 × 4 = 12 ଟି ଛୋଟ କ୍ଷେତ୍ରରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ଏଠାରେ \(\frac{1}{4}\) × \(\frac{1}{3}\) = \(\frac{1}{12}\)

(c) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 2ଟି ଧାଡ଼ି ଓ 5ଟି ସ୍ତମ୍ଭ, 2 × 5 = 10 ଟି ଛୋଟ କ୍ଷେତ୍ରରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ଏଠାରେ \(\frac{1}{5}\) × \(\frac{1}{2}\) = \(\frac{1}{10}\)

(d) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 5ଟି ଧାଡ଼ି ଓ ଟି ସ୍ତମ୍ଭ, 5 × 6 = 30 ଟି ଛୋଟ କ୍ଷେତ୍ରରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ଏଠାରେ \(\frac{1}{6}\) × \(\frac{1}{5}\) = \(\frac{1}{30}\)

Question 2.
ବର୍ତ୍ତମାନ \(\frac{1}{12}\) × \(\frac{1}{18}\) ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
Solution:

ଏଠାରେ ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରରେ ଏହି ଭଗ୍ନସଂଖ୍ୟାଦ୍ୱୟକୁ ଦର୍ଶାଇବା କଷ୍ଟକର । ତେଣୁ ଉପରୋକ୍ତ କ୍ଷେତ୍ରରେ ଆମ କ’ଣ କଲୁ ତାହା ଅନୁଧ୍ୟାନ କରି ଏହାର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର । ଏଠାରେ ଏକ ପୂରା କ୍ଷେତ୍ରକୁ ଧାଡ଼ି ଓ ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରାଯାଇଛି ।

ଧାଡ଼ି ସଂଖ୍ୟା ପ୍ରଥମ ସଂଖ୍ୟା (ଗୁଣ୍ୟ)ର ହର ଯାହା, ଏହି କ୍ଷେତ୍ରରେ 18 ଓ ସ୍ତମ୍ଭ ସଂଖ୍ୟା ଦ୍ଵିତୀୟ ସଂଖ୍ୟା (ଗୁଣ୍ୟ)ର ହର ଯାହା, ଏହି କ୍ଷେତ୍ରରେ 12 । ତେଣୁ ପୂରା ଚିତ୍ରକୁ 18 × 12 ସମାନ ଅଂଶରେ ବିଭକ୍ତ କରାଯାଇଛି ।

ତେଣୁ \(\frac{1}{18}\) × \(\frac{1}{2}\) = \(\frac{1}{18 \times 12}\) = \(\frac{1}{216}\)
ଦୁଇଟି ଏକ ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣଫଳ ହେଉଛି =
ଯେଉଁ ଭଗ୍ନସଂଖ୍ୟାର ଲବ 1 (ଏକ) ତାହାକୁ ଏକକ ଭଗ୍ନସଂଖ୍ୟା କୁହାଯାଏ ।
ତାହାକୁ ଆମେ ଦର୍ଶାଇ ପାରିବା \(\frac{1}{b} \times \frac{1}{d}=\frac{1}{b \times d}\)

Question 3.
ନିମ୍ନଲିଖତ ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର । ଭଗ୍ନସଂଖ୍ୟାଗୁଡ଼ିକୁ ଚିତ୍ରରେ ଉପସ୍ଥାପନା କରିବା ପାଇଁ ଏବଂ ଗୁଣନ ପ୍ରକ୍ରିୟା ସମ୍ପନ୍ନ ପାଇଁ ଏକକ ବର୍ଗଚିତ୍ର ବ୍ୟବହାର କର ।
(a) \(\frac{2}{3}\) × \(\frac{4}{5}\)
(b) \(\frac{1}{4}\) × \(\frac{2}{3}\)
(c) \(\frac{3}{5}\) × \(\frac{1}{2}\)
(d) \(\frac{4}{6}\) × \(\frac{3}{5}\)
Solution:
(a) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 5ଟି ଧାଡ଼ି ଓ 3ଟି ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ସମୁଦାୟ ବର୍ଗକ୍ଷେତ୍ର = 5 × 3 = 15
ଓ 2 × 4 = 8 ଟି ଛୋଟ ଛାୟାଯୁକ୍ତ କ୍ଷେତ୍ର ।
ସମୁଦାୟ ଛାୟାଯୁକ୍ତ ଅଂଶ = \(\frac{8}{15}\)
ଏଠାରେ \(\frac{2}{3}\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

(b) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 3ଟି ଧାଡ଼ି ଓ 4ଟି ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ସମୁଦାୟ ବର୍ଗକ୍ଷେତ୍ର = 4 × 3 = 12
ଓ 1 × 2 = 2 ଟି ଛୋଟ ଛାୟାଯୁକ୍ତ କ୍ଷେତ୍ର ।
ସମୁଦାୟ ଛାୟାଯୁକ୍ତ ଅଂଶ = \(\frac{2}{12}\) = \(\frac{1}{6}\)
ଏଠାରେ \(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

(c) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 2ଟି ଧାଡ଼ି ଓ 5ଟି ସ୍ତମ୍ଭରେ ଦ୍ବିଭକ୍ତ କରାଯାଇଛି ।
ସମୁଦାୟ ବର୍ଗକ୍ଷେତ୍ର = 2 × 5 = 10
ଓ 3 × 1 = 3 ଟି ଛୋଟ ଛାୟାଯୁକ୍ତ କ୍ଷେତ୍ର ।
ସମୁଦାୟ ଛାୟାଯୁକ୍ତ ଅଂଶ = \(\frac{3}{10}\)
ଏଠାରେ \(\frac{3}{5}\) × \(\frac{1}{2}\) = \(\frac{3}{10}\)

(d) ଏକ ବର୍ଗ ଏକକ ବିଶିଷ୍ଟ ବର୍ଗଚିତ୍ରକୁ 5ଟି ଧାଡ଼ି ଓ 6ଟି ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରାଯାଇଛି ।
ସମୁଦାୟ ବର୍ଗକ୍ଷେତ୍ର = 5 × 6 = 30
ଓ 4 × 3 = 12 ଟି ଛୋଟ ଛାୟାଯୁକ୍ତ କ୍ଷେତ୍ର ।
ସମୁଦାୟ ଛାୟାଯୁକ୍ତ ଅଂଶ = \(\frac{12}{30}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)
ଏଠାରେ \(\frac{4}{6}\) × \(\frac{3}{5}\) = \(\frac{12}{30}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

ଲବ ଓ ହରର ଗୁଣନ :

Question 1.
ବର୍ତ୍ତମାନ \(\frac{5}{12}\) × \(\frac{7}{18}\) ର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କରିବା ।
Solution:


ପ୍ରଥମେ ଏକ କ୍ଷେତ୍ର (ସମଗ୍ର ଅଂଶ)କୁ 18 ଧାଡ଼ି ଓ 12ଟି ସ୍ତମ୍ଭରେ ବିଭକ୍ତ କରି 12 × 18 ସମାନ ଅଂଶରେ ବିଭକ୍ତ କର ।
ପ୍ରଥମେ ଏହି ଚିତ୍ରରୁ 12 × 18 ର \(\frac{7}{18}\) ଅଂଶର କୋଠରି ସଂଖ୍ୟା \(\frac{7}{12 \times 18}\) ଅଂଶ ହେବ ।
ତା’ପରେ ଆମେ ଏହି ଫଳାଫଳକୁ 5 ଦ୍ଵାରା ଗୁଣନ କଲେ \(\frac{5 \times 7}{12 \times 18}\)ଗୁଣଫଳ ପାଇବା ।
ତେଣୁ \(\frac{5}{12}\) × \(\frac{7}{8}\) = \(\frac{5 \times 7}{12 \times 18}\) = \(\frac{35}{216}\)
ଏଥୁରୁ ଆମେ ଜାଣିଲେ ଯେ, \(\frac{a}{b} \times \frac{c}{d}\) = \(\frac{\mathrm{a} \times \mathrm{c}}{\mathrm{~b} \times \mathrm{d}}\)
ଏହି ସୂତ୍ରଟି ପ୍ରଥମେ ବ୍ରହ୍ମଗୁପ୍ତଙ୍କଦ୍ଵାରା 628 ଖ୍ରୀଷ୍ଟାବ୍ଦରେ ତାଙ୍କଦ୍ୱାରା ଲିଖ୍ତ ବ୍ରହ୍ମସ୍ଫୁଟ ସିଦ୍ଧାନ୍ତରେ ଏହିଭଳି ଏକ ସାଧାରଣ ଗାଣିତିକ ଉକ୍ତିରେ କୁହାଯାଇଥିଲା ।

ଗୁଣକ କିମ୍ବା ଗୁଣ୍ୟ ଏକ ପୂର୍ଣ୍ଣସଂଖ୍ୟା ହୋଇଥିଲେ ମଧ୍ୟ ଉପରୋକ୍ତ ସୂତ୍ରଟି ପ୍ରୟୋଗ କରିବା । ଆମେ ଯେକୌଣସି ପୂର୍ଣ୍ଣସଂଖ୍ୟାକୁ ହର 1 (ଏକ) ନେଇ ଏକ ଭଗ୍ନସଂଖ୍ୟା ଭାବେ ପୁନଃଲିଖନ କରିପାରିବା ।
ଉଦାହରଣ ସ୍ୱରୂପ 3 × \(\frac{3}{4}\) କୁ \(\frac{3}{1}\) × \(\frac{3}{4}\) ଭାବେ ଲେଖୁପାରିବା ।
ତେଣୁ \(\frac{3 \times 3}{1 \times 4}\) = \(\frac{9}{4}\)

ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣନ : ଲଘିଷ୍ଠ (ସର୍ବନିମ୍ନ) ଆକାରକୁ ସରଳୀକରଣ :
Question 1.
ନିମ୍ନଲିଖତ ଭଗ୍ନସଂଖ୍ୟାଗୁଡ଼ିକୁ ଗୁଣନ କର ଏବଂ ଗୁଣଫଳକୁ ଲଘିଷ୍ଠ ଆକାରରେ ପ୍ରକାଶ କର ।
\(\frac{12}{7}\) × \(\frac{5}{24}\)
Solution
ପ୍ରଥମେ ଭଗ୍ନସଂଖ୍ୟାର ଲବ (12 ଏବଂ 5) ଓ ହର (7 ଏବଂ 24)କୁ ଗୁଣନ କରିବା ପରିବର୍ତ୍ତେ ଆମେ ନିମ୍ନୋକ୍ତ ଉପାୟରେ ସରଳ କରି ଲଘିଷ୍ଠ କରିପାରିବା ।

ଉଭୟ ଗୋଲ୍ ବୁଲାଯାଇଥିବା ସଂଖ୍ୟାର ଏକ ସାଧାରଣ ଗୁଣନୀୟକ 12 ଅଛି ।
ଯେତେବେଳେ ଭଗ୍ନସଂଖ୍ୟାର ଲବ ଓ ହରକୁ ସାଧାରଣ ଗୁଣନୀୟକଦ୍ଵାରା ଭାଗ କରାଯାଏ ସେତେବେଳେ ଭଗ୍ନସଂଖ୍ୟାର ମୂଲ୍ୟ ଅପରିବର୍ତିତ ରହିଥାଏ ।

ଏହି କ୍ଷେତ୍ରରେ ଆମେ ଗୋଲ ବୁଲାଯାଇଥିବା ସଂଖ୍ୟାମାନଙ୍କୁ 12 ଦ୍ଵାରା ଭାଗ କରିପାରିବା ।

ଆସ ଆଉ ଏକ ଗୁଣନରେ ସମାନ ପଦ୍ଧତି ଅବଲମ୍ବନ କରିବା ।

ଭଗ୍ନସଂଖ୍ୟାକୁ ଗୁଣନ କରିବା ସମୟରେ ପ୍ରଥମେ ଲବ ଓ ହରରେ ଥ‌ିବା ସଂଖ୍ୟାମାନଙ୍କୁ ସେମାନଙ୍କର ସାଧାରଣ ଗୁଣନୀୟକଦ୍ଵାରା ଭାଗ କରି ଏବଂ ତାପରେ ଲବ୍ଧ ଲବ ଓ ହରକୁ ପୂର୍ବ ପଦ୍ଧତିରେ ଗୁଣନ କରିବା । ଏହାକୁ ସାଧାରଣ ଉତ୍ପାଦକଗୁଡ଼ିକୁ ବାତିଲ କରିବା ପଦ୍ଧତି କୁହାଯାଏ । (ଭଗ୍ନସଂଖ୍ୟାର ଲବ ଓ ହରକୁ ସାଧାରଣ ଗୁଣନୀୟକଦ୍ୱାରା କାଟି ଦିଆଯାଏ ।)

ନିଜେ କରି ଦେଖ :

Question 1.
ଏକ ପାଣିଟାଙ୍କିକୁ ଏକ ନଳ ଦ୍ଵାରା ପୂର୍ଣ କରାଯାଏ । ଯଦି ନଳଟି 1 ଘଣ୍ଟା ପାଇଁ ଖୋଲା ରହେ, ତେବେ ଟାଙ୍କିର \(\frac{7}{10}\) ଅଂଶ ପୂରଣ ହୁଏ । ଯଦି ନଳଟି ତଳଲିଖତ ସମୟ ପାଇଁ ଖୋଲା ରହେ ତେବେ ଟାଙ୍କିର କେତେ ଅଂଶ ପୂର୍ଣ୍ଣ ହେବ ?
(a) \(\frac{1}{3}\) ଘଣ୍ଟାରେ
(b) \(\frac{2}{3}\) ଘଣ୍ଟାରେ
(c) \(\frac{3}{4}\) ଘଣ୍ଟାରେ
(d) \(\frac{7}{10}\) ଘଣ୍ଟାରେ
(e) ଟାଙ୍କିଟି ପୂର୍ଣ୍ଣ ହେବା ପାଇଁ କେତେ ସମୟ ଲାଗିବ ?
Solution:
(a) 1 ଘଣ୍ଟାରେ ଟାଙ୍କିର ପୂର୍ଣ୍ଣହୁଏ = \(\frac{7}{10}\) ଅଂଶ ।
\(\frac{1}{3}\) ଘଣ୍ଟାରେ ପୂର୍ବ ହୁଏ = \(\frac{7}{10}\) × \(\frac{1}{3}\) = \(\frac{7}{30}\) ଅଂଶ ।
∴ ଯଦି ନଳଟି \(\frac{1}{3}\) ଘଣ୍ଟା ସମୟ ପାଇଁ ଖୋଲା ରହିଲେ ତେବେ ଟାଙ୍କିର \(\frac{7}{30}\)

(b) 1 ଘଣ୍ଟାରେ ଟାଙ୍କିର ପୂର୍ଣ୍ଣହୁଏ = \(\frac{7}{10}\) ଅଂଶ ।
\(\frac{2}{3}\) ଘଣ୍ଟାରେ ପୂର୍ଣ୍ଣ ହୁଏ = \(\frac{7}{10}\) × \(\frac{2}{3}\) = \(\frac{14}{30}\) = \(\frac{7}{15}\) ଅଂଶ ।
∴ ଯଦି ନଳଟି \(\frac{2}{3}\) ଘଣ୍ଟା ସମୟ ପାଇଁ ଖୋଲା ରହିଲେ ତେବେ ଟାଙ୍କିର \(\frac{7}{15}\) ଅଂଶ ପୂର୍ଷ ହେବ ।

(c) 1 ଘଣ୍ଟାରେ ଟାଙ୍କିର ପୂର୍ଷହୁଏ = \(\frac{7}{10}\) ଅଂଶ ।
\(\frac{3}{4}\) ଘଣ୍ଟାରେ ପୂର୍ଣ ହୁଏ = \(\frac{7}{10}\) × \(\frac{3}{4}\) = \(\frac{21}{40}\) ଅଂଶ ।
∴ ଯଦି ନଳଟି \(\frac{3}{4}\) ଘଣ୍ଟା ସମୟ ପାଇଁ ଖୋଲା ରହିଲେ ତେବେ ଟାଙ୍କିର \(\frac{21}{40}\) ଅଂଶ ପୂର୍ଷ ହେବ ।

(d) 1 ଘଣ୍ଟାରେ ଟାଙ୍କିର ପୂର୍ଣହୁଏ = \(\frac{7}{10}\) ଅଂଶ ।
\(\frac{7}{10}\) ଘଣ୍ଟାରେ ପୂର୍ବ ହୁଏ = \(\frac{7}{10}\) × \(\frac{7}{10}\) = \(\frac{49}{100}\) ଅଂଶ ।
∴ ଯଦି ନଳଟି \(\frac{7}{10}\) ଘଣ୍ଟା ସମୟ ପାଇଁ ଖୋଲା ରହିଲେ ତେବେ ଟାଙ୍କିର \(\frac{49}{100}\) ଅଂଶ ପୂର୍ଣ୍ଣ ହେବ।

(e) 1 ଘଣ୍ଟାରେ ଟାଙ୍କିର ପୂର୍ଣ୍ଣହୁଏ = \(\frac{7}{10}\) ଅଂଶ ।
ଟାଙ୍କିଟି ପୂର୍ଣ ହେବ = \(\frac{\frac{1}{7}}{10}\) = \(\frac{10}{7}\) = 1\(\frac{3}{7}\) ଘଣ୍ଟାରେ ।
∴ ଟାଙ୍କିଟି ପୂର୍ଣ୍ଣ ହେବା ପାଇଁ 1\(\frac{3}{7}\) ଘଣ୍ଟା ସମୟ ଲାଗିବ ।

Question 2.
ସରକାର ରାସ୍ତା ନିର୍ମାଣ ପାଇଁ ରେବତୀଙ୍କ ଜମିର \(\frac{1}{6}\) ଅଂଶ ନେଇଛନ୍ତି । ବର୍ତ୍ତମାନ ରେବତୀ ପାଖରେ
ମୋଟ ଜମିର କେତେ ଅଂଶ ଅଛି ? ସେ ତାଙ୍କ ପାଖରେ ଥିବା ବଳକା ଜମିର ଅଧା ଝିଅ କ୍ରିଷ୍ଣାକୁ ଏବଂ \(\frac{1}{3}\)
ଅଂଶ ତାଙ୍କ ପୁଅ ଅନିଲକୁ ଦେଇ ଅବଶିଷ୍ଟ ଜମି ନିଜପାଇଁ ରଖୁଲେ ?
(a) କ୍ରିଷ୍ଣା ସମୁଦାୟ ଜମିର କେତେ ଅଂଶ ପାଇଥିଲା ?
(b) ଅନିଲ ସମୁଦାୟ ଜମିର କେତେ ଅଂଶ ପାଇଥିଲା ?
(c) ରେବତୀ ସମୁଦାୟ ଜମିର କେତେ ଅଂଶ ନିଜ ପାଇଁ ରଖୁଥିଲେ ?
Solution:
ମନେକର ସମୁଦାୟ ଜମିକୁ 1 ଧରାଯାଉ ।
ସରକାର ରାସ୍ତା ନିର୍ମାଣ ପାଇଁ ରେବତୀଙ୍କ ଜମିରୁ \(\frac{1}{6}\) ଅଂଶ ନେଇଛନ୍ତି ।
ରେବତୀଙ୍କ ପାଖରେ ବଳକା ଜମି ରହିଲା = 1 – \(\frac{1}{6}\) = \(\frac{6-1}{6}\) = \(\frac{5}{6}\)
∴ ରେବତୀଙ୍କ ପାଖରେ ବଳକା \(\frac{5}{6}\) ଅଂଶ ଜମି ରହିଲା ।

(a) ରେବତୀଙ୍କ ପାଖରେ ଜମି ରହିଲା = \(\frac{5}{6}\) ଅଂଶ ।
ସେ ତାଙ୍କ ଝିଅ କ୍ରିଷ୍ଣାଙ୍କୁ ଅଧା ଅଂଶ ଦେଇଥିଲେ ।
କ୍ରିଷ୍ଣା ପାଇଥିଲେ = \(\frac{1}{2}\) × \(\frac{5}{6}\) = \(\frac{5}{12}\) ଅଂଶ ।
∴ କ୍ରିଷ୍ଣା ସମୁଦାୟ ଜମିର \(\frac{5}{12}\) ଅଂଶ ପାଇଥିଲା ।

(b) ରେବତୀଙ୍କ ପାଖରେ ଜମି ରହିଲା \(\frac{5}{6}\)
ସେ ଅନିଲକୁ ଦେଲେ = \(\frac{1}{3}\)ଅଂଶ ।
ଅନିଲ ପାଇଥିଲେ = \(\frac{1}{3}\) × \(\frac{5}{6}\) = \(\frac{5}{18}\) ଅଂଶ ।
∴ ଅନିଲ ସମୁଦାୟ ଜମିର \(\frac{5}{18}\) ଅଂଶ ପାଇଥିଲା ।

(c) ସରକାର, କ୍ରିଷ୍ଣା ଓ ଅନିଲ ଜମି ନେଇଥିଲେ
= \(\frac{1}{6}\) + \(\frac{5}{12}\) + \(\frac{5}{18}\) = \(\frac{31}{36}\) ଅଂଶ ।
ରେବତୀ ପାଖରେ ରହିଲା = 1 – \(\frac{31}{36}\) = \(\frac{5}{36}\) ଅଂଶ ।
∴ ରେବତୀ ସମୁଦାୟ ଜମିର \(\frac{5}{36}\) ଅଂଶ ନିଜ ପାଇଁ ରଖୁଥଲେ ।

Question 3.
ଏକ ଆୟତକ୍ଷେତ୍ରର ଦୁଇ ସଂଲଗ୍ନ ବାହୁ (ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥ) ଯଥାକ୍ରମେ 3\(\frac{3}{4}\) ଫୁଟ ଓ 9\(\frac{3}{5}\) ଫୁଟ ହେଲେ ଏହି କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ କେତେ ?
Solution:
ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ = 3\(\frac{3}{4}\) ଫୁଟ = \(\frac{15}{4}\) ଫୁଟ
ପ୍ରସ୍ଥ 9\(\frac{3}{5}\) ଫୁଟ = \(\frac{48}{5}\) ଫୁଟ ।
ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ
= \(\frac{15}{4}\) × \(\frac{48}{5}\) = (3 × 12) ବର୍ଗଫୁଟ = 36 ବର୍ଗଫୁଟ ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ 36 ବର୍ଗଫୁଟ ।

Question 4.
ରହିମ୍ ତାଙ୍କ ବଗିଚାରେ ଗୋଟିଏ ଧାଡ଼ିରେ 4ଟି ଚାରା ଲଗାଇଛନ୍ତି । ଦୁଇଟି ଚାରା ମଧ୍ୟରେ ଦୂରତା \(\frac{3}{4}\) ମିଟର ହେଲେ ପ୍ରଥମ ଓ ଶେଷ ଚାରା ମଧ୍ୟରେ ଦୂରତା କେତେ ? (ସୂଚନା : ଦୁଇଟି ଚାରା ମଧ୍ୟରେ ଦୂରତା \(\frac{3}{4}\) ମିଟର ସହିତ ଚାରୋଟି ଚାରାର୍ ରଫ୍ ଚିତ୍ର ଅଙ୍କନ କର ।)
Solution:

ପ୍ରଥମ ଓ ଶେଷ ଚାରା ମଧ୍ୟରେ ଦୂରତା = ଦୂରତା ସଂଖ୍ୟା × ଦୁଇଟି ଚାରା ମଧ୍ୟରେ ଦୂରତା
= 3 × \(\frac{3}{4}\) ମିଟର = \(\frac{3 \times 3}{4}\) = \(\frac{9 }{4}\) ମିଟର = 2\(\frac{1}{4}\) ମିଟର ।
∴ ପ୍ରଥମ ଓ ଶେଷ ଚାରା ମଧ୍ୟରେ ଦୂରତା 2\(\frac{1}{4}\) ମିଟର ।

Question 5.
କେଉଁଟି ଅଧ୍ବକ ଭାରୀ : 500 ଗ୍ରାମର \(\frac{12}{15}\) କିମ୍ବା 4 କିଲୋଗ୍ରାମ୍ବର \(\frac{3}{20}\)
Solution:
500 ଗ୍ରାମର \(\frac{12}{15}\) = \(\frac{12}{15}\) × 500 = 400 ଗ୍ରାମ୍ ।
4 କିଲୋଗ୍ରାମ୍ = 4000 ଗ୍ରାମ୍ ।
4 କିଲୋଗ୍ରାମ୍‌ \(\frac{3}{20}\) = \(\frac{3}{20}\) × 4000 = 600 ଗ୍ରାମ୍ ।
400 ଗ୍ରାମ୍ < 600 ଗ୍ରାମ୍ ।
∴ 500 ଗ୍ରାମ୍ର \(\frac{12}{15}\) ଠାରୁ 4 କିଲୋଗ୍ରାମର \(\frac{3}{20}\) ଭାରୀ ।

Question 6.
ଗୋଟିଏ ଟ୍ରାକ୍ଟର 1 ଘଣ୍ଟାରେ ଏକ ଜମିକୁ ହଳ କରିପାରେ ଏବଂ ଏଥ‌ିପାଇଁ 2 ଲିଟର ଡିଜେଲ ଦରକାର ହୁଏ । ତେବେ 1 ଲିଟର ଡିଜେଲରେ ଜମିର କେତେ ଅଂଶ ହଳ ହୋଇପାରିବ ?
Solution:
ଗୋଟିଏ ଟ୍ରାକ୍ଟର 1 ଘଣ୍ଟାରେ ଏକ ଜମିକୁ ହଳ କରିପାରେ ।
1 ଘଣ୍ଟାରେ ଲାଗେ 2 ଲିଟର ଡିଜେଲ । 1 ଲିଟର ଡିଜେଲରେ ହଳ କରିବ 1 × \(\frac{1}{2}\) = \(\frac{1}{2}\)
ତେବେ 1 ଲିଟର ଡିଜେଲରେ ଜମିର \(\frac{1}{2}\) ଅଂଶ ହଳ ହୋଇପାରିବ ।

ଗୁଣଫଳ ସର୍ବଦା ଗୁଣନ କରାଯାଉଥିବା ସଂଖ୍ୟା ଅପେକ୍ଷା ବଡ଼ କି ?
ଯେତେବେଳେ କୌଣସି ସଂଖ୍ୟାକୁ 1 ଦ୍ଵାରା ଗୁଣନ କରାଯାଏ, ଗୁଣଫଳ ଅପରିବର୍ତ୍ତିତ ରହେ ଅର୍ଥାତ୍ ଗୁଣଫଳ ସେହି ସଂଖ୍ୟା ହୁଏ । ଆମେ ଗୁଣନ ପାଇଁ ଏକ ସଂଖ୍ୟା ଯୋଡ଼ା ନେବା ଯେଉଁଠାରେ କୌଣସି ସଂଖ୍ୟା 1 (ଏକ) ହୋଇ ନ ଥ‌ିବ ।
∴ 3 × 5 = 15
ଏଠାରେ ଗୁଣଫଳ 15 ଉଭୟ 3 ଏବଂ 5 ଠାରୁ ବଡ଼ ।

କିନ୍ତୁ ଯେତେବେଳେ ଆମେ \(\frac{1}{4}\) ଓ 8 କୁ ଗୁଣନ କରିବା,ଏହାର ଗୁଣଫଳ ଉଭୟ ସଂଖ୍ୟାଠାରୁ ବଡ଼ ହେବ କି ? \(\frac{1}{4}\) x 8 = 2
ଏହି ଗୁଣନରେ ଗୁଣଫଳ 2, ଯାହା \(\frac{1}{4}\) ଠାରୁ ବଡ଼ କିନ୍ତୁ 8 ଠାରୁ ସାନ ।

Question 1.
\(\frac{3}{4}\) ଓ \(\frac{2}{5}\) କୁ ଗୁଣନ କଲେ କ’ଣ ହେବ, ଆସ ପରୀକ୍ଷା କରିବା ।
Solution:
\(\frac{3}{4}\) × \(\frac{2}{5}\) = \(\frac{6}{20}\)
∴ ଗୁଣଫଳ \(\frac{6}{20}\) କୁ \(\frac{3}{4}\) ଏବଂ ସହିତ ତୁଳନା କରିଲେ
ଆମେ \(\frac{3}{4}\) କୁ \(\frac{15}{20}\) ଓ \(\frac{2}{5}\) କୁ \(\frac{8}{20}\) ଭଗ୍ନାଂଶ ରୂପେ ପ୍ରକାଶ କରିପାରିବା ।
∴ ଗୁଣଫଳ \(\frac{6}{20}\) ଉଭୟ \(\frac{15}{20}\) ଓ \(\frac{8}{20}\) ଠାରୁ ସାନ ହେବ ।

ପରିସ୍ଥିତି	ଗୁଣନ	ସମ୍ପର୍କ
ପରିସ୍ଥିତି – 1	ଉଭୟ ସଂଖ୍ୟାଠାରୁ 1 ଅଧୂକ । ଉଦାହରଣ : \(\frac{4}{3}\) × 4	ଗୁଣଫଳ \(\frac{16}{3}\) ଉଭୟ ସଂଖ୍ୟାଠାରୁ ବଡ଼ ।
ପରିସ୍ଥିତି – 2	ଉଭୟ ସଂଖ୍ୟାର ମୂଲ୍ୟ 0 ଓ 1 ମଧ୍ୟରେ । ଉଦାହରଣ : \(\frac{3}{4}\) × \(\frac{2}{5}\)	ଗୁଣଫଳ \(\frac{3}{10}\) ଉଭୟ ସଂଖ୍ୟାଠାରୁ କ୍ଷୁଦ୍ରତର ।
ପରିସ୍ଥିତି – 3	ଗୋଟିଏ ସଂଖ୍ୟା 0 ଓ 1 ମଧ୍ଯରେ ଓ ଅନ୍ୟ ସଂଖ୍ୟାଟି 1 ଠାରୁ ବଡ଼ । ଉଦାହରଣ \(\frac{3}{4}\) × 5	ଗୁଣଫଳ \(\frac{15}{4}\) ପୂର୍ବ ସଂଖ୍ୟାଠାରୁ କ୍ଷୁଦ୍ରତର ଓ 0 ଓ 1 ମଧ୍ୟରେ ଥ‌ିବା ସଂଖ୍ୟାଠାରୁ ବୃହତ୍ତର ।

Question 2.
ଯେତେବେଳେ ଗୁଣନ କରାଯାଉଥିବା ସଂଖ୍ୟାଯୋଡ଼ି ମଧ୍ୟରୁ ଗୋଟିଏ ସଂଖ୍ୟା 0 ଏବଂ 1 ମଧ୍ଯରେ ଥାଏ, ଗୁଣଫଳ ଅନ୍ୟ ସଂଖ୍ୟା ଅପେକ୍ଷା ________________ (ଅଧିକ / କମ୍) ହୋଇଥାଏ ।
Solution:
କମ୍ ହୋଇଥାଏ ।

Question 3.
ଯେତେବେଳେ ଗୁଣନ କରାଯାଉଥିବା ସଂଖ୍ୟା ଯୋଡ଼ି ମଧ୍ୟରୁ ଗୋଟିଏ ସଂଖ୍ୟା 1 (ଏକ) ଠାରୁ ଅଧିକ ହୋଇଥାଏ, ଗୁଣଫଳ ଅନ୍ୟ ସଂଖ୍ୟା ଅପେକ୍ଷା _________________ (ଅଧିକ / କମ୍) ହୋଇଥାଏ ।
Solution:
ଅଧିକ ହୋଇଥାଏ ।

ଗୁଣନର କ୍ରମ :

ଆମେ ଜାଣିଛୁ ଯେ \(\frac{1}{2}\) × \(\frac{1}{4}\) = \(\frac{1}{8}\)
ବର୍ତ୍ତମାନ \(\frac{1}{4}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\)

ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ ପ୍ରସ୍ଥ ଅଦଳବଦଳ ହେଲେ ମଧ୍ୟ କ୍ଷେତ୍ରଫଳ ସମାନ ରହିଥାଏ ।

ଗୁଣନରେ କ୍ରମ ବଦଳିଲେ ମଧ୍ୟ ଗୁଣଫଳ ସମାନ ରହେ । ଏଣୁ \(\frac{a}{b} \times \frac{c}{d}\) = \(\frac{c}{d} \times \frac{a}{b}\)
ଏହି ନିୟମ ବ୍ରହ୍ମଗୁପ୍ତଙ୍କ ଭଗ୍ନସଂଖ୍ୟାର ଗୁଣନ ପାଇଁ ଉଦ୍ଦିଷ୍ଟ ସୂତ୍ରରେ ମଧ୍ୟ ଦେଖାଯାଏ ।

8.2 ଭଗ୍ନସଂଖ୍ୟାମାନଙ୍କର ହରଣ :

Page No. (189 to 192)

ଆମେ ଭଗ୍ନସଂଖ୍ୟାକୁ ହରଣ କରିବା ପାଇଁ ହେଲେ ହରଣ ଚିହ୍ନକୁ ଗୁଣନ ସମସ୍ୟାରେ ପରିଣତ କରିବା

Question 1.
1 ÷ \(\frac{2}{3}\) = କେତେ ?
Solution:
ଏହାକୁ ଗୁଣନ ସମସ୍ୟାରେ ପରିଣତ କରି ଲେଖୁବା ହେବ = \(\frac{2}{3}\) × ? = 1
ବା \(\frac{2}{3}\) ସହ କେତେ ଗୁଣନ କଲେ ଗୁଣଫଳ 1 ପାଇବା ।
ଯଦି କୌଣସି ପ୍ରକାରେ 2 ଓ 3 କୁ କଟାଯାଇପାରିବ, ତେବେ ଆମ ପାଖରେ । (ଏକ) ରହିବ ।

ଆସ ଆଉ ଏକ ସମସ୍ୟାର ସମାଧାନ କରିବା ପାଇଁ ଚେଷ୍ଟା କରିବା ।

Question 2.
3 ÷ \(\frac{2}{3}\) = କେତେ ? ଏହ । \(\frac{2}{3}\) × ?= 3 ଭକ୍ତି ସହ ସମାନ ଅଟେ ।
Solution:
ଆମେ ଜାଣୁ ଯେ, 1 ପାଇବା ପାଇଁ \(\frac{2}{3}\) କୁ କେଉଁ ସଂଖ୍ୟା ସହିତ ଗୁଣନ କରିବାକୁ ପଡ଼ିବ । 3 ପାଇବା ପାଇଁ ଆମକୁ କେବଳ ସେଥୁରେ 3 ଗୁଣନ କରିବାକୁ ପଡ଼ିବ ।

Question 3.
\(\frac{1}{5}\) ÷ \(\frac{1}{2}\) = କେତେ ?
Solution:

Question 4.
\(\frac{2}{3}\) ÷ \(\frac{3}{5}\) = କେତେ ?
Solution:

Question 5.
ଆଲୋଚନା କରିବା :
Solution:
1. ପ୍ରଥମେ ଖୋଜି ଦେଖୁବା, ଭାଜକଦ୍ଵାରା କେଉଁ ସଂଖ୍ୟା ଗୁଣନ କଲେ 1 ମିଳୁଛି, ଆମେ ଦେଖୁଲେ ଯେ ଭାଜକ ଭଗ୍ନସଂଖ୍ୟାର ଲବକୁ ହର ଓ ହରକୁ ଲବ ରୂପେ ଲେଖୁଲେ ଆବଶ୍ୟକୀୟ ଭଗ୍ନସଂଖ୍ୟାଟି
ମିଳୁଛି । ଏଠାରେ ସଂଖ୍ୟାର ଭାଜକ \(\frac{3}{5}\) ପାଇଁ ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{5}{3}\)। ଆମେ \(\frac{5}{3}\) କୁ \(\frac{3}{5}\) ର ବ୍ୟକ୍ରମ / ପ୍ରତିଲୋମୀ ବୋଲି କହିଥାଉ । ଅର୍ଥାତ୍ ଯେତେବେଳେ ଆମେ ଏକ ଭଗ୍ନସଂଖ୍ୟାକୁ ତାହାର ବ୍ୟତ୍‌କ୍ରମ ବା ପ୍ରତିଲୋମୀ ସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନ କରୁ ଭାଗଫଳ 1 ମିଳିବ । ତେଣୁ ପ୍ରଥମେ ଆମେ ଭାଜକର ବ୍ୟୁତ୍‌କ୍ରମକୁ ଜାଣିବା ।

2. ଆମେ ଭାଗ୍ୟକୁ ଭାଜକର ବ୍ୟୁତ୍‌କ୍ରମ ସହିତ ଗୁଣନ କରି ଭାଗଫଳ ପାଇବା । ସଂକ୍ଷେପରେ ଦୁଇଟି ଭଗ୍ନସଂଖ୍ୟା ଭାଗ କରିବା ପାଇଁ
– ଭାଜକର ବ୍ୟକ୍ରମ ଜାଣିବା
– ଭାଗଫଳ ପାଇବା ପାଇଁ ଭାଜ୍ୟକୁ ଭାଜକର ବ୍ୟକ୍ରମ ସହିତ ଗୁଣନ କରିବା ।
ତେଣୁ =\(\frac{a}{b} \div \frac{c}{d}\) = \(\frac{a}{b} \times \frac{d}{c}\) = \(\frac{\mathrm{a} \times \mathrm{d}}{\mathrm{~b} \times \mathrm{c}}\)
ଯେପରି ଭଗ୍ନସଂଖ୍ୟାର ଯୋଗ, ବିୟୋଗ ଓ ଗୁଣନ ପାଇଁ ପଦ୍ଧତି ଏବଂ ସୂତ୍ର ଆମେ ପୂର୍ବରୁ ଶିଖୁଥିଲେ, ସେହିପରି ଏହି ପଦ୍ଧତି ଭଗ୍ନସଂଖ୍ୟାର ଭାଗକ୍ରିୟା ପାଇଁ ଏକ ସାଧାରଣ ସୂତ୍ର ଅଟେ । ପ୍ରଥମେ ଏହା ବ୍ରହ୍ମଗୁପ୍ତଙ୍କଦ୍ଵାରା ଲିଖ୍ତ ବ୍ରହ୍ମସ୍ଫୁଟ ସିଦ୍ଧାନ୍ତ (628 ଖ୍ରୀଷ୍ଟାବ୍ଦରେ) ସ୍ପଷ୍ଟ ଭାବରେ ଉଲ୍ଲେଖ କରାଯାଇଥିଲା ।

Question 6.
ଉଦାହରଣ ସ୍ଵରୂପ (ବ୍ରହ୍ମଗୁପ୍ତଙ୍କ ସୂତ୍ର ବ୍ୟବହାର କରି ସମାଧାନ କର)
Solution:
\(\frac{2}{3}\) ÷ \(\frac{3}{5}\) କୁ ସମୀକ୍ଷା କଲେ ଆମେ ଲେଖୁପାରିବା
\(\frac{2}{3}\) ÷ \(\frac{3}{5}\) = \(\frac{2}{3}\) × \(\frac{5}{3}\) \(\frac{2 \times 5}{3 \times 3}\) = \(\frac{10}{9}\)

ଭାଜ୍ୟ, ଭାଜକ ଓ ଭାଗଫଳ :
ଆମେ ଯେତେବେଳେ ଦୁଇଟି ପୂର୍ଣ୍ଣସଂଖ୍ୟାକୁ ଭାଗକରୁ, ମନେକର 6 + 3
6 ÷ 3 = 2, ଏଠାରେ ଭାଗଫଳ = 2
ଏଠାରେ ଭାଗଫଳ ଭାଜ୍ୟଠାରୁ କମ୍, 6 ÷ 3 = 2, 2 < 6
କିନ୍ତୁ ଯେତେବେଳେ ଆମେ 6 କୁ \(\frac{1}{4}\) ଦ୍ଵାରା ଭାଗକରୁ 6 ÷ \(\frac{1}{4}\) = 24
ଏଠାରେ ଭାଗଫଳ, ଭାଜ୍ୟଠାରୁ ବଡ଼ 24 > 6
ଯେତେବେଳେ ଆମେ \(\frac{1}{8}\) କୁ \(\frac{1}{4}\) ଦ୍ଵାରା ଭାଗ କଲେ ପାଇବା, \(\frac{1}{8}\) ÷ \(\frac{1}{4}\) = \(\frac{1}{2}\)
ଏଠାରେ ମଧ୍ୟ ଭାଗଫଳ, ଭାଜ୍ୟଠାରୁ ବଡ଼ ।

8.3 ଭଗ୍ନସଂଖ୍ୟା ସମ୍ପର୍କିତ କିଛି ପ୍ରଶ୍ନ :

Page No. (193 to 201)

Question 1.
ଉଦାହରଣ – 3 : ଲୀନା 5 କପ୍ ଚା ତିଆରି କରିଥିଲା । ସେ ଏଥିପାଇଁ \(\frac{1}{4}\) ଲିଟର କ୍ଷୀର ବ୍ୟବହାର କରିଥିଲା । ତେବେ ପ୍ରତ୍ୟେକ କପ୍ ଚା’ ପାଇଁ କେତେ କ୍ଷୀର ଆବଶ୍ୟକ ହୋଇଛି ?
Solution:

ଲୀନା \(\frac{1}{4}\) ଲିଟର କ୍ଷୀରରେ 5 କପ୍ ଚା’ ତିଆରି କରେ ।
ତେଣୁ 1 କପ୍ ଚା’ ପାଇଁ ଦରକାର ହେଉଥ‌ିବା କ୍ଷୀରର ପରିମାଣ = \(\frac{1}{4}\) ÷ 5
ଆମେ ଏହାକୁ ଗୁଣନ କ୍ରିୟାରେ ଲେଖୁପାରିବା
= 5 × (ପ୍ରତ୍ୟେକ କପ୍’ ଚା’ ପାଇଁ ଆବଶ୍ୟକ କ୍ଷୀର) = \(\frac{1}{4}\)
ବ୍ରହ୍ମଗୁପ୍ତଙ୍କ ପଦ୍ଧତି ଅନୁସାରେ ଆମେ ନିମ୍ନଲିଖୁତ ଉପାୟରେ ଭାଗକ୍ରିୟା କରିଥାଉ ।

(i) ଭାଜକ 5 ର ବ୍ୟକ୍ରମ ହେଉଛି \(\frac{1}{5}\) (ପ୍ରଥମେ ନିର୍ଣ୍ଣୟ କରିବାକୁ ହୁଏ)

(ii) ଏହି ବ୍ୟକ୍ରମକୁ ଭାଗ୍ୟ \(\frac{1}{4}\) ଦ୍ଵାରା ଗୁଣନ କଲେ ପାଇବୁ \(\frac{1}{5}\) × \(\frac{1}{4}\) = \(\frac{1}{20}\) (ଭାଗଫଳ)
ତେଣୁ ପ୍ରତ୍ୟେକ କପ୍ ଚା’ ପାଇଁ \(\frac{1}{20}\) ଲିଟର କ୍ଷୀର ଆବଶ୍ୟକ ହୋଇଛି ।

Question 2.
ଉଦାହରଣ – 4 : 7\(\frac{1}{2}\) ବର୍ଗ ଏକକ କ୍ଷେତ୍ରଫଳ ବିଶିଷ୍ଟ କ୍ଷେତ୍ରକୁ \(\frac{1}{5}\) ଏକକ ବର୍ଗାକୃତି ଇଟାରେ ଆଚ୍ଛାଦିତ କର । ତେବେ ସେହି କ୍ଷେତ୍ରକୁ ଆଚ୍ଛାଦିତ କରିବା ପାଇଁ କେତୋଟି ବର୍ଗକୃତି ଇଟା ଦରକାର ?
Solution:
ପ୍ରତ୍ୟେକ ବର୍ଗାକୃତି ଇଟାର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{5}\) × \(\frac{1}{5}\) = \(\frac{1}{25}\) ବର୍ଗ ଏକକ ।
ଦିଆଯାଇଥିବା କ୍ଷେତ୍ରର ମୋଟ କ୍ଷେତ୍ରଫଳ = 7\(\frac{1}{2}\) ବର୍ଗ ଏକକ = \(\frac{15}{2}\) ବର୍ଗ ଏକକ ।
ଇଟା ସଂଖ୍ୟା = କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ÷ ପ୍ରତ୍ୟେକ ଇଟାର କ୍ଷେତ୍ରଫଳ = \(\frac{15}{2}\) ÷ \(\frac{1}{25}\)
ଏଠାରେ ଭାଜକର ବ୍ୟୁତ୍‌କ୍ରମ 25 ଅଟେ ।
ଭାଜ୍ୟ \(\frac{15}{2}\) ସହିତ ଏହି ବ୍ୟକ୍ରମକୁ ଗୁଣନ କଲେ ଆମେ ପାଇବା, 25 × \(\frac{15}{2}\) = \(\frac{25 \times 15}{2}\) = \(\frac{375}{2}\) ଖଣ୍ଡ ଇଟା ।

Question 3.
ଉଦାହରଣ – 5 : ଗୋଟିଏ କୁଣ୍ଡ ଚାରୋଟି ନଳଦ୍ଵାରା ସଂଯୁକ୍ତ । ପ୍ରଥମ ଦଳଦ୍ଵାରା ଏକ ଦିନରେ କୁଣ୍ଡଟି ପୂର୍ଷ ହୋଇପାରିବ । ଦ୍ଵିତୀୟଟି ଏହାକୁ ଅଧା ଦିନରେ ପୂରଣ କରିପାରିବ । ତୃତୀୟ ନଳ ଏହାକୁ ଦିନର ଏକ ଚତୁର୍ଥାଂଶ ସମୟରେ ପୂରଣ କରିପାରିବ । ଚତୁର୍ଥ ନଳଟି ଏକ ଦିନର ଏକ ପଞ୍ଚମାଂଶ ସମୟରେ କୁଣ୍ଡଟିକୁ ପୂରଣ କରିପାରିବ । ଯଦି ସମସ୍ତ ଚାରୋଟି ନଳ ଏକ ସମୟରେ ଖୋଲିଦିଆଯାଏ, ତେବେ କୁଣ୍ଡଟି କେତେ ସମୟରେ ପୂର୍ଣ୍ଣ ହେବ ?
Solution:
ଗୋଟିଏ ଦିନରେ ନଳଦ୍ଵାରା କେତୋଟି କୁଣ୍ଡ ପୂରଣ ହୋଇପାରିବ ଦେଖୁବା ।
ପ୍ରଥମ ନଳ । ଦିନରେ ପୂରଣ କରିବ = 1 ÷ 1 = 1ଟି ।
ଦ୍ଵିତୀୟ ନଳ । ଦିନରେ ପୂରଣ କରିବ = 1 ÷ \(\frac{1}{2}\) = 2 ଟି ।
ତୃତୀୟ ନଳ । ଦିନରେ ପୂରଣ କରିବ = 1 ÷ \(\frac{1}{4}\) = 4 ଟି ।
ଚତୁର୍ଥ ନଳ 1 ଦିନରେ ପୂରଣ କରିବ = 1 ÷ 1 ÷ \(\frac{1}{5}\) = 5 ଟି ।
ଗୋଟିଏ ଦିନ ଚାରୋଟି ନଳ ଏକାଠି ଏହିଭଳି କେତୋଟି କୁଣ୍ଡ ପୂରଣ କରିପାରିବେ ?
= 1 ଟି + 2 ଟି + 4 ଟି + 5 ଟି = 12 ଟି
ତେଣୁ ଚାରୋଟି ନଳଦ୍ଵାରା କୁଣ୍ଡଟିକୁ ପୂରଣ କରିବାପାଇଁ ସମୟ ଲାଗିବ \(\frac{1}{12}\) ଦିନ ।

ଭଗ୍ନଅଂଶର ସମ୍ପର୍କ :


ଏଠାରେ ଥ‌ିବା ବର୍ଗଚିତ୍ର ମଧ୍ୟରେ କେତେକ ରେଖା ଅଙ୍କନ କରାଯାଇଛି ଓ କେତେକ ଅଂଶକୁ ରେଖାଙ୍କିତ କରାଯାଇଛି ।
ରେଖାଙ୍କିତ ଅଂଶଟି ସମ୍ପୂର୍ଣ୍ଣ ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର କେତେ ଅଂଶ ?
ଏହି ସମସ୍ୟାଟିକୁ ବିଭିନ୍ନ ଉପାୟରେ ସମାଧାନ କରାଯାଇପାରିବ ।
ଏଠାରେ ଗୋଟିଏ ଉପାୟ ଦିଆଗଲା ।
ମନେକର ସମଗ୍ର ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ 1 (ଏକ) ବର୍ଗ ଏକକ ।
ଆମେ ଚିତ୍ର (ଖ)ରେ ବର୍ଗଚିତ୍ରର ଡାହାଣ ପାର୍ଶ୍ଵ କୋଣରେ ଏକ ଛୋଟ ବର୍ଗଚିତ୍ର ଦେଖୁଛୁ । ତାହା ବଡ଼ ବର୍ଗକ୍ଷେତ୍ରର । \(\frac{1}{4}\) ଅଂଶରେ ଅଛି ।
ବଡ଼ ବର୍ଗକ୍ଷେତ୍ରର ଲାଲ ଅଂଶର କ୍ଷେତ୍ରଫଳ \(\frac{1}{4}\) ବର୍ଗ ଏକକ ହେବ ।
ଏହି ରଙ୍ଗୀନ ବର୍ଗଚିତ୍ରକୁ ଅନୁଧ୍ୟାନ କରିବା ।
ଏହା ଭିତରେ ଦୁଇଟି ବଡ଼ ତ୍ରିଭୁଜ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକ ବଡ଼ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ ଲାଲ ବର୍ଗଚିତ୍ରର ଅଧା ।
(ଗ) ତେଣୁ ହଳଦିଆ ରଙ୍ଗ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) × \(\frac{1}{4}\) = \(\frac{1}{8}\) ବର୍ଗଏକକ ।

Question 1.
ବର୍ତ୍ତମାନ ହଳଦିଆ ତ୍ରିଭୁଜର କେତେ ଅଂଶ ରେଖାଙ୍କିତ ହୋଛି ?

Solution:
ଯଦି ତ୍ରିଭୁଜ ମଧ୍ୟସ୍ଥ କ୍ଷୁଦ୍ର ବର୍ଗଚିତ୍ରଟିକୁ ଦୁଇ ସମାନ ଭାଗ କରିବ । ତେବେ ହଳଦିଆ ତ୍ରିଭୁଜଟି ସମାନ ଚାରି ଭାଗରେ ବିଭକ୍ତ ହେବ ଏବଂ ଏଥୁରୁ 3 ସମାନ ଭାଗ ରେଖାଙ୍କିତ ହୋଇଥିବାର ଦେଖୁପାରିବ । ଏଣୁ ରେଖାଙ୍କିତ ଅଂଶଟି ଏହି ହଳଦିଆ ରଙ୍ଗ ତ୍ରିଭୁଜର \(\frac{3}{4}\) ଅଂଶ ।
ତେଣୁ ରେଖାଙ୍କିତ ଅଂଶର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{8}\) ର \(\frac{3}{4}\) = \(\frac{1}{8}\) × \(\frac{3}{4}\) = \(\frac{3}{32}\) ବର୍ଗ ଏକକ ।
∴ ରେଖାଙ୍କିତ ଅଂଶଟି ସମଗ୍ର ବର୍ଗଚିତ୍ରର \(\frac{3}{32}\) ଅଂଶ ଅଧିକାର କରିଛି ।

Question 2.
ତଳେ ଦିଆଯାଇଥବା ପ୍ରତ୍ୟେକ ଚିତ୍ରରେ ରେଖାଙ୍କିତ ଅଂଶଟି ସମଗ୍ର ବର୍ଗକ୍ଷେତ୍ରର କେତେ ଅଂଶରେ ରହିଛି, ତାହା ନିଶ୍ଚୟ କର ।

Solution:
ଛାତ୍ରଛାତ୍ରୀମାନେ ନିଜେ ଅଭ୍ୟାସ କରନ୍ତୁ ।

Question 3.
ଜଣେ କୃପଣର ଦାନ : ନିମ୍ନଲିଖତ ସମସ୍ୟାଟିକୁ ଭାସ୍କରାଚାର୍ଯ୍ୟ (ଦ୍ଵିତୀୟ ଭାସ୍କର)ଙ୍କ 1150 ଖ୍ରୀଷ୍ଟାବ୍ଦରେ ଲିଖ୍ତ ପୁସ୍ତକ ଲୀଳାବତୀରୁ ଅନୁବାଦ କରାଯାଇଛି : ହେ ବୁଦ୍ଧଜନେ ! ଜଣେ କୃପଣ ଲୋକ ଭିକାରୀକୁ ଏକ ଡ୍ରାମାର \(\frac{1}{5}\) ଅଂଶର \(\frac{1}{16}\) ଅଂଶର \(\frac{1}{4}\) ଅଂଶର \(\frac{1}{2}\) ଅଂଶର \(\frac{2}{3}\) ଅଂଶର \(\frac{3}{4}\) ଅଂଶ ଦାନ କଲେ । ଯଦି ଅଛି, ତେବେ ମୋତେ କୁହ ସେହି କୃପଣ ଲୋକଟି ଭିକାରୀକୁ ତୁମର ଭଗ୍ନସଂଖ୍ୟା ଗଣିତରେ ଦକ୍ଷତା କେତେ କଉଡ଼ି ଦାନ କରିଥିଲେ ?
Solution:
ତତ୍‌କାଳୀନ ସମୟରେ ଡ୍ରାମା ଏକ ରୂପା ମୁଦ୍ରାକୁ ବୁଝାଏ ଏବଂ ଏହି କାହାଣୀରେ କୁହାଯାଇଛି ଯେ ଏକ ଡ୍ରାମାର ମୂଲ୍ୟ 1280 କଉଡ଼ି ସହିତ ସମାନ ।
ଆସ ଦେଖୁବା ସେହି କୃପଣ ଲୋକଟି ଏକ ଡ୍ରାମର କେତେ ଅଂଶ ଦାନ କରିଥିଲେ ।
ଡ୍ରାମାର ଦେଇଥିବା ଅଂଶ = \(\frac{1}{5}\) × \(\frac{1}{16}\) × \(\frac{1}{4}\) × \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) = 6
(କଉଡ଼ି – ଏହା ପୂର୍ବକାଳ ପ୍ରଚଳିତ ଏକ ମୁଦ୍ରା)
= 1280 × \(\frac{1}{5}\) × \(\frac{1}{16}\) × \(\frac{1}{4}\) × \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) = 1280 × \(\frac{6}{7680}\) × \(\frac{7680}{7680}\) = 1
ଏହାକୁ ଲଘିଷ୍ଠ ଆକାରରେ ସରଳ କଲେ ଆମେ ପାଇବା \(\frac{1}{7680}\) = \(\frac{1}{1280}\)
∴ ସେ ମାତ୍ର ଗୋଟିଏ କଉଡ଼ି ଦାନ ଦେଇଥିଲେ ।
ବିକଳ୍ପ ଉତ୍ତର : ଏଠାରେ ଆମେ ଜାଣିଲୁ । (ଏକ) ଡ୍ରାମା = 1280 କଉଡ଼ି ।
ଅର୍ଥାତ୍ ସେ ପ୍ରଥମେ 1280 କଉଡ଼ିର \(\frac{1}{5}\) = 1280 × \(\frac{1}{5}\) = 256
256 ର \(\frac{1}{16}\) = 256 × \(\frac{1}{16}\) = 16, 16 ର \(\frac{1}{4}\) = 16 × \(\frac{1}{4}\)= 4
4 ର \(\frac{1}{2}\) = 4 × \(\frac{1}{2}\) = 2, 2 ର \(\frac{2}{3}\) = 2 × \(\frac{2}{3}\) = \(\frac{4}{3}\), \(\frac{4}{3}\) ର \(\frac{3}{4}\) = \(\frac{4}{3}\) × \(\frac{3}{4}\) = 1
ଯଦି ଆମେ ଧରିନେବା 1 ସୁନା ଦିନାର = 12 ରୁପା ଡ୍ରାମା ।
1 ରୁପା ଡ୍ରାମା = 4 ତମ୍ବା ପାନା, 1 ତମ୍ବା ପାନା = 6 ମଶାକା ଏବଂ 1 ତମ୍ବା ପାନା = 30 କଉଡ଼ି ।
1 ଡମ୍ବା ପାନା = ସୁନା ଦିନାର \(\frac{1}{48}\) ସୁନା ଦିନାର (\(\frac{1}{12}\) × \(\frac{1}{4}\))
1 କଉଡ଼ି = \(\frac{1}{30}\) ତମ୍ବା ପାନା, 1 କଉଡ଼ି = \(\frac{1}{1440}\) ସୁନା ଦିନାର ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକୁ ସମାଧାନ କର ଓ ମୂଲ୍ୟ ନିରୂପଣ କର ।

Solution:

Question 2.
ତଳେ ଦିଆଯାଇଥବା ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନ ପାଇଁ ସମାଧାନ ନିମନ୍ତେ ଦିଆଯାଇଥିବା ଭକ୍ତି ମଧ୍ୟରୁ ଠିକ୍ ଉକ୍ତିଟି ବାଛ । ତା’ପରେ ସରଳ କର ।
(a) ମାରିଆ ସ୍କୁଲ ବ୍ୟାଗ୍ଗୁଡ଼ିକରେ ଲେସ୍ ଲଗାଇ ସଜାଇବା ପାଇଁ 8 ମିଟର ଲେସ୍ କିଣିଥିଲା । ସେ ପ୍ରତ୍ୟେକ ବ୍ୟାଗରେ \(\frac{1}{4}\) ମିଟର ଲେସ୍ ଲଗାଇ ସବୁ ଲେସ୍ ଶେଷ କରିଥିଲେ । ସେ କେତୋଟି ବ୍ୟାଗରେ ଲେସ୍ ଲଗାଇଥିଲେ ?
(i) 8 × \(\frac{1}{4}\)
(ii) \(\frac{1}{8}\) × \(\frac{1}{4}\)
(iii) 8 ÷ \(\frac{1}{4}\)
(iv) \(\frac{1}{4}\) ÷ 8
Solution:
ମାରିଆ ପାଖରେ ଥିଲା ଲେସ୍ = 8 ମିଟର ।
ପ୍ରତ୍ୟେକ ବ୍ୟାଗରେ ଲାଗେ = \(\frac{1}{4}\) ମିଟର ଲେସ୍ ।
8 ମିଟର ଲେସ୍ ଲାଗିବ = 8 ÷ \(\frac{1}{4}\)ବ୍ୟାଗରେ ।
∴ (iii) ଟି ଠିକ୍ ଉତ୍ତର ।
ଯେହେତୁ \(\frac{1}{4}\) ର ପ୍ରତିଲୋମାନୁପାତୀ 4 ⇒ 8 ÷ \(\frac{1}{4}\) = 8 × 4 = 32ଟି ବ୍ୟାଗରେ ।
∴ ସେ 32ଟି ବ୍ୟାଗରେ ଲେସ୍ ଲଗାଇଥିଲେ ।

(b) 8 ଟି ବ୍ୟାଜ୍ ତିଆରି କରିବା ପାଇଁ \(\frac{1}{2}\) ମିଟର ରିବନ ଫିତା ବ୍ୟବହୃତ ହୁଏ । ପ୍ରତ୍ୟେକ ବ୍ୟାଜ୍
ତିଆରି ପାଇଁ କେତେ ଲମ୍ବର ରିବନ ଫିତା ଦରକାର ?
(i) 8 × \(\frac{1}{2}\)
(ii) \(\frac{1}{2}\) ÷ \(\frac{1}{8}\)
(iii) 8 ÷ \(\frac{1}{2}\)
(iv) \(\frac{1}{2}\) ÷ 8
Solution:
8ଟି ବ୍ୟାଜ୍ ତିଆରି ପାଇଁ ରିବନ୍ ଫିତା ଦରକାର = \(\frac{1}{2}\) ମିଟର ।
1ଟି ବ୍ୟାଜ୍ ତିଆରି ପାଇଁ ରିବନ ଫିତା ଆବଶ୍ୟକ = \(\frac{1}{2}\) ÷ 8 ମିଟର ।
∴ (iv) ଟି ଠିକ୍ ଉତ୍ତର ।
ଯେହେତୁ 8ର ପ୍ରତିଲୋମାନୁପାତୀ \(\frac{1}{8}\) ⇒ \(\frac{1}{2}\) ÷ 8 = \(\frac{1}{2}\) × \(\frac{1}{8}\) = \(\frac{1}{16}\) ମିଟର ।
∴ ପ୍ରତ୍ୟେକ ବ୍ୟାଜ୍ ତିଆରି ପାଇଁ \(\frac{1}{16}\) ମିଟର ରିବନ ଫିତା ଦରକାର ।

(c) ଜଣେ ରୁଟି ପ୍ରସ୍ତୁତକାରୀଙ୍କୁ ଗୋଟିଏ ରୁଟି ତିଆରି ପାଇଁ \(\frac{1}{6}\) କିଲୋଗ୍ରାମ୍ ମଇଦା ଦରକାର ହୁଏ । ତାଙ୍କ ପାଖରେ 5 କିଲୋଗ୍ରାମ୍ ମଇଦା ଅଛି । ସେ କେତୋଟି ରୁଟି ତିଆରି କରିପାରିବେ ?
(i) 5 × \(\frac{1}{6}\)
(ii) \(\frac{1}{6}\) ÷ 5
(iii) 5 ÷ \(\frac{1}{6}\)
(iv) 5 × 6
Solution:
\(\frac{1}{6}\) କିଲୋଗ୍ରାମ୍‌ରେ ତିଆରି ହୁଏ = 1 ଟି ରୁଟି ।
5 କିଲୋଗ୍ରାମ୍ରେ ତିଆରି ହେବ = 5 ÷ \(\frac{1}{6}\) ରୁଟି ।
∴ (iii) ଟି ଠିକ୍ ଉତ୍ତର ।
ଯେହେତୁ \(\frac{1}{6 }\) ର ପ୍ରତିଲୋମାନୁପାତୀ 6 ⇒ 5 ÷ \(\frac{1}{6}\) = 5 × 6 = 30ଟି ରୁଟି ।
∴ ସେ 30ଟି ରୁଟି ତିଆରି କରିପାରିବେ ।

Question 3.
ଯଦି 12ଟି ରୁଟି ତିଆରି କରିବା ପାଇଁ \(\frac{1}{4}\) କିଲୋଗ୍ରାମ୍ ଅଟା ଦରକାର ହୁଏ । ତେବେ ସେହିପରି ଟି ରୁଟି ତିଆରି କରିବାକୁ କେତେ କି.ଗ୍ରା. ଅଟା ଦରକାର ହେବ ?
Solution:
12ଟି ରୁଟି ତିଆରି କରିବା ପାଇଁ ଅଟା ଦରକାର = \(\frac{1}{4}\) କିଲୋଗ୍ରାମ୍ ।
1 ଟି ରୁଟି ତିଆରି କରିବା ପାଇଁ ଅଟା ଦରକାର = \(\frac{1}{4}\) ÷ 12 କିଲୋଗ୍ରାମ୍ ।
ଯେହେତୁ 12 ର ପ୍ରତିଲୋମାନୁପାଢୀ \(\frac{1}{12}\) ⇒ \(\frac{1}{4}\) ÷ 12 = \(\frac{1}{4}\) × \(\frac{1}{12}\) \(\frac{1}{48}\) = କିଲୋଗ୍ରାମ୍ ।
6 ଟି.ରୁଟି ତିଆରି କରିବା ପାଇଁ ଅଟା ଦରକାର = \(\frac{1}{48}\) × 6 = \(\frac{1}{8}\) କିଲୋଗ୍ରାମ୍ ।
∴ 6ଟି ରୁଟି ତିଆରି କରିବାକୁ \(\frac{1}{8}\) କି.ଗ୍ରା. ଅଟା ଦରକାର ହେବ ।

Question 4.
9ମ ଶତାବ୍ଦୀରେ ଶ୍ରୀଧରାଚାର୍ଯ୍ୟ, ତାଙ୍କଦ୍ୱାରା ଲିଖୁତ ପାଟୀଗଣିତ ପୁସ୍ତକରେ ଏହି ସମସ୍ୟାଟିକୁ ଉଲ୍ଲେଖ କରିଥିଲେ । ବନ୍ଧୁ ‘ଚିନ୍ତା କରିବା ଓ ପରେ କହିବା – 1 ÷ \(\frac{1}{6}\) , 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) ଓ 1 ÷ \(\frac{1}{2}\)‘ କୁ ମିଶାଇଲେ ମିଶାଣଫଳ କେତେ ହେବ ? ବନ୍ଧୁ କ’ଣ ଉତ୍ତର ଦେବା ଉଚିତ ?
Solution:
ଯୋଗ କରିଲେ = 1 ÷ \(\frac{1}{6}\) + 1 ÷ \(\frac{1}{10}\) + 1 ÷ \(\frac{1}{13}\) + 1 ÷ \(\frac{1}{9}\) + 1 ÷ \(\frac{1}{2}\)
ଯେହେତୁ \(\frac{1}{6}\) ର ପ୍ରତିଲୋମାନୁପାତୀ 6, \(\frac{1}{10}\) ର 10,
\(\frac{1}{13}\) ର 13, \(\frac{1}{9}\) ର 9, \(\frac{1}{2}\) ର 2
∴ ମିଶାଣଫଳ 1 ÷ \(\frac{1}{6}\) + 1 ÷ \(\frac{1}{10}\) + 1 ÷ \(\frac{1}{13}\) + 1 ÷ \(\frac{1}{9}\) + 1 ÷ \(\frac{1}{2}\)‘
= 1 × 6 + 1 × 10 + 1 × 13 + 1 × 9 + 1 × 2 = 6 + 10 + 13 + 9 + 2 = 40
∴ ବନ୍ଧୁ 40 ଉତ୍ତର ଦେବା ଉଚିତ ।

Question 5.
ମୀରା 400 ପୃଷ୍ଠା ବିଶିଷ୍ଟ ଏକ ଉପନ୍ୟାସ ପଢୁଛନ୍ତି । ସେ ଗତିକାଲି ବହିର \(\frac{1}{5}\) ପୃଷ୍ଠା ଏବଂ ଆଜି ବହିର
\(\frac{3}{10}\) ପୃଷ୍ଠା ପଢ଼ିଛନ୍ତି । ଉପନ୍ୟାସଟି ଶେଷ କରିବା ପାଇଁ ତାଙ୍କୁ ଆଉ କେତେ ପୃଷ୍ଠା ପଢ଼ିବାକୁ ପଡ଼ିବ ?
Solution:
ମୀରା 400 ପୃଷ୍ଠା ବିଶିଷ୍ଟ ଏକ ଉପନ୍ୟାସ ପଢୁଛନ୍ତି ।
ସେ ଗତକାଲି ବହିର \(\frac{1}{5}\) ପୃଷ୍ଠା ପଢ଼ିଛନ୍ତି ।
ସେ ଗତକାଲି ପଢ଼ିଛନ୍ତି = 400 × \(\frac{1}{5}\) = 80 ପୃଷ୍ଠା ।
ସେ ଆଜି ବହିର \(\frac{3}{10}\) ପୃଷ୍ଠା ପଢ଼ିଛନ୍ତି ।
ସେ ଆଜି ପଢ଼ିଛନ୍ତି = 400 × \(\frac{3}{10}\) = 120 ପଣ୍ଡା ।
∴ ସେ ଦୁଇ ଦିନରେ ପଢ଼ିଛନ୍ତି = 80 + 120 = 200 ପୃଷ୍ଠା |
ବାକି ରହିଛି ପଢ଼ିବାକୁ = 400 – 200 = 200 ପୃଷ୍ଠା ।
∴ ଉପନ୍ୟାସଟି ଶେଷ କରିବା ପାଇଁ ତାଙ୍କୁ ଆଉ 200 ପୃଷ୍ଠା ପଢ଼ିବାକୁ ପଡ଼ିବ ।

Question 6.
ଗୋଟିଏ କାର 1 ଲିଟର ପେଟ୍ରୋଲ ବ୍ୟବହାର କରି 16 କିଲୋମିଟର ଚାଲିଥାଏ । 2\(\frac{3}{4}\) ଲିଟର ପେଟ୍ରୋଲ ବ୍ୟବହାର କରି କେତେ କିଲୋମିଟର ଚାଲିପାରିବ ?
Solution:
1 ଲିଟର ପେଟ୍ରୋଲରେ ଯାଏ = 16 କିଲୋମିଟର ।
2\(\frac{3}{4}\) ଲିଟର ପେଟ୍ରୋଲରେ ଯାଏ = 16 × 2\(\frac{3}{4}\) = 16 × \(\frac{11}{4}\) = 4 × 11 = 44 କି.ମି.
∴ 2\(\frac{3}{4}\) ଲିଟର ପେଟ୍ରୋଲ ବ୍ୟବହାର କରି କାଟି 44 କିଲୋମିଟର ଚାଲିପାରିବ ।

Question 7.
ଅମୃତ ପାଲ ତାଙ୍କ ଅବକାଶ ଛୁଟିରେ ଏକ ପର୍ଯ୍ୟଟନ ସ୍ଥଳକୁ ଯିବାପାଇଁ ସ୍ଥିର କଲେ । ଯଦି ସେ ଟ୍ରେନ୍‌ରେ ଯିବେ, ତେବେ ତାଙ୍କୁ ସେଠାରେ ପହଞ୍ଚିବାକୁ 5\(\frac{1}{6}\) ଘଣ୍ଟା ସମୟ ଲାଗିବ । ଯଦି ସେ ବିମାନରେ ଯିବେ ତେବେ ତାଙ୍କୁ \(\frac{1}{2}\) ଘଣ୍ଟା ସମୟ ଲାଗିବ । ବିମାନରେ ଗଲେ ସେ କେତେ ସମୟ ବଞ୍ଚାଇ ପାରିବେ ?
Solution:
ଯଦି ସେ ଟ୍ରେନ୍‌ରେ ଯିବେ ସମୟ ଲାଗିବ = 5\(\frac{1}{6}\) ଘଣ୍ଟା = \(\frac{31}{6}\) ଘଣ୍ଟା ।
ଯଦି ସେ ବିମାନରେ ଯିବେ ସମୟ ଲାଗିବ = \(\frac{1}{2}\) ଘଣ୍ଟା ।
ବିମାନରେ ଗଲେ ସମୟ ବଞ୍ଚିପାରିବ = \(\frac{31}{6}\) – \(\frac{1}{2}\) ଘଣ୍ଟା = \(\frac{31}{3}\) – \(\frac{3}{6}\) = \(\frac{31 – 3}{6}\) = \(\frac{28}{6}\) = \(\frac{14}{3}\) = 4\(\frac{2}{3}\) ଘଣ୍ଟା
∴ ବିମାନରେ ଗଲେ ସେ 4\(\frac{2}{3}\) ଘଣ୍ଟା ସମୟ ବଞ୍ଚାଇ ପାରିବେ ।

Question 8.
ସରିତାଙ୍କ ଜେଜେ ମା’ ଏକ କେକ୍ ତିଆରି କରିଥିଲେ । ସରିତା ଓ ତା’ର ସମ୍ପର୍କୀୟ ଭାଇମାନେ କେଜ୍‌ର \(\frac{4}{5}\) ଅଂଶ ଖାଇଦେଲେ । ବଳକା କେକ୍‌ ସରିତା ଓ ତା’ର ତିନି ସାଙ୍ଗଙ୍କ ମଧ୍ୟରେ ସମାନ ଭାବରେ ବାଣ୍ଟିଦେଲା । ତେବେ ପ୍ରତ୍ୟେକ ସାଙ୍ଗ କେକ୍‌ର କେତେ ଅଂଶ ପାଇଲେ ?
Solution:
ମନେକର କେକ୍‌କୁ 1 ଧରାଯାଉ ।
ସରିତା ଓ ତା’ ସମ୍ପର୍କୀୟ ଭାଇମାନେ କେକ୍‌ର \(\frac{4}{5}\) ଅଂଶ ଖାଇଦେଲେ ।
ଆଉ କେକ୍ ରହିଲା = 1 – \(\frac{4}{5}\) = \(\frac{4}{5}\) = ଅଂଶ ।
\(\frac{1}{5}\) ଅଂଶ କେକ୍‌କୁ ତିନି ଜଣ ସାଙ୍ଗଙ୍କ ମଧ୍ୟରେ ସମାନ ଭାବରେ ବାଣ୍ଟିଦିଆଗଲା ।
∴ ପ୍ରତ୍ୟେକ ସାଙ୍ଗ ପାଇବେ = \(\frac{1}{5}\) ÷ 3 ଅଂଶ ।
ଯେହେତୁ 3 ର ପ୍ରତିଲୋମାନୁପାତୀ \(\frac{1}{3}\) ⇒ \(\frac{1}{5}\) ÷ 3 = \(\frac{1}{5}\) × \(\frac{1}{3}\) = \(\frac{1}{15}\)
∴ ପ୍ରତ୍ୟେକ ସାଙ୍ଗ କେକ୍‌ର \(\frac{1}{15}\) ଅଂଶ ପାଇଲେ ।

Question 9.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ବିକଳ୍ପ ଉତ୍ତର ମଧ୍ୟରୁ ଠିକ୍ ଉକ୍ତିଗୁଡ଼ିକୁ ବାଛ । ଯେଉଁ ଉକ୍ତିଗୁଡ଼ିକ (\(\frac{565}{465}\) × \(\frac{707}{676}\)) ର ଗୁଣଫଳକୁ ଠିକ୍ ଭାବେ ବର୍ଣ୍ଣନା କରୁଛି ।
(a) > \(\frac{565}{465}\)
(b) < \(\frac{565}{465}\)
(c) > \(\frac{707}{676}\)
(d) < \(\frac{707}{676}\)
(e) > 1
(f) < 1
Solution:
\(\frac{565}{465}\) × \(\frac{707}{676}\) > 1 → ବିକଳ୍ପ (e)
\(\frac{565}{465}\) × \(\frac{707}{676}\) > \(\frac{565}{465}\) → ବିକଳ୍ପ (a)
\(\frac{565}{465}\) × \(\frac{707}{676}\) > \(\frac{707}{676}\) → ବିକଳ୍ପ (c)
∴ ସଠିକ୍ ଉତ୍ତର ହେଉଛି (a), (c) ଓ (e)

Question 10.
ସମସ୍ତ ବର୍ଗଚିତ୍ରର କେତେ ଅଂଶ ରେଖାଙ୍କିତ ହୋଇଛି ?
(ଯହିଁରେ ଚିହ୍ନିତ ବିନ୍ଦୁଟି ସାନ ବର୍ଗଚିତ୍ରର ମଧ୍ୟବିନ୍ଦୁ)

Solution:
ଏଠାରେ ସମସ୍ତ ବର୍ଗଚିତ୍ରକୁ 4ଟି ଛୋଟ ବର୍ଗକ୍ଷେତ୍ର ଅଂଶରେ ଭାଗ ହେଉଛି ।
ଡାହାଣ ପାର୍ଶ୍ଵର ତଳେ ଥ‌ିବା ଛୋଟ ବର୍ଗଚିତ୍ରରେ ରେଖାଙ୍କିତ ଅଂଶ ଦେଖୁଛୁ ।
ଯାହା ସମସ୍ତ ବର୍ଗକ୍ଷେତ୍ରର \(\frac{1}{4}\) ଅଂଶ ।
ଏହା ମଧ୍ଯରେ ହଳଦିଆ ରଙ୍ଗର ତ୍ରିଭୁଜ ଅଂଶ = \(\frac{1}{2}\) × \(\frac{1}{4}\) = \(\frac{1}{8}\) ଅଂଶ ।
∴ ସମସ୍ତ ବର୍ଗଚିତ୍ରର ରେଖାଙ୍କିତ ଅଂଶ = \(\frac{3}{4}\) × \(\frac{1}{8}\) = \(\frac{3}{32}\)

Question 11.
ଖାଦ୍ୟ ସନ୍ଧନାରେ ପିମ୍ପୁଡ଼ିମାନଙ୍କର ଏକ ଦଳ ବାହାରି ପଡ଼ିଲେ । ସେମାନେ ଖାଦ୍ୟ ଖୋଜିବା ସମୟରେ ପ୍ରତ୍ୟେକ ଛକସ୍ଥାନରେ ସମାନ ଭାବରେ ଭାଗ ହୋଇ ଚାଲନ୍ତି ଏବଂ ଦୁଇଟି ଖାଦ୍ୟ ଥ‌ିବା ସ୍ଥାନକୁ ଯଥା- ଗୋଟିଏ ଆମ୍ବ ଗଛ ନିକଟରେ ଏବଂ ଅନ୍ୟଟି ଆଖୁ କ୍ଷେତ ନିକଟରେ ପହଞ୍ଚନ୍ତି । ତେବେ ଦଳର କେତେ ଅଂଶ ପ୍ରତ୍ୟେକ ଖାଦ୍ୟ ଉତ୍ସ ନିକଟରେ ପହଞ୍ଚିଗଲେ ?

Solution:
ଖାଦ୍ୟ ଉତ୍ସରେ ଶେଷ ହେଉଥ‌ିବା 7ଟି ରାସ୍ତା ଅଛି ।
ଆମ୍ବଗଛ ନିକଟକୁ 5ଟି ରାସ୍ତା ଏବଂ ଆଖୁଗଛ ନିକଟକୁ 2ଟି ରାସ୍ତା ଅଛି ।
ପ୍ରତ୍ୟେକ ଲାଲ ବିନ୍ଦୁ (ଛକ)ରେ ପିମ୍ପୁଡ଼ିମାନେ ଦୁଇଭାଗରେ ବିଭକ୍ତ ହେଲେ ।
ଖାଦ୍ୟ ଉତ୍ସରେ ପହଞ୍ଚିବା ପାଇଁ ପ୍ରତ୍ୟେକ ଦଳର ଅଂଶ = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{16}\) ଅଂଶ ।
ତେଣୁ, ଆମ୍ବଗଛ ନିକଟରେ ପହଞ୍ଚିବା ପାଇଁ 5ଟି ରାସ୍ତାର ପିମ୍ପୁଡ଼ିଙ୍କ ସଂଖ୍ୟା 5 × \(\frac{1}{16}\) = \(\frac{5}{16}\) ଅଂଶ।
∴ ଆଖୁଗଛ ନିକଟରେ ପହଞ୍ଚିବା ପାଇଁ 2ଟି ରାସ୍ତାରେ ପିମ୍ପୁଡ଼ି ସଂଖ୍ୟା 2 × \(\frac{1}{16}\) = \(\frac{2}{16}\) = \(\frac{1}{8}\) ଅଂଶ ।

Question 12.
ସରଳ କର ।
(a) 1 – \(\frac{1}{2}\)
(b) (1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\))
(c) (1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\))
(d) (1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\))
Solution:

For better exam readiness, students can include Class 7 Science MCQ and Curiosity Chapter 4 The World of Metals and Non-metals MCQ in their daily study routine.

Class 7 Science Chapter 4 The World of Metals and Non-metals MCQ

The World of Metals and Non-metals MCQ Class 7

MCQ On The World of Metals and Non-metals Class 7

Question 1.
Which of the following objects is lustrous as well as a good conductor of electricity?
(a) Eraser
(b) Wooden block
(c) Silver coin
(d) Glass tumbler
Answer:
(c) The silver coin, is lustrous as well as a good conductor of electricity, as silver is a metal.

Question 2.
An ironsmith heats an iron rod before hammering it into a desired shape. Which property of metals is mainly utilised in this process?
(a) Sonority
(b) Ductility
(c) Malleability
(d) Lustre
Answer:
(c) Malleability

Question 3.
A student needs a material to make thin wires for an electrical circuit. Apart from being a good conductor of electricity, which of the following properties should the material possess?
(a) Malleability
(b) Hardness
(c) Ductility
(d) Sonority
Answer:
(c) Ductility

Question 4.
Which of the following objects produces a ringing sound when dropped from a certain height?

Answer:
(b) A metal spoon produces a ringing sound when dropped from a certain height as it is sonorous, while the piece of coal, wooden block and plastic bottle produce dull sounds.

Question 5.
On a winter evening, Nitin’s mother cooked soup and asked him to taste the salt. He noticed that the metal spoon became very hot, while the wooden spoon stayed cool, even though both were placed in the same soup. This is because

(a) the metal spoon was smaller than the wooden spoon.
(b) metal is a good conductor of heat, while wood is a poor conductor of heat.
(c) the metal spoon was lustrous, while the wooden spoon was non-lustrous.
(d) metal is a poor conductor of heat, while wood is a good conductor of heat.
Answer:
(b) Metal is a good conductor of heat; thus, the heat from the hot soup transfers easily throughout the metal spoon. In contrast, wood is a poor conductor of heat; therefore, the wooden spoon stayed cool.

Question 6.
On a Sunday morning, Amar’s father called an electrician to their home to repair the electric heater. Amar noticed that the electrician was wearing rubber gloves and shoes while working. What could be the possible reason for this?
(a) Rubber provides a better grip on his tools while working.
(b) Rubber is a poor conductor of electricity; thus, it protects the electrician against electric shock.
(c) Rubber is a good conductor of electricity; hence, it increases the flow of electric current in the heater.
(d) Rubber helps keep the electrician’s hands and feet warm while working with electrical wires.
Answer:
(b) Rubber is a poor conductor of electricity; thus, it protects the electrician against electric shock.

Question 7.
While returning from school with her father, Reena saw a small furniture shop where she noticed an old, damaged iron almirah with several dark brown patches on it. Her father explained that this is due to the rusting of iron. Select the option that represents the correct condition(s) required for the rusting of iron.
(a) Presence of air (oxygen) only
(b) Presence of hydrogen only
(c) Presence of water only
(d) Presence of both air and water
Answer:
(d) Presence of both air and water

Question 8.
An element X is burned in air. To the product formed, water is added and tested with red and blue litmus papers. The following results are obtained:
(i) Blue litmus paper turns red.
(ii) Red litmus paper remains red.
Which of the following statements is true about element X and the nature of its oxide?
(a) X is a non-metal that forms a basic oxide.
(b) X is a metal that forms an acidic oxide.
(c) X is a metal that forms a basic oxide.
(d) X is a non-metal that forms an acidic oxide.
Answer:
(d) An acidic oxide turns blue litmus paper red and does not change the colour of red litmus paper. Non-metals generally form acidic oxides. Hence, X must be a non-metal.

Question 9.
Consider the Venn diagram given alongside. Here, A represents the group of properties shown by metals only, B represents the group of properties shown by non-metals only, and the overlapping part C represents the properties shown by both metals and non-metals. Which of the following properties can be written in the overlapping part?

(a) Sonority
(b) Good conductors of heat and electricity
(c) Formation of oxides
(d) Malleability and Ductility
Answer:
(c) Both metals and non-metals form oxides when burned in the presence of air (oxygen).

Question 10.
A metal X is so soft that it can be cut with a knife and is also stored under kerosene. Which of the following statements is true regarding metal X?
(a) Metal X does not react with oxygen or water.
(b) The oxide of metal X turns moist blue litmus red.
(c) The oxide of metal X turns moist red litmus blue.
(d) No heat is generated when metal X reacts with oxygen and water.
Answer:
(c) It is given that metal X is soft and can be cut with the help of a knife. Therefore, metal X can be sodium or potassium. Both these metals react vigorously with oxygen and water, and a lot of heat is generated in the reactions. Storing these metals under kerosene prevents their exposure to air and moisture. Metallic oxides are basic in nature; hence, they turn moist red litmus blue.

Question 11.
During a science fair, students were asked to design a cooking pan that heats quickly but has a handle that remains cool. Which combination of materials would be the most suitable?
(a) Iron pan with an iron handle
(b) Aluminium pan with a plastic handle
(c) Plastic pan with an aluminium handle
(d) Copper pan with a copper handle
Answer:
(b) Aluminium is a good conductor of heat, while plastic is a poor conductor of heat and therefore it remains cool.

Question 12.
During a music class, Rohan explored different musical instruments such as the veena, sitar, violin, and guitar. He noticed that there are metal wires that play an important role in producing a variety of sounds. Which of the following sets of properties of metals is used here?
(a) Lustre and ductility
(b) Sonority and ductility
(c) Ductility and malleability
(d) Lustre and malleability
Answer:
(b) Ductility – so that the metals are drawn into wires. Sonority – so that the metal wires produce sweet sound when plucked or bowed.

Question 13.
Students of class 7 were instructed to test the electrical conductivity of different materials using a simple circuit. They were provided with an aluminium foil, an iron nail, an eraser and a piece of dry wood. Observations made are listed in the following table:

Select the option that correctly fills the numbered blanks:
(a) (i) – Yes, (ii) – Yes, (iii) – Poor, (iv) – No, (v) – Good
(b) (i) – No, (ii) – Yes, (iii) – Poor, (iv) – No, (v) – Good
(c) (i) – Yes, (ii) – Yes, (iii) – Good, (iv) – Yes, (v) – Poor
(d) (i) – Yes, (ii) – Yes, (iii) – Good, (iv) – No, (v) – Poor
Answer:
(d) The material that makes the bulb glow is a good conductor of electricity, while the material that does not make the bulb glow is a poor conductor of electricity.

Question 14.
Which of the following metals are soft?
(i) Sodium
(ii) Iron
(iii) Gold
(iv) Potassium
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (iii) and (iv)
Answer:
(c) Sodium and potassium are so soft that they can be cut with a knife.

Question 15.
Mehak loved dancing, so she joined a dance class at her school. While practising, she heard the pleasant ringing sound of her ghungroos. After the class ended, she heard the ringing sound of the school bell. Based on this scenario, which of the following statements is/are correct?
(i) Both the ghungroos and the school bell produced sound as they were malleable.
(ii) Both the ghungroos and the school bell produced sound as they were sonorous.
(iii) The ringing sound was produced because metals are good conductors of heat.
(iv) Both the ghungroos and the school bell produced sound because they were made up of non-metals.
(a) (i) only
(b) (ii) only
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(b) Both the ghungroos and the school bell were made up of metals, and hence, they produced ringing sounds due to a property called sonority.

Question 16.
A science teacher of class 7 listed the names of the following objects on a whiteboard to explain the different properties of materials:
(i) An aluminium spoon
(ii) A copper wire
(iii) A wooden spoon
(iv) An eraser
(v) A silver coin
During class discussion, a few students of her class gave the following statements:
Priya: Object (i) produces a ringing sound when struck against a hard surface, while no ringing sound is produced by object (iii).
Ravina: Both (i) and (v) objects are good conductors of heat and electricity.
Raj: Object (iii) is a good conductor of heat.
Rohan: Electric circuits contain (iii) because the material is malleable, ductile, a good conductor of electricity, and economical.
Which of these students gave the correct statement(s)?
(a) Priya and Ravina only
(b) Raj only
(c) Priya only
(d) Priya, Ravina, and Rohan
Answer:
(a) Object (iii), i.e., the wooden spoon is a poor conductor of heat.

Question 17.
Which of the following statements is/are incorrect regarding corrosion?
(i) Corrosion of iron objects is called rusting of iron.
(ii) The formation of black coating on both copper and silver objects is due to corrosion.
(iii) Corrosion results in the formation of a green coating on copper objects and a black coating on silver objects.
(iv) Only iron undergoes corrosion, as it is a highly reactive metal.
(a) (i) and (iii)
(b) (iii) only
(c) (ii) and (iv)
(d) (iv) only
Answer:
(c) (ii) and (iv)

Question 18.
Iron nails are placed in four different jars: A, B, C and D. In which of the following will rusting of iron take place?

(a) C only
(b) B and C only
(c) A, B and D
(d) C and D only
Answer:
(b) Rusting will take place in jars B and C because both air (oxygen) and water are essential for the rusting of iron to take place. Jar A has oil over water, which cuts off the supply of air, while in Jar D, both water and air are missing.

Question 19.
The following reactions show the burning of magnesium and sulphur in air (oxygen):
Magnesium + Oxygen → X
Sulphur + Oxygen → Y
Which of the following statements is/are incorrect regarding X and Y?
(i) X is an acidic oxide and turns blue litmus paper red.
(ii) Y is a basic oxide and turns red litmus paper blue.
(iii) X is a basic oxide, while Y is an acidic oxide.
(iv) Both X and Y do not change the colour of any of the litmus papers.
(a) (iii) only
(b) (i) and (ii) only
(c) (i), (ii), and (iv)
(d) (iv) only
Answer:
(c) X, i.e., magnesium oxide is a basic oxide and turns red litmus paper blue, while Y, i.e., sulphur dioxide is an acidic oxide and turns blue litmus paper red.

The World of Metals and Non-metals Class 7 Assertion and Reason Questions

The following questions consist of two statements: Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

1. (A): All alloys are mixtures of two metals.
(R): Alloys are important for daily use as utensils and tools due to their improved strength and resistance to corrosion.
Answer:
(d) Alloys can be mixtures of two or more metals or a metal and a non-metal.

2. (A): A piece of copper becomes flat on beating with a hammer.
(R): Copper is ductile.
Answer:
(b) A piece of copper becomes flat on beating with a hammer because it is malleable.

3. (A): Nitrogen is used in the manufacturing of fertilisers.
(R): Nitrogen is a metal.
Answer:
(c) Nitrogen is a non-metal.

4. (A): The application of paint or oil on the surface of iron objects helps in preventing rusting.
(R): The protective coating of paint or oil acts as a barrier and prevents the direct exposure of iron to air (oxygen) and moisture, which are essential for rusting.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

5. (A): Non-metal oxides turn moist red litmus blue.
(R): Non-metal oxides are acidic in nature.
Answer:
(d) Non-metal oxides are acidic in nature and turn moist blue litmus red.

6. (A): Copper is used in making electrical wires.
(R): Copper is a lustrous non-metal.
Answer:
(c) Copper is ductile, thus; it can be drawn into thin wires. It is a lustrous metal, and is also a good conductor of electricity.

The World of Metals and Non-metals Class 7 Fill in the Blanks

1. ………… metal exists in liquid state at room temperature.
Answer: Mercury

2. Copper vessels are used for boiling water because copper is a ………… (good/poor) conductor of heat.
Answer: good

3. Materials such as plastic, glass, wood, rubber, and paper are ………… (good/poor) conductors of electricity.
Answer: poor

4. An iron object kept in dry air does not rust because the absence of ………… prevents the rusting.
Answer: moisture/water

5. Rusting of iron can be prevented by applying a protective layer of ………… metal on iron.
Answer: zinc

6. Metals react with oxygen to form ………… (acidic/basic) oxides, while non-metals react with oxygen to form ………… (acidic/basic) oxides.
Answer: basic, acidic

7. Sulphur burns in air (oxygen) to form …………. On dissolving it in water,……….. is formed.
Answer: sulphur dioxide, sulphurous acid

8. ………… is essential in everyday life because it is the building block of all life forms.
Answer: Carbon

The World of Metals and Non-metals Class 7 True or False

Write true or false. Rewrite the false statements correctly.

1. Generally, metals are non-lustrous, while nonmetals are lustrous.
Answer: False: Generally, metals are lustrous, while non-metals are non-lustrous.

2. All metals are hard and solid at room temperature.
Answer: False: Not all metals are hard and solid at room temperature. Mercury is a liquid; sodium and potassium are so soft that they can be cut with a knife.

3. An iron object undergoes rusting when it comes into contact with dry air.
Answer: False: An iron object undergoes rusting when it comes into contact with both air and water.

4. Due to corrosion, black deposits are formed on the surface of a silver object.
Answer: True

5. Phosphorus is stored in water as it catches fire when exposed to atmospheric air.
Answer: True

6. An iodine solution can be applied to wounds as an antiseptic.
Answer: True

The World of Metals and Non-metals Class 7 Match the Following

Question 1.

Answer:

Column I	Column II
(i) Malleability	D. Beaten into thin sheets
(ii) Sonority	C. Produces a ringing sound
(iii) Ductility	B. Drawn into thin wires
(iv) Good conductor of heat	A. Allow heat to pass through it

Question 2.

Answer:

	Column I		Column II
(i)	A metal oxide	B.	Basic in nature
(ii)	Rusting of iron	C.	Can be prevented by galvanisation
(iii)	A non-metal oxide	D.	Acidic in nature
(iv)	Plastic covering on screwdrivers	A.	A. Poor conductor of electricity, thus prevents electric shock

For better exam readiness, students can include Class 7 Science MCQ and Curiosity Chapter 3 Electricity: Circuits and Their Components MCQ in their daily study routine.

Class 7 Science Chapter 3 Electricity: Circuits and Their Components MCQ

Electricity: Circuits and Their Components MCQ Class 7

MCQ On Electricity: Circuits and Their Components Class 7

Question 1.
Why are multiple cells used in a torch?
(a) To increase the size of the torch
(b) To make it heavier
(c) To provide more energy for longer operation
(d) To make it look modern
Answer:
(c) To provide more energy for longer operation

Question 2.
In a typical electric cell, the metal cap acts as ………(i)………, and the metal disc at the base acts as ……….(ii)………..?
(a) (i) – negative terminal; (ii) – positive terminal
(b) (i) – positive terminal; (ii) – negative terminal
(c) (i) – insulator; (ii) – conductor
(d) (i) – filament; (ii) – terminal
Answer:
(b) In a typical electric cell, the metal cap is the positive terminal, and the metal disc at the base is the negative terminal.

Question 3.
Which of these is the correct arrangement of electric cells in a battery?

Answer:
(d)

Question 4.
In the given figure, the LED has two terminals: a longer leg and a shorter leg. To ensure proper functioning of the LED, the longer leg should be connected to…………….

(a) positive terminal of the battery.
(b) negative terminal of the battery.
(c) any terminal of the battery.
(d) both the terminals of the battery.
Answer:
(a) positive terminal of the battery.

Question 5.
The main difference between an LED and an incandescent bulb is that
(a) LEDs use filaments; incandescent bulbs do not.
(b) Incandescent bulbs use filaments; LEDs do not.
(c) Both use filaments, but in different materials.
(d) LEDs always need AC current.
Answer:
(b) An incandescent bulb glows because its filament gets heated, whereas an LED produces light without a filament.

Question 6.
A wildlife volunteer observes that a pigeon sitting on a high-voltage wire remains unharmed, but another bird gets electrocuted when it lands on a nearby pole and touches a wire at the same time. Which statement best explains this observation?
(a) Larger birds attract more current than smaller birds.
(b) A bird is safe on a single live wire, but birds get electrocuted when they touch two different wires or a wire and another conductor at the same time, creating a complete path for current.
(c) The feet of some species of birds are insulated, while others are not.
(d) Power lines carry current only during the day; the electrocution happened because the line was active at that moment.
Answer:
(b) A bird on one live wire is safe because no complete circuit forms. If it touches a wire and the pole (or two wires) together, a complete path forms and current flows through its body.

Question 7.
What will be the order in which the bulbs A, B, and C will glow when the switch is moved to the ON position?

(a) First A, then B, and then C
(b) First B, then C, and then A
(c) First A, then B and C together
(d) All the bulbs will glow together
Answer:
(d) When the switch is turned ON, the circuit becomes closed, so current flows through the whole circuit at the same time. Hence A, B, and C glow together.

Question 8.
Select the option that shows the correct direction of electric current in the circuit.

Answer:
(b) In a simple circuit, conventional current flows from the positive terminal of the cell (metal cap), passes through the wire and the bulb, and returns to the negative terminal (metal disc/base).

Question 9.
Which of the following precautions need not be taken while using electric gadgets/appliances/ circuits?
(a) We should never touch a lighted electric bulb connected to the mains.
(b) We should never experiment with the electric supply from the mains or a generator or an inverter.
(c) We should never use just any wire or strip of metal in place of a fuse.
(d) We should never turn the switch in ON position.
Answer:
(d) This is not a safety precaution. Turning the switch to the ON position is perfectly safe and necessary to operate electric appliances.

Question 10.
In the given circuit, four identical bulbs P, Q, R and S are connected to a battery as shown. Choose the correct statement.

(a) If bulb R fuses, bulbs P and Q go off.
(b) If bulb P fuses, bulb R goes off while bulb S stays lit.
(c) If bulb Q fuses, bulb R goes off while bulb S stays lit.
(d) If bulb S fuses, then the rest of the bulbs stay lit.
Answer:
(d) If bulb S fuses, the other bulbs P, Q and R still have complete paths in the circuit, so they remain lit.

Question 11.
In the LED symbol, the triangle points in the direction in which current can flow. What do the two small arrows near the LED symbol indicate?
(a) The LED stores electricity.
(b) The LED emits light.
(c) The LED has a filament.
(d) The LED is an insulator.
Answer:
(d) The LED is an insulator.

Question 12.
Which of the following terminal arrangements will never make a battery?

Answer:
(c)

Question 13.
A student made an incomplete circuit with a small gap as shown in the image. He placed a few different materials in the gap and noted the intensity of the light produced by the bulb.

Material	Intensity of Light
Glass	Lamp does not glow
I.ead	Dim
Steel	Bright
Copper	Very bright

What can be concluded based on the activity?
(a) Steel should be used to make insulating cover for electric wires
(b) Copper is the best material to conduct electricity
(c) Lead is the best material for making wires
(d) Glass can be used to store electricity
Answer:
(b) Copper is the best material to conduct electricity

Question 14.
Electrical wires, switches, connectors of plugs, and sockets are made of ………….(i)…….., while materials like rubber, plastics, and ceramics are used as ……..(ii)………. to prevent electric shocks.
(a) (i) – Insulators, (ii) – Conductors
(b) (i) – Conductors, (ii) – Insulators
(c) (i) – Conductors, (ii) – Conductors
(d) (i) – Insulators, (ii) – Non-conductors
Answer:
(b) (i) – Conductors, (ii) – Insulators

Question 15.
A simple circuit has a cell, a bulb, and two wires. The bulb is not glowing. Four students suggest the following possible reasons:
(i) There is a small gap between one wire and the bulb holder.
(ii) The bulb filament inside is broken, even though the glass looks fine.
(iii) The plastic covering of the wire is not painted, so current cannot flow.
(iv) The wires are connected without removing insulation.
Which of the above statements correctly describe situations that can break a circuit?
(a) (i), (ii), and (iv)
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) (iii) and (iv) only
Answer:
(a) The plastic covering is an insulator; whether it is painted or not does not affect current flow. Current flows through the metal wire inside.

Question 16.
A student performs two activities using an incandescent bulb and an LED.

• Activity I: An incandescent bulb is connected to a single cell.
• Activity II: An LED is connected to a single cell.

(i) For an incandescent bulb to glow, the circuit only needs to be closed; its terminals can be connected in either direction.
(ii) For an LED to glow, the direction of connection matters; it allows current only in one direction.
(iii) In Activity I, reversing the connections breaks the circuit, so the bulb should not glow.
(iv) These activities show that all electrical components work only in one fixed direction.
Which of the given statements are correct?
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii), (iii), and (iv)
(d) (i), (ii), and (iv)
Answer:
(a) In Activity I, even if the connections are reversed, the incandescent bulb will still glow as long as the circuit is closed. Only some components like an LED work in a fixed direction of current.

Question 17.
In which of the following setups, the bulb will glow?

(a) (i) and (ii) only
(b) (i) and (iv) only
(c) (ii) and (iv) only
(d) (i), (ii) and (iv)
Answer:
(b) Circuit is complete in circuits shown in options (i) and (iv).

Question 18.
Which of the following statements are correct?
(i) A switch can be placed anywhere in a circuit.
(ii) The positive terminal of an incandescent lamp must be connected to the positive terminal of the cell.
(iii) The longer leg of the LED is the positive terminal.
(a) (i) only
(b) (iii) only
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) In incandescent lamps, it does not matter which of its terminals connects to the positive or negative terminal of the cell.

Question 19.
Smita was trying to connect the circuit in different ways practically, and she was also drawing the circuit diagrams to try out other possible connections. Some circuits tried out by Smita are listed in table below. Identify the correct pair(s) of the circuit and its corresponding circuit diagram.

(a) (i)-(A), (ii)-(B)
(b) (ii)-(B), (iii)-(C)
(c) (i)-(A), (iv)-(D)
(d) (iv)-(D) only
Answer:
(c) (i)-(A), (iv)-(D)

Electricity: Circuits and Their Components Class 7 Assertion and Reason Questions

The following questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

1. (A): A cell has two terminals, one positive and the other negative.
(R): The metal cap of the cell represents the negative terminal.
Answer:
(c) The metal cap of the cell represents the positive terminal.

2. (A): Two cells will always form a working battery.
(R): The positive terminal of one cell should be connected to the negative terminal of the other to form a battery.
Answer:
(d) Two cells will not always work as a battery; they must be connected properly, i.e., the positive terminal of one cell should be connected to the negative terminal of the other.

3. (A): A bulb glows brighter if more cells are added.
(R): More cells provide more electricity for the circuit.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

4. (A): An incandescent bulb glows because of its filament.
(R): The filament of an incandescent lamp is connected to two thick wires.
Answer:
(b) Both statements are true, but the reason does not explain the assertion. The incandescent bulb glows because its filament becomes hot and glows when current passes through it.

5. (A): When the switch is in the OFF position, the lamp does not glow.
(R): In the OFF position, the switch creates a closed path in the circuit.
Answer:
(c) In OFF position, the switch breaks the circuit (creates an open circuit), so no current flows and the lamp does not glow.

6. (A): Plastic can replace a metal pin in a circuit.
(R): Plastic is an insulator and does not allow the current to pass through it.
Answer:
(d) Plastic cannot replace a metal pin in a circuit because plastic does not conduct electricity, so the circuit would remain open.

Electricity: Circuits and Their Components Class 7 Fill in the Blanks

1. An electric cell is a portable source of ………….
Answer:
electrical energy

2. In incandescent lamps, the ………… gets hot and glows to produce light.
Answer:
filament

3. LEDs have two terminals; the ………… terminal is attached to the longer wire, while the ………… terminal is attached to the shorter wire.
Answer:
positive, negative

4. The direction of electric current in a closed electrical circuit is taken to be from the ………… to the ………… terminal of the electric cell.
Answer:
positive, negative

5. An electric ………… is a device used to either break an electric circuit or complete it at our will.
Answer:
switch

6. A representation of an electrical circuit using symbols is called its ………….
Answer:
circuit diagram

7. Batteries generally provide …………, whereas the electricity from power plants is ………….
Answer:
direct current (DC), alternating current (AC)

8. The human body behaves like a ………… of electricity because body fluids can allow current to pass through.
Answer:
conductor

Electricity: Circuits and Their Components Class 7 True or False

Question 1.
The positive terminal of a cell is connected to the positive terminal of another cell in a battery.
Answer: False In a battery, the positive terminal of one cell is connected to the negative terminal of the next cell.

Question 2.
The current generated through a battery is known as alternating current (AC).
Answer: False A battery provides direct current (DC).

Question 3.
An incandescent lamp will only glow if its positive terminal is connected to the positive terminal of the cell.
Answer: False An incandescent lamp glows as long as the circuit is complete, irrespective of which lamp terminal is connected to the positive terminal of the battery.

Question 4.
In a circuit diagram, an incandescent lamp and an LED are represented by the same symbol.
Answer: False In a circuit diagram, an incandescent lamp and an LED are represented by different symbols.

Question 5.
Ceramics are poor conductors (insulators) of electricity.
Answer: True

Question 6.
Rubber and plastic are used as covering on wires because they are good conductors of electricity.
Answer: False Rubber and plastic are used as coverings because they are insulators (poor conductors) of electricity.

Electricity: Circuits and Their Components Class 7 Match the Following

Question 1.
Match column I with column II.

Answer:

Column I	Column II
(i) Electric cell	D. Supplies electrical energy to the circuit
(ii) Battery	C. A combination of two or more cells
(iii) Switch	F. Completes or breaks the circuit to control current
(iv) Connecting wire	A. Provides a conducting path for current in a circuit
(v) Incandescent bulb filament	E. Heats up and glows when current flows through it
(vi) LED	B. Produces light only when connected in the correct direction

Question 2.

Answer:
(i)-D, (ii)-C, (iii)-A, (iv)-B

For better exam readiness, students can include Class 7 Science MCQ and Curiosity Chapter 2 Exploring Substances Acidic Basic and Neutral MCQ in their daily study routine.

Class 7 Science Chapter 2 Exploring Substances Acidic Basic and Neutral MCQ

Exploring Substances Acidic Basic and Neutral MCQ Class 7

MCQ On Exploring Substances Acidic Basic and Neutral Class 7

Choose the correct option.

Question 1.
A student understands that a litmus solution has a purple colour in distilled water. What colour is likely to be observed when it is placed in an acidic solution?
(a) Blue
(b) Green
(c) Purple
(d) Red
Answer:
(d) Red

Question 2.
A scientist tests a soil sample from an uncultivated field and concludes that the soil is highly basic in nature. She suggests that adding organic matter in the soil could improve the quality of the soil. How does organic matter improve the quality of the soil?
(a) It releases acids which neutralises the soil.
(b) It traps water vapour and increases the moisture content.
(c) It makes the soil lighter so that it can be easily transported.
(d) It acts as food for the organisms already present in the soil.
Answer:
(a) It releases acids which neutralises the soil.

Question 3.
Which of the following set of substances contain acids?
(a) Grapes, lime water
(b) Vinegar, soap
(c) Curd, milk of magnesia
(d) Curd, vinegar
Answer:
(d) Curd, vinegar

Question 4.
In a science experiment, a student was provided with two solutions, P and Q, and the natures of both the solutions were tested via the litmus test as shown in the image alongside:
Based on the experiment, which of the following statements cannot be true?

(a) Solution P could be an orange juice, while solution Q could be a soap solution.
(b) Solution P cannot turn the red rose indicator red.
(c) Solution P does not change the colour of the turmeric indicator, while solution Q turns the turmeric indicator red.
(d) Solution P could be a curd, while solution Q could be a detergent solution.
Answer:
(b) Solution P is acidic as it turns blue litmus paper red. Whereas solution Q is basic, as it turns red litmus paper blue. Hence, solution P turns the red rose indicator to a shade of red.

Question 5.
Pulkit was experimenting in the science lab to understand acid-base indicators. Initially, when he added a few drops of red litmus solution to the lime water solution, the colour of the solution turned blue. When he added an excess of lemon juice to this test tube, the colour of the solution eventually changed from blue to red. Which of the following statements is the correct conclusion to the above observation?
(a) The resultant solution is basic, as lemon juice and lime water both are bases.
(b) The resultant solution is neutral, which is why the blue litmus solution turned red.
(c) The resultant solution is acidic, as lemon juice is acidic in nature.
(d) The resultant solution is neutral, as lemon juice is neutral in nature.
Answer:
(c) When a few drops of red litmus solution were added to the lime water solution, the colour of the solution turned blue as lime water is basic in nature. However, upon the addition of an excess of lemon juice, the colour of the solution eventually changed from blue to red. Hence, the resultant solution became acidic, as lemon juice is acidic in nature.

Question 6.
A neutralisation reaction is represented as follows:
Base + X → Y + Water
Based on the above reaction, which of the following statements is true?
(a) X can turn blue litmus paper red.
(b) X is salt, while Y is acid.
(c) X can turn the red rose indicator green.
(d) Neither X nor Y changes the colour of either red or blue litmus paper.
Answer:
(a) The given reaction is represented as follows:

Hence, X, i.e., the acid, can turn blue litmus paper red. Y represents the salt formed in the reaction. X, being an acid, cannot turn the red rose indicator green. Y being neutral, does not change the colour of red or blue litmus paper.

Question 7.
During a class discussion, a class 7 student, Vamika, claimed that any sour substance will turn the turmeric indicator red. Her statement is
(a) Correct, as only acidic substances turn the turmeric indicator red.
(b) Incorrect, as only neutral substances turn the turmeric indicator red.
(c) Incorrect, as only basic substances turn the turmeric indicator red.
(d) Correct, as both acidic as well as basic substances turn the turmeric indicator red.
Answer:
(c) The claim made by the student is incorrect, as acidic substances are sour in taste and they cannot turn the turmeric indicator red. Rather, the basic substances which are bitter in taste, turn the turmeric indicator red.

Question 8.
On the occasion of Mother’s Day, Kartik wrote a message for her mother on a white sheet of paper. First, he applied a turmeric paste all over the paper. Once the paste dried, he wrote his message using a solution X . From the given options, what could be solution X?
(a) Sugar solution
(b) Soap solution
(c) Lemon juice
(d) Amla juice
Answer:
(b) From the given solutions, solution X could be a soap solution, as being basic in nature, it turns the turmeric indicator red. Hence, the message written by Kartik will appear red.

Question 9.
Sheena, a class 7 student, was bitten by an ant while playing in the playground. Her father suggested her to apply a substance, P, to get quick relief from the pain. Select the option that correctly identifies substance P and the reason for applying P.
(a) P could be curd, as it is acidic in nature.
(b) P could be lime water, as it is neutral in nature.
(c) P could be vinegar, as it is acidic in nature.
(d) P could be moist baking soda, as it is basic in nature.
Answer:
(d) When an ant bites, it injects formic acid into the skin. The effect of an ant bite is therefore neutralised by applying a mild basic substance onto the skin. Therefore P could be moist baking soda as it is a mild base, and when rubbed onto the skin, provides relief to the affected area by neutralising the injected acid.

Question 10.
Pawan’s mother used amla juice while cooking to make the gravy sour. Here, the amla juice contains which of the following substances?
(a) Ascorbic acid
(b) Lactic acid
(c) Oxalic acid
(d) Tartaric acid
Answer:
(a) Ascorbic acid

Question 11.
A solution X changes the colour of the red rose indicator to a shade of red. While a solution Y changes the colour of turmeric paper to red. Solutions X and Y can be, respectively:
(a) Soap solution and detergent solution
(b) Salt solution and vinegar
(c) Tamarind juice and curd
(d) Amla juice and baking soda solution
Answer:
(d) Since the solution X turns the red rose indicator to a shade of red, it must be acidic in nature. As solution Y changes the colour of the turmeric paper to red, it must be basic in nature.

Question 12.
The colour of the red rose extract changes to P in a lemon juice solution, and to Q in a lime water solution. Based on this information, select the option that correctly identifies P and Q.
(a) P – Green, Q – Pink
(b) P – Yellow, Q – Red
(c) P – Red, Q – Green
(d) P – Red, Q – Blue
Answer:
(c) The colour of the red rose extract changes to to a shade of red in a lemon juice solution, as it is acidic. While the colour of the red rose extract changes to green in lime water, as it is basic.

Question 13.
A reaction between two solutions, A and B, results in the formation of salt and water along with the evolution of heat. From the given options, select the correct pair of A and B.
(a) Lemon juice and amla juice
(b) Distilled water and sugar solution
(c) Vinegar and baking soda solution
(d) Soap solution and lime water
Answer:
(c) Since the reaction between solutions A and B results in the formation of salt and water, one of these solutions will be acidic, while the other will be basic. Among the given options, only option (c) consists of an acid (vinegar) and a base (baking soda solution).

Question 14.
Ratika, a class 7 student, tested a solution X using blue and red litmus papers as shown in the given picture:

Which of the following statements is true based on this activity?
(a) Solution X must be basic in nature, and it will change the colour of the turmeric indicator to red.
(b) Solution X must be neutral in nature, and it will retain the odour of an onion strip.
(c) Solution X must be basic in nature, and it will change the colour of the red rose extract to red.
(d) Solution X must be acidic in nature, and it will not cause any change in the colour of the turmeric indicator.
Answer:
(d) Since solution X turns blue litmus paper red and does not change the colour of red litmus paper, it must be acidic in nature. Acidic solutions do not cause any change in the colour of the turmeric indicator.

Question 15.
Sanjana, a class 7 student, dipped three turmeric paper strips into three different solutions as shown below:

Which of the following statements is correct?
(a) The strips dipped in solutions A and B will turn red.
(b) The strip dipped in solution C will turn red.
(c) The strip dipped in solution B will turn red.
(d) The strips dipped in solutions A and C will turn red.
Answer:
(c) Only the strip dipped in solution B will turn red, as soap solution is basic in nature and turmeric paper turns red in basic solutions.

Question 16.
When a substance X is added to the red rose extract solution, the colour change occurs as shown in the image given alongside. Based on the given colour change, select the correct statements about X.

(i) Substance X could be lemon juice, as it is acidic in nature.
(ii) Substance X could be curd, as it is basic in nature.
(iii) Substance X also turn the turmeric indicator red.
(iv) Substance X does not affect the turmeric indicator.
(v) Substance X could be a detergent solution, as it is basic in nature.
(a) (ii) and (iii)
(b) (iii) and (v)
(c) (i) and (iv)
(d) (iv) and (v)
Answer:
(b) Since substance X turns the red rose indicator green, it is basic in nature. Hence, it could not be lemon juice or curd, as both are acidic, and will turn the red rose indicator to a shade of red. However, X could be a detergent solution, as a detergent solution, being basic, will turn the red rose indicator green. The solution X will therefore also turn the turmeric indicator red.

Question 17.
In daily life, neutralisation can be observed in which of the following situations?
(i) Applying moist baking soda on an ant bite
(ii) Using lime water to treat acidic soil
(iii) Adding vinegar to make pickles sour
(iv) Treating basic soil with compost or manure
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iv)
(d) (i), (iii) and (iv)
Answer:
(c) An ant bite injects formic acid thus, moist baking soda can be applied to neutralise it. Acidic soil due to fertilisers can be neutralised by treating it with lime water (a base). Basic soil can be neutralised by adding compost or manure, which release acids. In contrast, adding vinegar to pickles only increases acidity, so no neutralisation reaction is taking place.

Question 18.
Which of the following statement(s) is/are correct regarding a neutralisation reaction?
(i) An acid reacts with a base to form salt and water.
(ii) Heat is generally released during the process.
(iii) Heat is required for the reaction to take place.
(iv) When an excess acid is mixed with a base, the resultant solution is neutral in nature.
(a) (i) only
(b) (i) and (ii) only
(c) (i) and (iii) only
(d) (i), (ii) and (iv)
Answer:
(b) The reaction in which an acid and a base combine to form a corresponding salt and water, with the release of heat, is called a neutralisation reaction. When an excess acid is mixed with a base, the resultant solution is acidic in nature.

Question 19.
Which of the following statements are correct?
(i) A soap solution turns the red rose indicator to a shade of red.
(ii) No heat is evolved in a neutralisation reaction.
(iii) Turmeric indicator cannot be used to distinguish between amla juice and sugar solution.
(iv) The colour of a blue litmus solution changes to red when a few drops of lemon juice are added to it.
(a) (ii) and (iv)
(b) (i) and (iii)
(c) (iii) and (iv)
(d) (i) and (ii)
Answer:
(c) A soap solution turns the red rose indicator green. Heat is evolved in a neutralisation reaction.

Question 20.
During a class discussion, four students of class 7 were discussing the nature of various substances, and they gave the following statements:
Arun: Lemon juice and amla juice, being acidic, turn the turmeric indicator red.
Arjun: A detergent solution is neutral, and therefore does not affect either red or blue litmus paper.
Akriti: Soap solution, being basic, turns red litmus paper blue.
Aman: Curd and tamarind juice, being acidic, turn the blue litmus paper red.
Students who gave the correct statements are:
(a) Arjun and Aman
(b) Akriti and Aman
(c) Arun and Arjun
(d) Arjun and Akriti
Answer:
(b) Lemon juice and amla juice, being acidic, do not change the colour of turmeric indicator. A detergent solution is basic; therefore, it turns red litmus paper blue. However, it will not affect the blue litmus paper.

Question 21.
Observe the following flowcharts:

Based on the above flowcharts, select the correct statements from the following:
(i) When sufficient quantities of solutions X and Y are mixed, no heat is evolved in the reaction.
(ii) Solution X can be a baking soda solution, and it turns the turmeric indicator red.
(iii) Solution X is acidic, while solution Y is basic in nature.
(iv) Solution X is basic, while solution Y is acidic in nature.
(v) Solution Y can be lime water, and it also turns the red rose indicator green.
(a) (i) and (ii) only
(b) (i), (iii) and (v)
(c) (ii) and (iv) only
(d) (iii) and (v) only
Answer:
(d) Since solution X does not affect the red litmus paper, but turns the blue litmus paper red, it is acidic in nature. Hence, solution X cannot be a baking soda solution, as it is basic in nature. Whereas, solution Y turns red litmus paper blue, but does not affect the blue litmus paper, hence it is basic in nature. Therefore, solution Y can be lime water, and it also turns the red rose indicator green.

Exploring Substances Acidic Basic and Neutral Class 7 Assertion and Reason Questions

The following questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): Turmeric paper indicator can be used to test both acidic and basic solutions.
(R): The yellow colour of turmeric paper turns red in basic solutions, but remains unchanged in acidic and neutral solutions.
Answer:
(d) Turmeric paper indicator can only be used to test basic solutions.

Question 2.
(A): Applying moist baking soda on an ant bite will relieve the pain caused by the injected formic acid.
(R): Baking soda is a base that neutralises the formic acid.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Question 3.
(A): Lemon juice turns blue litmus paper red.
(R): Lemon juice is basic in nature.
Answer:
(c) Lemon juice is acidic in nature, thus, it turns blue litmus paper red.

Question 4.
(A): Onion is an olfactory indicator.
(R): The smell of an onion strip is retained in a soap solution.
Answer:
(c) The smell of an onion strip is not retained in a basic solution like soap solution.

Question 5.
(A): Sugar solution is acidic in nature.
(R): A sugar solution does not change the colour of red or blue litmus paper.
Answer:
(d) Sugar solution is neutral in nature.

Question 6.
(A): The colour of the red rose extract changes to a shade of red in a vinegar solution.
(R): Red rose extract is a visual, natural indicator.
Answer:
(b) Both A and R are true but R is not the correct explanation of A.

Question 7.
(A): Turmeric indicator turns red in distilled water.
(R): Distilled water is neutral in nature.
Answer:
(d) Turmeric indicator does not change its colour in distilled water.

Exploring Substances Acidic Basic and Neutral Class 7 Fill in the Blanks

1. Turmeric indicator can distinguish between neutral and ………… substances.
Answer: basic

2. Blue litmus turns red in an/a …………solution.
Answer: acidic

3. Indicators whose odour change in an acidic or basic medium are called ………… indicators.
Answer: olfactory

4. …………acid is present in vinegar.
Answer: Acetic

5. The colour of red rose extract changes to …………in a basic solution, and to …………in an acidic solution.
Answer: green, red

6. …………are the substances that generally have a bitter taste and feel soapy (slippery) to the touch.
Answer: Bases

7. Turmeric and litmus are ………… acid-base indicators.
Answer: natural

8. When an acidic solution is mixed with a basic solution, they …………each other forming …………and water.
Answer: neutralise, salt

9. The colour of the turmeric paper changes to …………in a lime water solution.
Answer: red

10. In acidic soil, the colour of the hydrangea flowers is …………, while in basic soil, the colour of these flowers is ………….
Answer: blue, pink/red

11. The onion smell is not retained in a/an ………..solution.
Answer: basic

Exploring Substances Acidic Basic and Neutral Class 7 True or False

1. Applying vinegar to an ant bite is an effective home remedy.
Answer:
False: An ant bite injects formic acid into the skin. Vinegar is also acidic, therefore, it will only worsen the effect of formic acid. A mild base like moist baking soda should be applied onto the skin.

2. The acidic factory waste can be neutralised by adding an acid-base indicator before releasing it into the lake.
Answer:
False: The acidic factory waste can be neutralised by adding basic substances before releasing it into the lake.

3. The acidic nature of the soil can be treated by adding organic manure to it.
Answer:
False: Organic matter releases acidic substances, which will increase the acidity of the soil. It should be treated by adding a base such as lime water.

4. The colour of the red rose indicator changes to red when a few drops of tamarind juice are added to it.
Answer:
True

5. Lemon juice tastes sour because it is basic in nature.
Answer:
False: Lemon juice tastes sour because it is acidic in nature.

6. A neutralisation reaction is accompanied by the release of heat.
Answer:
True

7. Lime water is acidic in nature, while lemon juice is basic in nature.
Answer:
False: Lime water is basic in nature, while lemon juice is acidic in nature.

8. A washing powder solution turns red litmus blue, while an orange juice turns blue litmus red.
Answer:
True

9. Litmus is a synthetic indicator obtained from lichens.
Answer:
False: Limtus is a natural indicator obtained from lichens.

Exploring Substances Acidic Basic and Neutral Class 7 Match the Following

Question 1.
Match column I with column II.

Answer:

	Column I		Column II
(i)	Lemon	C.	Citric acid
(ii)	Curd	D.	Lactic acid
(iii)	Tamarind	B.	Tartaric acid
(iv)	Vinegar	A.	Acetic acid

Question 2.

Answer:

	Column I		Column II
(i)	Turmeric indicator	D.	Turns red in basic solutions
(ii)	Red rose extract	C.	Turns green in basic solutions
(iii)	Blue litmus paper	A.	Cannot distinguish between basic and neutral solutions
(iv)	Red litmus paper	B.	Turns blue in basic solutions

Students can include Class 7 Curiosity Extra Questions and Class 7 Science Chapter 12 Earth, Moon, and the Sun Extra Question Answer in their daily self-study routine.

Class 7 Science Chapter 12 Earth, Moon, and the Sun Extra Questions

Class 7 Science Chapter 12 Extra Questions on Earth, Moon, and the Sun

Earth, Moon, and the Sun Class 7 Very Short Question Answer

Question 1.
(i) State the direction of the Earth’s rotation.
(ii) What is the time taken by the Earth to complete one rotation?
Answer:
(i) The Earth rotates from west to east.
(ii) The Earth requires approximately 24 hours to complete one full rotation on its axis.

Question 2.
Meera observed shadows of trees in her garden and found that the shadows were longest in the morning. What causes this change in shadow length?
Answer:
Shadows change length because the Earth rotates, causing the Sun’s position to shift during the day. In the morning, the Sun is low on the horizon, so shadows appear longer, while at noon, when the Sun is overhead, shadows become shorter.

Question 3.
Why do constellations appear to shift their position in the sky over different months?
Answer:
Constellations appear to shift their position in the sky over different months because of the Earth’s revolution around the Sun. As a result, different parts of the night sky are visible at different times of the year.

Question 4.
Why are seasons in the Northern Hemisphere opposite to those in the Southern Hemisphere?
Answer:
Seasons in the Northern Hemisphere are opposite to those in the Southern Hemisphere due to the tilt of the Earth’s axis. When the Northern Hemisphere tilts towards the Sun and experiences summer, the Southern Hemisphere tilts away and experiences winter, and vice versa.

Question 5.
(i) Arjun saw that sometimes the Moon completely blocks the Sun’s light during the day. Which phenomenon explains this?
(ii) Differentiate between a partial lunar eclipse and a total lunar eclipse.
Answer:
(i) When the Moon completely blocks the Sun’s light during the day, the phenomenon is called a solar eclipse.
(ii) When the Moon passes completely into the Earth’s shadow, it appears dark red in colour. This phenomenon is known as a total lunar eclipse.
When only a part of the Moon enters the Earth’s dark shadow, that portion appears dark or reddish, while the remaining part of the Moon, which is outside the shadow, remains bright and visible. This phenomenon is known as a partial lunar eclipse.

Question 6.
One night, a child notices that a group of stars seems to change position in one hour, but one particular star appears almost fixed in the sky. Explain why this happens.
Answer:
The apparent change in position of a group of stars within one hour is due to the rotation of the Earth. The star that appears almost fixed is the Pole Star, which lies nearly along the Earth’s axis of rotation above the North Pole, so its position appears almost unchanged while other stars seem to move around it.

Question 7.
Raghav travels from Mumbai to Kolkata and finds that the Sun seems to rise earlier in Kolkata on the same date. Explain.
Answer:
Kolkata lies to the east of Mumbai. Since the Earth rotates from west to east, places located further east receive sunlight earlier. Therefore, sunrise is observed earlier in Kolkata than in Mumbai on the same date.

Question 8.
Riya says, “If the Earth stopped rotating but kept revolving around the Sun, we would still have normal days and nights.” Is this statement correct? Give a reason.
Answer:
No, this statement is not correct. Normal day and night occur due to the Earth’s rotation on its axis. If the Earth stopped rotating, one side would face the Sun continuously for months (continuous day) while the other side would remain in darkness for months (continuous night), so normal days and nights would not occur.

Question 9.
On a particular day, a student in Delhi says, “Today is the longest day of the year for us,” while another student in Sydney says, “Today is our shortest day.” Explain why this happens.
Answer:
Delhi is in the Northern Hemisphere, while Sydney is in the Southern Hemisphere. When the Northern Hemisphere is tilted towards the Sun and experiences its longest day (around 21 June), the Southern Hemisphere is tilted away from the Sun and experiences its shortest day.

Question 10.
Christmas in Australia is celebrated in the summer season. Explain how this is possible.
Answer:
Australia lies in the Southern Hemisphere. Around 25 December, the Southern Hemisphere is tilted towards the Sun, experiencing summer. Therefore, Christmas in Australia is celebrated during the summer season.

Earth, Moon, and the Sun Class 7 Short Question Answer

Question 1.
Ravi fixed a camera at a particular spot and took pictures of the night sky at different times. He observed that the stars and constellations were not in the same place in all the pictures.
(i) In which direction do you think the stars have moved?
(ii) Are the stars and constellations really moving in the night sky?
(iii) If not, then why do they appear to change their position?
Answer:
(i) The stars appear to move from east to west.
(ii) No, the stars and constellations are not actually moving in the night sky.
(iii) They appear to change position because the Earth rotates on its axis, making the sky seem to move.

Question 2.
Consider a region on Earth where daytime lasts for many hours in summer but becomes very short in winter.
(i) What could be the reason for such extreme variation in day length?
(ii) How do the Earth’s axial tilt and its motion around the Sun influence seasons in such regions?
Answer:
(i) Such extreme variation in day length occurs in regions near the poles. It is caused by the tilt of the Earth’s axis.
(ii) The Earth’s axis is tilted, and it maintains this tilt while revolving around the Sun. Because of this, different regions of the Earth receive different intensities and durations of sunlight during the year. When a region is tilted towards the Sun, it receives more intense sunlight for a longer time, and when tilted away, it receives less intense sunlight for a shorter time. These differences in sunlight cause the change of seasons in different regions.

Question 3.
(i) Describe the causes of a solar eclipse.
(ii) Why is a solar eclipse visible only from a small part of the Earth?
Answer:
(i) A solar eclipse occurs when the Moon comes between the Earth and the Sun and blocks sunlight from reaching parts of the Earth, forming a shadow.
(ii) A solar eclipse is visible only from a small part of the Earth because the Moon’s shadow is narrow and covers only a limited region on the Earth’s surface.

Question 4.
A student lives near the Equator and records the length of day and night at different times of the year.
(i) What does the student observe about the duration of day and night in different months? Give a reason for your answer.
(ii) Does the apparent motion of the Sun mean that the Sun is actually moving around the Earth in one day? Explain briefly.
Answer:
(i) The student observes that the length of day and night remains nearly equal throughout the year. This happens because the Equator receives almost the same amount of sunlight in all seasons, and the illuminated half of the Earth always divides the Equator into nearly equal halves.
(ii) No, the apparent motion of the Sur across the sky does not mean that the Sun moves around the Earth. It only appears to move because the Earth rotates from west to east on its axis.

Question 5.
(i) Explain why the North Pole can have continuous daylight for about six months and continuous darkness for about six months.
(ii) Why are the seasonal changes not very prominent in places close to the Equator compared to places farther north or south?
Answer:
(i) The Earth’s axis is tilted, and the Earth maintains this tilt while revolving around the Sun. Because of this tilt, the North Pole remains tilted towards the Sun for about six months, during which the Sun does not set and there is continuous daylight. During the other six months, the North Pole is tilted away from the Sun, so sunlight does not reach it, resulting in continuous darkness.
(ii) Seasonal changes are not very prominent near the Equator because the intensity of sunlight and the length of day and night change very little during the year. The Sun remains almost overhead, resulting in nearly uniform heating throughout the year.

Question 6.
Arjun lives in Delhi (Northern Hemisphere), and his cousin Sara lives in Melbourne, Australia (Southern Hemisphere). They video call each other twice: once in June and once in December.
(i) In June, Arjun says it is very hot and the days are long, while Sara says it is cold and the days are shorter. Explain why they experience opposite seasons.
(ii) Which main features of the Earth together mainly cause the seasons?
Answer:
(i) In June, the Northern Hemisphere is tilted towards the Sun, so places like Delhi receive more intense sunlight for longer hours and experience summer. At the same time, the Southern Hemisphere is tilted away from the Sun, so places like Melbourne receive less sunlight and experience winter.
(ii) The seasons are mainly caused by the tilt of the Earth’s axis, its revolution around the Sun, and the spherical shape of the Earth.

Question 7.
During a total solar eclipse, some regions of the Earth still experience a partial solar eclipse. Explain why this happens.
Answer:
During a total solar eclipse, the Moon comes between the Sun and the Earth and casts its shadow on the Earth’s surface. The darkest part of the Moon’s shadow falls on a small region of the Earth. Observers present in this region are unable to see any part of the Sun and, therefore, experience a total solar eclipse. At the same time, in the surrounding regions of the Earth, the Moon blocks only a portion of the Sun. Hence, observers in these areas experience a partial solar eclipse.

Earth, Moon, and the Sun Class 7 Long Question Answer

Question 1.
(i) Mention the two main types of motion that the Earth shows and briefly describe each.
(ii)Explain how one of these motions is related to the change of day and night.
(iii) Explain how one of these motions, along with the tilt of the Earth’s axis, is related to the change of seasons.
(iv) Describe the changes that take place in the night sky over the course of a year.
Answer:
(i) Rotation: The spinning of the Earth about its own axis, an imaginary line passing through the North Pole and South Pole.
Revolution: The motion of the Earth around the Sun in a nearly circular orbit, taking about 365 days to complete one round.
(ii) The rotation of the Earth is related to the change of day and night. As the Earth rotates, the side facing the Sun experiences day, while the side turned away from the Sun experiences night.
(iii) Revolution, together with the tilted axis, causes seasons. When a hemisphere is tilted towards the Sun, it gets more direct sunlight and experiences summer; when it is tilted away, it gets less sunlight and experiences winter.
(iv) The night sky changes over the course of a year because, at a fixed time, different constellations are seen in different months. This happens because the Earth revolves around the Sun, and our viewing direction into space at night slowly changes. As the Earth moves along its orbit, we look out towards different parts of space at night, so some constellations move out of view and new constellations come into view at the same time of night in different seasons.

Question 2.
(i) What is meant by summer solstice and winter solstice in the Northern Hemisphere?
(ii) On which approximate dates do the longest day and the shortest day occur in the Northern Hemisphere?
(iii) Why are the summer and winter solstices experienced at different times in the Northern and Southern Hemispheres?
Answer:
(i) In the Northern Hemisphere, the summer solstice is the time of the year when the day is the longest and the night is the shortest. This happens when the Northern Hemisphere is tilted most towards the Sun. The winter solstice is the time when the day is the shortest and the night is the longest, when the Northern Hemisphere is tilted most away from the Sun.
(ii) In the Northern Hemisphere, the longest day of the year occurs around 21 June (summer solstice), and the shortest day occurs around 22 December (winter solstice).
(iii) The summer and winter solstices are opposite in the two hemispheres because when one hemisphere is tilted towards the Sun, the other is tilted away from it.

Question 3.
Draw a neat, labelled diagram of the Earth showing its tilted axis and the rays of sunlight falling on it during June. Use the diagram to explain the seasons in the Northern and Southern Hemispheres.
Answer:

In June, the Northern Hemisphere is tilted towards the Sun.

• Sunlight is more intense: The incident rays of sunlight fall directly and are concentrated over a smaller area.
• Days are longer: The hemisphere receives sunlight for more than 12 hours, resulting in extended daylight duration.

The Southern Hemisphere, on the other hand, is tilted away from the Sun during June.

• Sunlight is less intense: The incident rays of sunlight fall at a slanting angle and are spread over a larger area.
• Days are shorter: The hemisphere receives sunlight for fewer than 12 hours, resulting in reduced daylight duration.
  Together, the intense sunlight and the longer duration of daylight make the Northern Hemisphere experience summer and winter in the Southern Hemisphere.

Question 4.
(i) Draw a labelled diagram of a solar eclipse showing the arrangement of the Sun, Earth and the Moon.
(ii) Describe a total solar eclipse and a partial solar eclipse.
Answer:
(i)

(ii) When the Moon comes between the Sun and the Earth in a way that obstructs the light from the Sun from reaching us. In places where the Moon completely covers the Sun, people see a total solar eclipse. For a few minutes, the day turns into darkness. In places where the Moon blocks only some part of the Sun, a partial solar eclipse is seen.

Earth, Moon, and the Sun Class 7 Case Based Questions

Read the following passages and answer the questions that follow:

Question 1.
Aryan lives in Kolkata, and his cousin Meher lives in Norway, close to the North Pole. In June, Aryan notices that in Kolkata, the Sun rises early and sets late, but there is still a clear night. Meher messages him that in her town, it is bright even at midnight for many days. In December, Aryan feels that days have become shorter and nights longer, but Meher experiences several days when the Sun does not rise at all. Later that year, Aryan’s class watches a solar eclipse using special filters, while a lunar eclipse is watched directly from the school playground.

(i) If Meher lived in Singapore, which is near the Equator, how would the length of day and night compare throughout the year?
Answer:
If Meher lived in Singapore near the Equator, the length of day and night would be almost equal (about 12 hours) for the whole year.

(ii) On some days of the year, Aryan notices that day and night in Kolkata are nearly equal in length. Such a day is called:
(a) Summer solstice
(b) Equinox
(c) Eclipse
(d) Winter solstice
Answer:
(b) Equinox
OR
(ii) Meher’s report of “bright even at midnight” in June and “no sunrise” for many days in December suggests that:
(a) Her town does not rotate with the rest of the Earth.
(b) Her region can remain facing the Sun or turned away from the Sun for long stretches in a year.
(c) The Sun moves closer to her town only in June.
(d) The Moon blocks the Sun every day in December.
Answer:
(b) As Meher’s town is close to the North Pole, which remains tilted towards the Sun during summer, causing sunlight to be received for a longer time during June. Whereas during winters, it remains tilted away from the Sun, causing longer nights.

(iii) Why does Aryan’s class use special filters for watching the solar eclipse, but not for the lunar eclipse?
Answer:
During a solar eclipse, the Sun’s light can still be strong enough to damage the eyes, so special filters or safe viewing methods are needed. In lunar eclipse, we only see the Sun’s reflected light coming from the Moon, which is not intense enough to harm the eyes, so it can be watched directly.

Question 2.
Ravi wakes up early in the morning and goes to the nearby valley to practice Yoga. His uncle has taught him different asanas and their benefits, including Surya Namaskar, which he does every day facing the east, so that the rays of the rising Sun fall on his face. Ravi’s uncle travelled to the United States to give a talk on Yoga and the Patañjali Yoga-sūtra. One day, Ravi called his uncle after coming home from school. To his surprise, while it was daytime in India, it was already night in the United States.

(i) To get the rays of the rising Sun, Ravi faces east while doing Surya Namaskar. What motion of the Earth makes the Sun appear to rise in the east?
Answer:
The rotation of the Earth from west to east makes the Sun appear to rise in the east.

(ii) If the Earth were to move in the opposite direction (east to west), how would the apparent motion of the Sun change?
Answer:
If the Earth rotated in the opposite direction (east to west), the Sun would appear to rise in the west and set in the east.

(iii) Why was it night in the United States while it was daytime in India?
Answer:
It was night in the U.S. while it was daytime in India because the Earth rotates on its axis, and different parts of the Earth face the Sun at different times. So when it is daytime in India, the United States is oriented away from the Sun and experiences night.
OR
(iii) If the Earth stopped rotating, what would happen to the cycle of day and night?
Answer:
If the Earth stopped rotating, one side would experience continuous day while the other side would have continuous night for months. There would be no 24 hours day-night cycle.

Earth, Moon, and the Sun Class 7 Picture Based Questions

Observe the given pictures and answer the questions that follow:

Question 1.
The following diagram shows the Earth at four positions in its orbit around the Sun.

(i) At which labelled positions in the Northern Hemisphere are the days longer than the nights? Name the corresponding seasons in India.
Answer:
In the Northern Hemisphere, the days are longer than the nights at position C (June); at this time, India experiences summer.
OR
(i) At which labelled positions in the Northern Hemisphere are the nights longer than the days? Name the corresponding seasons in India.
Answer:
The nights are longer than the days at position A (December); at this time, India experiences winter.

(ii) Looking at the March and September positions, on about 21 March and 23 September, how long does daytime last, and what are these days called in the Northern Hemisphere?
Answer:
On or around 21 March and 23 September, daytime lasts for about 12 hours (and night also for about 12 hours) at most places on the Earth. In the Northern Hemisphere, these days are called the spring and the autumn equinox, respectively.

(iii) Using the diagram, explain clearly why the Northern and Southern Hemispheres experience opposite seasons at the same time of the year.
Answer:
The Earth’s axis remains tilted in the same direction as the Earth goes around the Sun. When the Northern Hemisphere is tilted towards the Sun, it gets more direct sunlight and longer days, so it experiences summer, while the Southern Hemisphere is tilted away and has winter.

Half a year later, the situation is reversed, the Northern Hemisphere is tilted away from the Sun and experiences winter, and the Southern Hemisphere is tilted towards the Sun and has summer. Therefore, the two hemispheres experience opposite seasons at the same time.

Question 2.
The two pictures (A) and (B) show the Earth’s axis tilted while sunlight falls on it. Four locations are marked as Positions 1, 2, 3 and 4. The bright part of each circle shows the region experiencing day, and the dark part shows night.

(i) In picture (A), compare Positions 1 and 2. Which position has a longer day and shorter night?
Answer:
In picture (A), Position 1 has a longer day and shorter night than Position 2.
OR
(i) In picture (B), compare Positions 3 and 4. Which position has a longer day and shorter night?
Answer:
In picture (B), Position 3 has a longer day than Position 4.

(ii) In picture (B), imagine a city at Position 3 and another city at the Equator. Whose length of daytime will change more from one season to another?
Answer:
The city at Position 3 will have a bigger change in the length of daytime from one season to another than the city on the Equator. This is because places farther from the Equator receive more varied sunlight during the year as the tilted Earth goes around the Sun, while the Equator receives almost the same amount of sunlight and nearly equal day and night in all seasons.

(iii) Which of the following conclusions can best be drawn by comparing the two pictures?
(a) Every place on the Earth has equal day and night in picture (A), but not in picture (B).
(b) Places farther from the Equator (like Positions 1 and 3) show a bigger change in the length of daytime between the two seasons than places on the Equator.
(c) Position 2 will always have a longer day than Position 3 in every season of the year.
(d) The tilt of the Earth’s axis does not affect the length of day and night; only the distance from the Sun matters.
Answer:
(b) Places farther from the Equator (like Positions 1 and 3) show a bigger change in the length of daytime between the two seasons than places on the Equator.

Earth, Moon, and the Sun Class 7 Activity Based Questions

Read the following activity and answer the questions that follow:

1. Objective: To model how the relative positions of the Sun, Earth and Moon cause solar and lunar eclipses.
Materials required: A bright torch or table lamp, one large ball, one smaller ball fixed on a pencil or stick, and a dark room.

Procedure:

• Step 1: Go to a dark room and place the torch or lamp on a table so that it shines steadily. This represents the Sun.
• Step 2: Ask one student to hold the large ball at some distance in front of the lamp. This ball represents the Earth.
• Step 3: Another student holds the small ball (Moon) on a stick between the lamp and the large ball, all in a straight line. Observe the shadow of the small ball falling on the large ball.
• Step 4: Now move the small ball so that the large ball comes between the lamp and the small ball. Observe the shadow of the large ball on the small ball.
• Step 5: Repeat Steps 3 and 4 slowly, watching carefully how the direction of light and the shadows change.

(i) Identify the type of eclipse being modelled in step 3 of the activity.
Answer:
In Step 3, the Moon is between the Sun and the Earth, so it models a solar eclipse.
OR
(i) Identify the type of eclipse being modelled in step 4 of the activity.
Answer:
In Step 4, the Earth is between the Sun and the Moon, so it models a lunar eclipse.

(ii) In the eclipse shown by Step 3, can people all over the Earth see the eclipse? Explain clearly who can see it and who cannot see it.
Answer:
No, people all over the Earth cannot see the eclipse shown in Step 3. In a solar eclipse, only those people who are in the small region of the Earth where the Moon’s shadow falls can see the Sun covered. People outside this narrow shadow region only see partial solar eclipse or do not see eclipse at all.

(iii) In the eclipse shown by Step 4, can people all over the Earth see the eclipse? Explain.
Answer:
No, only the people on the nighttime side of the Earth can see the eclipse shown in Step 4, provided the Moon is above their horizon, and the sky is clear.

Earth, Moon, and the Sun Extra Questions for Practice

Question 1.
Which pair correctly matches the event and the date in the Northern Hemisphere?
(a) Summer solstice: around 22 December
(b) Winter solstice: around 21 June
(c) Summer solstice: around 21 June
(d) Equinox: around 21 June
Answer:
(c) In the Northern Hemisphere, the longest day occurs around 21 June, known as the summer solstice.

Question 2.
In a long-exposure photo taken in the Northern Hemisphere, star trails form circles. The centre of these circles indicates:
(a) The direction of Earth’s revolution.
(b) The approximate position of the Pole Star.
(c) The position of the Sun at midnight.
(d) The path of the Moon around Earth.
Answer:
(b) In the Northern Hemisphere. Long-exposure photographs show stars forming circular trails because of the Earth’s rotation. The stars seem to move in circles around a nearly fixed point in the sky, which lies close to the Pole Star. Ilence, the centre of the circles shows the Pole Star’s approximate position.

Question 3.
Which of the following statements are correct about Earth’s daily motion?
(i) Earth completes one rotation in about 24 hours.
(ii) This motion causes regular changes between light and darkness.
(iii) All places on Earth experience daylight at the same time.
(iv) Observers on Earth feel as if the sky is moving around them.
(a) (ii) and (iv) only
(b) (i), (ii) and (iv)
(c) (i) and (iii) only
(d) (ii), (iii) and (iv)
Answer:
(b) (i), (ii) and (iv)

Question 4.
Which of the following statements are correct about the observation of stars in the night sky?
(i) Some stars appear to change their position over several hours.
(ii) All stars rise and set at exactly the same time every night.
(iii) The apparent movement of stars can be noticed by careful observation.
(iv) People at the same place see different stars at different times of the year.
(a) (i) and (iii) only
(b) (i), (iii) and (iv)
(c) (ii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Answer:
(b) (i), (iii) and (iv)

Questions 5 and 6 consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 5.
Assertion (A): When the Northern Hemisphere has longer days, the Southern Hemisphere must also have longer days.
Reason (R): Earth’s axis is tilted, so one hemisphere gets a longer duration of daylight while the other gets a shorter duration of daylight.
Answer:
(d) The assertion is false because when the Northern Hemisphere has longer days, the Southern Hemisphere has shorter days.

Question 6.
Assertion (A): Objects in the night sky may appear to move even if they are actually stationary.
Reason (R): The observer on Earth is moving along with the Earth.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Question7.
What is meant by the apparent motion of stars?
Answer:
It is the movement of stars across the sky as seen from Earth. This movement is not real but appears due to the rotation of the Earth about its axis.

Question 8.
Why does the Sun appear to rise in the east and set in the west?
Answer:
It is because the Earth rotates from west to east about its axis. Due to this rotation, the Sun seems to move across the sky in the opposite direction.

Question 9.
Explain the following:
(i) The equatorial regions experience nearly equal lengths of day and night throughout the year.
(ii) The seasons in the Southern Hemisphere are opposite to those in the Northern Hemisphere.
Answer:
(i) The equatorial regions experience nearly equal lengths of day and night throughout the year because the effect of Earth’s tilt is minimal there, and the illuminated half of the Earth always divides the Equator into nearly equal halves. As a result, the duration of sunlight received there remains almost the same throughout the year.

(ii) The seasons in the Southern Hemisphere are opposite to those in the Northern Hemisphere because Earth’s axis is tilted. When the Northern Hemisphere is tilted towards the Sun and receives longer, more direct sunlight, it experiences summer, while the Southern Hemisphere is tilted away and experiences winter. Later, when the Southern Hemisphere is tilted towards the Sun, the seasons are reversed.

Question 10.
Explain the terms equinox and solstice with suitable examples from the Northern Hemisphere.
Answer:
An equinox is a day when day and night are nearly equal in length at most places on Earth because neither hemisphere is tilted towards or away from the Sun. In the Northern Hemisphere, equinoxes occur around 21 March and 23 September. A solstice is a day when the length of day is either the longest or the shortest in a year in a hemisphere. In the Northern Hemisphere, the summer solstice occurs around 21 June (longest day) and the winter solstice occurs around 22 December (shortest day).

Question 11.
A student observes the same playground at 8 AM, 12 noon, and 5 PM.

• At 8 AM, the shadow of a pole is long and points west.
• At noon, the shadow becomes very short.
• At 5 PM, the shadow becomes long again but points east.

Based on these observations:

(i) Identify the factor responsible for this change.
Answer:
The change in the length and direction of the shadow is due to the apparent movement of the Sun in the sky. which is caused by the rotation of the Earth on its axis.

(ii) Explain why the shadow direction changes during the day.
Answer:
As the Earth rotates from west to east, the Sun appears to move from east to west across the sky. Therefore, the direction of the sunlight. changes during the day, causing the shadow to change its direction.

(iii) Predict the shadow length at sunset.
Answer:
The shadow will be very long, stretching in the eastward direction.
OR
(iii) State one conclusion about the position of the Sun at noon based on the shadow observed.
Answer:
At noon, the Sun is nearly overhead, which causes the shadow of the object to be the shortest.

Question 12.
(i) Define rotation and revolution of the Earth.
Answer:
Rotation is the spinning of the Earth on its own axis. Revolution is the movement of the Earth around the Sun along its orbit.

(ii) State the time taken by the Earth to complete one rotation and one revolution.
Answer:
The Earth takes about 24 hours to complete one rotation. It takes about 365 days and 6 hours to complete one revolution.

(iii) Mention one major effect of rotation and one major effect of revolution on Earth.
Answer:
Rotation causes day and night on Earth. Revolution, along with Earth’s axial tilt, causes scasons.

(iv) Why does the Pole Star appear almost stationary in the sky while other stars seem to move?
Answer:
The Earth’s axis of rotation points very close to the Pole Star in the Northern IIemisphere. Therefore, the Pole Star appears nearly stationary in the sky when viewed from the Earth, while all other stars appear to move around it.

BSE Odisha Guru Class 7 Social Science Notes

BSE Odisha Class 7 Social Science Notes

BSE Odisha 7th Class Social Science Notes

• Class 7 Social Science Chapter 1 Notes Odia Medium – ଭାରତର ଭୌଗୋଳିକ ବିବିଧତା
• Class 7 Social Science Chapter 2 Notes Odia Medium – ପାଣିପାଗକୁ ବୁଝିବା
• Class 7 Social Science Chapter 3 Notes Odia Medium – ଭାରତର ଜଳବାୟୁ
• Class 7 Social Science Chapter 4 Notes Odia Medium – ନୂତନ ଅଭ୍ୟୁଦୟ : ସହର ଓ ରାଜ୍ୟ
• Class 7 Social Science Chapter 5 Notes Odia Medium – ସାମ୍ରାଜ୍ୟର ଅଭ୍ୟୁଦୟ
• Class 7 Social Science Chapter 6 Notes Odia Medium – ପୁନର୍ଗଠନର ଯୁଗ
• Class 7 Social Science Chapter 7 Notes Odia Medium – ଗୁପ୍ତଯୁଗ : ନିରନ୍ତର ସୃଜନଶୀଳତାର ଏକ ନିଦର୍ଶନ
• Class 7 Social Science Chapter 8 Notes Odia Medium – ଭୂମି ପବିତ୍ର ହୁଏ କିପରି?
• Class 7 Social Science Chapter 9 Notes Odia Medium – ଶାସକଙ୍କ ଠାରୁ ଶାସିତଙ୍କ ପର୍ଯ୍ୟନ୍ତ : ସରକାରର ପ୍ରକାର ଭେଦ
• Class 7 Social Science Chapter 10 Notes Odia Medium – ଭାରତର ସମ୍ବିଧାନ : ଏକ ପରିଚୟ
• Class 7 Social Science Chapter 11 Notes Odia Medium – ଦ୍ରବ୍ୟ ବିନିମୟରୁ ମୁଦ୍ରା ପର୍ଯ୍ୟନ୍ତ
• Class 7 Social Science Chapter 12 Notes Odia Medium – ବଜାରକୁ ବୁଝିବା

Also Read BSE Odisha Class 7 Social Science Solutions

Students can include Class 7 Curiosity Extra Questions and Class 7 Science Chapter 11 Light: Shadows and Reflections Extra Question Answer in their daily self-study routine.

Class 7 Science Chapter 11 Light: Shadows and Reflections Extra Questions

Class 7 Science Chapter 11 Extra Questions on Light: Shadows and Reflections

Light: Shadows and Reflections Class 7 Very Short Question Answer

Question 1.
Riya noticed that her book, table, and chair were clearly visible even though they were not emitting light. How is she able to see these objects?
Answer:
Riya is able to see the book, table and chair because light from a luminous source (such as a bulb) falls on them and is reflected to her eyes. These objects do not produce their own light; they are visible due to reflected light.

Question 2.
Why is our shadow shorter at noon compared to in the morning or evening?
Answer:
In the morning and evening, the Sun is near the horizon, so sunlight comes from a small angle from the horizon and makes our shadows long. At noon, the Sun is almost right above us, so sunlight comes straight down, and our shadows are very short.

Question 3.
What are luminous and non-luminous objects?
Answer:
Luminous objects are those that emit their own light (e.g., Sun, candle flame, electric bulb). Nonluminous objects do not emit light and are seen only when they reflect light from a luminous source (e.g., Moon, book, chair).

Question 4.
You are given a transparent glass sheet. Suggest any two ways to make it translucent without breaking it.
Answer:
Two ways to make a transparent glass sheet translucent (without breaking it):

• Rub its surface with sandpaper to make it rough.
• Apply a thin coating of oil/grease/wax/paint on the glass surface.

Question 5.
You have three opaque strips with very small holes of different shapes as shown in figure below. If you obtain an image of the sun on a wall through these holes, will the image formed by these holes be the same or different?

Answer:
When the image of the Sun is obtained on a wall through small holes of different shapes, the images formed will all appear the same, that is circular. The shape of the hole only changes brightness and sharpness, not the basic circular shape of the Sun’s image.

Question 6.
What are the three things required to form a shadow?
Answer:
Three things required to form a shadow are a source of light, an opaque object, and a screen/surface (wall, floor, paper etc.)

Question 7.
Which object produces a shadow of the same shape irrespective of its orientation with respect to the light source?
Answer:
A spherical object produces a circular shadow in every position because its outline remains the same in all orientations with respect to the light source and the screen.

Question 8.
Why does a rubber ball cast a shadow but a glass ball does not?
Answer:
A rubber ball is opaque, so it blocks light and forms a clear shadow. A glass ball is transparent, so it allows most light to pass through it and therefore does not form a distinct shadow.

Question 9.
Three torches A, B and C shown in figure alongside are switched on one by one. The light from which of the torches will not form a shadow of the ball on the screen.

Answer:
A shadow is formed only when an object blocks the path of light falling on a screen. If the light from a torch does not directly fall on the screen, then no shadow can be produced. In this case, the light from torch C does not reach the screen, so it will not form any shadow of the ball on the screen.

Question 10.
Look at the figure given below.

Will there be any difference in the shape of the shadow formed on the screen in A and B?
Answer:
No, the shape of the shadow formed in both the cases will be circular.

Question 11.
A small mirror can still show the complete image of a large tree. How is this possible?
Answer:
A plane mirror forms an image using the light reflected from the object. Even a small mirror can reflect light coming from different parts of a large tree into our eyes. Therefore, the whole tree can be seen in the mirror, though the image appears smaller because the tree is far away.

Question 12.
Write one difference and one similarity in the nature of the image formed by a plane mirror and a pinhole camera.
Answer:
Similarity: In both cases, the image formed is laterally inverted.
Difference: A plane mirror forms a vertically erect image, while a pinhole camera forms a vertically inverted image.

Light: Shadows and Reflections Class 7 Short Question Answer

Question 1.
A student covered a torch with red cellophane sheet to obtain red light. Using the red light she obtains a shadow of an opaque object. She repeats this activity with green and blue light. Will the colour of the light affect the shadow? Explain.
Answer:
The colour of light does not change the shadow. The shadow remains dark because it is the region where light is blocked by the opaque object. Only the colour of the illuminated area around the shadow changes according to the colour of the torch light (red/green/blue).

Question 2.
State any two observations that show light travels in a straight line.
Answer:

• A candle flame can be seen through a straight tube, but it cannot be seen through a bent tube.
• When light is passed through two or three pinholes, a bright spot is obtained on the screen only when the pinholes are exactly in one line; if any one pinhole is shifted, the spot disappears.

Question 3.
A sheet of plywood, a piece of muslin cloth and that of a transparent glass, all of the same size and shape, were placed at A one by one in the arrangement shown in the figure. Will a shadow be formed in each case? If yes, how will the shadow on the screen be different in each case? Give reasons for your answer.

Answer:
Yes, shadows will differ because the three materials are opaque, translucent and transparent. A sheet of plywood will form a dark, well-defined shadow on the screen because plywood is opaque and does not allow light to pass through it. A piece of muslin cloth will form a faint and less sharp shadow because muslin cloth is translucent. It allows some light to pass through, so the shadow is lighter and blurred. A transparent glass sheet will not form a distinct shadow (it may be very faint) because glass is transparent and allows most of the light to pass through it.

Question 4.
How can the same object form shadows of different sizes and shapes?
Answer:
The size of a shadow depends on the distance between the light source, the object and the screen. When the object is moved closer to the light source, the shadow becomes larger; when it is moved closer to the screen, the shadow becomes smaller and clearer. The shape of the shadow can change when the object is tilted or turned because the outline of the object facing the light changes.

Question 5.
Observe the given illustration of two aeroplanes flying at different heights above the ground. Why is the shadow of the aeroplane visible on the ground when it is flying low, but hardly visible when it is flying very high in the sky? Explain scientifically.

Answer:
When an aeroplane is flying very high, it is far from the ground. Therefore, the shadow spreads over a large area and its edges become less sharp. As a result, the shadow becomes very faint and is hardly visible on the ground. When the aeroplane is flying low, it is closer to the ground. It blocks sunlight over a smaller area and forms a sharper, darker shadow. Hence, the shadow is clearly visible.

Question 6.
Answer the following questions.

(i) Does the image formed in a plane mirror change size when we move farther or closer to it?
Answer:
No. The image formed by a plane mirror remains the same size as the object.

(ii) Why cannot images formed by a plane mirror be obtained on a screen?
Answer:
A plane mirror forms a virtual image. Since the reflected rays do not actually meet, the image cannot be obtained on a screen.

(iii) If a person moves 1 m closer to a plane mirror, how much does the image move and in which direction?
Answer:
The image also moves 1 m closer to the mirror, i.e., it shifts by the same distance as the person moves.

Question 7.
Answer the following questions.

(i) What is a kaleidoscope? How does it work?
Answer:
A kaleidoscope is an optical instrument made using plane mirrors arranged inside a tube. Small coloured beads or pieces are kept at one end. Light entering the tube undergoes multiple reflections in the mirrors and forms repeated images of the coloured pieces, producing patterns seen through the viewing end.

(ii) Why does a kaleidoscope show a new pattern every time it is rotated?
Answer:
On rotating the kaleidoscope, the coloured pieces change their arrangement. This changes the pattern of multiple reflections, so a new pattern is seen each time.

Light: Shadows and Reflections Class 7 Long Question Answer

Question 1.
What is reflection of light? Describe an activity to show this effect.
Answer:
Reflection of light is the phenomenon in which light, after striking a smooth and polished surface, bounces back into the same medium.
Activity to show reflection of light: Take a plane mirror or a shiny steel plate and place it upright on a table. Allow light from a torch to fall on its surface. The light ray strikes the mirror and is reflected back from its surface, showing the reflection of light. This activity demonstrates that smooth and shiny surfaces reflect light.

Question 2.
During a science activity, Riya placed two plane mirrors perpendicular to each other on a table. She kept a bottle cap between them and observed three images of the bottle cap.
(i) What will happen to the number of images if the mirrors in this activity are brought closer at a smaller angle?
(ii) What will happen if the mirrors are kept parallel to each other? Give one real-life example where this effect can be seen.
(iii) Name one device that works on the principle of multiple reflections.
(iv) Can these images be obtained on a screen?

Answer:
(i) When the mirrors are brought closer by making the angle between them smaller, the number of images increases. This is because light undergoes more reflections between the mirrors.
(ii) If the mirrors are kept parallel to each other, a very large number of images are formed due to repeated reflections of light. This effect can be seen in barber shops or dressing rooms where mirrors are placed opposite each other.
(iii) A kaleidoscope works on the principle of multiple reflections of light.
(iv) No, these images cannot be obtained on a screen because plane mirrors form virtual images.

Question 3.
A periscope and a kaleidoscope use mirrors in different ways. Explain the arrangement of mirrors in each and how this arrangement helps in their use.
Answer:
A periscope consists of two plane mirrors placed inside a tube in a particular arrangement. The mirrors are kept parallel to each other and are inclined at an angle (45°) so that light from an object enters the top opening, reflects from the first mirror, then from the second mirror, and finally reaches the observer’s eyes. This arrangement allows light to change direction twice, enabling a person to see objects that are not directly visible, such as objects behind a wall or above eye level.

A kaleidoscope uses plane mirrors (generally three) arranged in a pattern inside a tube, with coloured beads or pieces at one end. Light enters the tube and undergoes multiple reflections between the three mirrors. Because of these repeated reflections, many images of the coloured objects are formed, creating beautiful and symmetrical patterns. This arrangement is useful for producing decorative designs.

Question 4.
Riya, a curious class 7 student, chooses a reverse periscope as her science project for the school exhibition. She wants to show how light can be redirected so that a person looking from a low position can see above a raised platform.

(i) How many times does light get reflected in this optical device?
Answer:
Light gets reflected two times in the reverse periscope, once from the top mirror and once from the bottom mirror.

(ii) What would happen if one of the mirrors inside the reverse periscope is tilted at the wrong angle?
Answer:
If a mirror is tilted wrongly, the light rays will not reach the eye properly. As a result, the image may look shifted, unclear, or may not be visible at all.

(iii) What is the principle behind the working of a reverse periscope?
Answer:
The principle behind the working of a reverse periscope is the reflection of light travelling in a straight line from plane mirrors.
When light from an object strikes the mirrors placed at 45° inside the reverse periscope, it gets reflected twice and changes direction. This allows a person to see over a raised platform without changing their position.

(iv) Is the image formed by the given reverse periscope laterally inverted? Explain with the help of a diagram.
Answer:
No, the image formed by a periscope is not laterally inverted. This is because the two successive reflections cancel each other’s lateral inversion.

Light: Shadows and Reflections Class 7 Picture Based Questions

Observe the given pictures and answer the questions that follow:

Question 1.
A wall clock is shown in the figure alongside.

(i) What will be the vertical mirror image time of the clock?
Answer:
In the figure, the clock shows 3:00. Its vertical mirror image time will be 9:00.

(ii) At what time will the clock and its mirror image be the same?
Answer:
The clock and its mirror image show the same time at 12:00 and 6:00.

(iii) Find the vertical mirror image times if the actual time in the above clock is 1:00, 2:00, 4:00, and 5:00. Observe the pattern and state the relation between clock and its mirror image time.
[Hint: Compare each time with its mirror image and notice what they add up to.]
Answer:
Vertical mirror image times:
1: 00 11: 00 2: 00
10: 00,100
Relation/pattern: Actual time + mirror image time =12: 00 (i.e., mirror time =12: 00- actual time).

OR

(iii) Find the time in vertical mirror image if the time in the clock is (a) 4: 45 (b) 10:10.
Answer:
Mirror time =11: 60- given time
(a) 11: 60-4: 45=7: 15
(b) 11: 60-10: 10=1: 50

Question 2.
Few words are given in the boxes as shown alongside:

(i) Identify the words whose mirror image changes in a vertical plane mirror.
Answer:
BOOK, LOOK, COOK, TOOK

(ii) Identify the words whose mirror image changes in a horizontal plane mirror.
Answer:
MOM, LOOK, WOW, TOOK

(iii) Write two properties of the image formed by a plane mirror.
Answer:
Two properties of the image formed by a plane mirror:

• The image is erect.
• The image is of the same size as the object.

OR

(iii) If the distance between a plane mirror and a person is 2 m, what will be the distance between the person and his image in the mirror?
Answer:
Distance between mirror and person =2m In a plane mirror, the image forms at the same distance behind the mirror.
So, image is 2 m behind the mirror.
Distance between person and image
= 2 m+2m=4m

Light: Shadows and Reflections Class 7 Activity Based Questions

Read the following activity and answer the questions that follow:

1. Objective: To demonstrate that light travels in a straight line and use this fact to explain why a pinhole camera forms an inverted image.
Materials required: Torch, two cardboard sheets, pin/needle, tape, a candle, a white sheet/wall as screen, 2 books/stands.

Procedure:

• Take two cardboard sheets and make a small hole at the centre of each using a pin. Name them Sheet A and Sheet B.
• In a dark room, fix Sheet A upright on a table using books so that it stands vertically. Place the torch behind Sheet A so that light passes through the hole.
• Keep Sheet B a little away from Sheet A. Adjust Sheet B until you can see a bright spot of light passing through both holes and falling on the wall/screen.
• Now shift Sheet B slightly sideways and observe what happens to the bright spot on the screen.
• Remove Sheet B and place a lighted candle (or any bright object) behind Sheet A. Observe the image formed on the screen through the pinhole. Note that the image appears inverted.
• Enlarge the pinhole slightly and observe the image again.

(i) When Sheet B is shifted slightly to one side, the bright spot on the screen disappears. Explain why this happens.
Answer:
The bright spot is seen only when both pinholes are in the same straight line. Light travels in a straight line, so when the second sheet is shifted, the holes are no longer aligned and light cannot pass through both pinholes to reach the screen. Hence, the bright spot disappears.

(ii) Describe the image formed on the wall/screen by the pinhole camera. Mention any two characteristics of the image.
Answer:
The image formed by a pinhole camera on the wall/screen is inverted (upside down). It is usually smaller than the object and it is also laterally inverted.

(iii) Predict what will happen if the pinhole is made larger. Mention the change in brightness and sharpness of the image.
Answer:
If the pinhole is made larger, more light enters, so the image becomes brighter. However, the image becomes less sharp (blurred) because rays from one point of the object spread over a larger area of the screen.

OR

(iii) Why is the image formed by the pinhole inverted?
Answer:
The image formed by a pinhole is inverted because light travels in straight lines. Rays of light from the top of the object pass through the small hole and reach the lower part of the screen, while rays from the bottom of the object reach the upper part of the screen. This crossing of light rays at the pinhole causes the image to appear upside down.

Light: Shadows and Reflections Extra Questions for Practice

Question 1.
A ball is placed between a torch and a wall. The torch and wall remain fixed. The ball is moved closer to the torch. What change will occur?
(a) The shadow becomes smaller and sharper
(b) The shadow becomes larger and blurred
(c) The shadow disappears completely
(d) The shadow shifts colour
Answer:
(b) When the ball is moved closer to the torch, it blocks a wider spread of rays, so the shadow on the wall becomes larger: The shadow also becomes blurred and the edges less sharp because the light spreads out more from the torch.

Question 2.
A shiny steel plate is placed between a torch and a wall. A dark patch is seen on the wall, and a bright spot appears elsewhere on the opposite wall. The best explanation is:
(a) The steel plate is translucent, so both shadow and light pass through.
(b) The steel plate blocks some light and emits its own light.
(c) The steel plate blocks and reflects the same light to another direction.
(d) Torchlight bends around the steel plate to form the bright spot.
Answer:
(c) The steel plate acts as an opaque object for the direct beam, so it creates a shadow (dark patch). At the same time, its shiny surface reflects part of the light, producing a bright spot on the opposite wall.

Question 3.
A student observes a beam of sunlight entering a dark room through a small hole. Which of the following statements are correct?
(i) Light travels in a straight line.
(ii) The beam spreads in all directions immediately upon entering the room.
(iii) A narrow beam is formed due to linear propagation of light.
(iv) This observation supports the idea of reflection.
(a) (i) and (iii) only
(b) (i), (ii) and (iii)
(c) (ii) and (iv) only
(d) (i), (iii) and (iv)
Answer:
(a) (i) and (iii) only

Question 4.
An object is placed in front of a screen with a light source behind it. Which of the following statements are correct?
(i) The object must block light to form a shadow.
(ii) Transparent objects form dark shadows.
(iii) Opaque objects form dark shadows.
(iv) Shadow size depends only on the position of the object and the light source.
(a) (i) and (iii) only
(b) (i), (iii) and (iv)
(c) (ii) and (iv) only
(d) (i), (ii) and (iv)
Answer:
(a) Shadow size depends on the position of the object, the light source and the screen.

Questions 5 and 6 consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 5.
Assertion (A): The image formed by a pinhole camera shows the colours of the object.
Reason (R): A pinhole camera forms an image due to reflection from a shiny surface inside the box.
Answer:
(c) A pinhole camera can show colours because light coming from the object carries colour information and forms the image on the sereen. But the image is not formed due to reflection inside the box: it is formed because light passes through a tiny hole and falls on an opposite screen.

Question 6.
Assertion (A): The left side of the object appears as the right side in the mirror image.
Reason (R): The image formed by a plane mirror is laterally inverted.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Question 7.
What is a luminous object? Give one example.
Answer:
A luminous object is an object that emits its own light. The Sun is an example of a luminous object.

Question 8.
State one observation that proves that light travels in a straight line.
Answer:
When light passes through two or more pinholes, a bright spot is seen on the screen only when the pinholes are in one straight line. This observation shows that light travels in a straight line.

Question 9.
Define reflection of light with one observation.
Answer:
Reflection of light is the change in direction of light when it falls on a shiny surface or a mirror. For example, when sunlight falls on a plane mirror or a polished steel plate, a bright spot is seen on a wall, showing that the light has been redirected.

Question 10.
In a torch-ball-screen setup, two different shadow sizes are obtained without changing the ball. How is this possible?
Answer:
The size of the shadow depends on both the size and position of the object relative to the light source and the screen. If the ball is moved closer to the torch, its shadow on the screen becomes larger. If the ball is moved closer to the screen (keeping the torch fixed), the shadow becomes smaller.

Question 11.
Meera stands in front of a plane mirror while getting ready for school. She moves a little closer to the mirror and then steps back. She also raises her right hand and notices the movement of the image.

(i) Why is Meera able to see her image in the plane mirror?
Answer:
Light from Meera’s body falls on the plane mirror and is reflected back into her eyes, allowing her to see her image.

(ii) What change does Meera observe in the image when she moves closer to or away from the mirror?
Answer:
The image also appears to move closer or farther by the same distance, because the image in a plane mirror is formed at the same distance behind the mirror as the object is in front.

(iii) Why does Meera see her raised right hand as the left hand in the mirror?
Answer:
A plane mirror causes lateral inversion, due to which the right side of the object appears as the left side in the image.
OR
(iii) Will the size of Meera’s image change if she moves nearer to the mirror? Give a reason.
Answer:
No, the size of the image remains the same because a plane mirror always forms an image equal in size to the object.

Question 12.
(i) Classify materials as transparent, translucent and opaque on the basis of transmission of light. Give an example of each.
Answer:
Materials are classified on the basis of how much light they allow to pass through them. Transparent materials allow light to pass almost completely, for example, clean water Translucent materials allow light to pass partially, for example, frosted glass. Opaque materials do not allow light to pass through at all, for example, cardboard.

(ii) Explain how these materials affect shadow formation.
Answer:
Opaque materials block light completely and form dark, clear shadows. Translucent materials block only part of the light, so they form lighter shadows. Transparent materials allow most of the light to pass. so they generally do not form a dark shadow, though some transparent objects may produce a faint shadow.

Students can use Class 7 Math Solution Odia Medium and Class 7 Maths Chapter 7 Question Answer Odia Medium ତିନୋଟି ପରସ୍ପର ଛେଦୀ ସରଳ ରେଖାର କଥା to check their answers after solving exercises.

7th Class Maths Chapter 7 Question Answer Odia Medium

Class 7 Maths Chapter 7 Odia Medium

Question 1.
ଯେତେବେଳେ ତିନୋଟି ଶୀର୍ଷବିନ୍ଦୁ ଏକ ସରଳରେଖା ଉପରିସ୍ଥ ହେବେ ସେତେବେଳେ କ’ଣ ହେବ ?
Solution:
ଯେତେବେଳେ ତିନୋଟି ଶୀର୍ଷବିନ୍ଦୁ ଏକ ସରଳରେଖା ଉପରିସ୍ଥ ହେବେ ସେତେବେଳେ ଏକ ସରଳରେଖା ଉତ୍ପନ୍ନ ହେବ ।

7.1 ସମବାହୁ ତ୍ରିଭୁଜ :

Page No. (149 to 151)

ଯେଉଁ ତ୍ରିଭୁଜର ତିନୋଟି ବାହୁ ଓ ତିନୋଟି କୋଣ ପରସ୍ପର ସମାନ, ତାହାକୁ ସମବାହୁ ତ୍ରିଭୁଜ କୁହାଯାଏ । ଏହାର ପ୍ରତ୍ୟେକ କୋଣ

Question 1.
ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯାହାର ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟ 4 ସେ. ମି. ।
Solution:
• ସୋପାନ – 1 : ଚିତ୍ରରେ ଦର୍ଶାଯାଇଥ‌ିବା ଭଳି A ବିନ୍ଦୁକୁ କେନ୍ଦ୍ରକରି, AB = 4 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ନେଇ ଏକ ବୃହତ ଚାପ ଅଙ୍କନ କର । ‘B’ ବିନ୍ଦୁଟି ଏହି ଚାପ ଉପରିସ୍ଥ ଯେକୌଣସି ବିନ୍ଦୁ ହୋଇପାରିବ । (ଚିତ୍ର-1)

• ସୋପାନ – 2 : ‘B’ କୁ କେନ୍ଦ୍ରକରି 4 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ନେଇ ଅନ୍ୟ ଏକ ବୃହତଚାପ ଅଙ୍କନ କର । ଚାପଦ୍ଵୟର ଛେଦବିନ୍ଦୁଟିକୁ ‘C’ ନାମରେ ନାମିତ କରାଯାଉ । (ଚିତ୍ର-2)

• ସୋପାନ – 3 : AC ଓ BCର ଦୈର୍ଘ୍ୟ 4 ସେ.ମି. ହେବ । AC ଓ BC କୁ ଯୋଗ କଲାପରେ ଆମେ ଆବଶ୍ୟକୀୟ ସମବାହୁ ତ୍ରିଭୁଜ ABC ପାଇବା । (ଚିତ୍ର-3)

7.2 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ଦତ୍ତ ଥିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ :

Page No. (151 to 163)

Question 1.
ତ୍ରିଭୁଜଟିଏ ଅଙ୍କନ କର ଯାହାର ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ 4 ସେ.ମି., 5 ସେ.ମି. ଏବଂ 6 ସେ.ମି. ।
Solution:
ଅଙ୍କନ ପ୍ରଣାଳୀ :
ପ୍ରଥମ ସୋପାନ :
6 ସେ.ମି. ଦୈର୍ଘ୍ୟର \(\overline{\mathrm{AC}}\) ରେଖାଖଣ୍ଡ ଅଙ୍କନ କର ।

ଦ୍ଵିତୀୟ ସୋପାନ :
A କୁ ବ୍ୟାସାର୍ଦ୍ଧ କରି 4 ସେ.ମି. (AB) ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ଚାପ ଅଙ୍କନ କର ।

ତୃତୀୟ ସୋପାନ :
C କୁ କେନ୍ଦ୍ର କରି 5 ସେ.ମି. BC ବ୍ୟାସାର୍କ ବିଶିଷ୍ଟ ଚାପ ଅଙ୍କନ କର, ଯେପରି ତାହା ପୂର୍ବରୁ ଅଙ୍କା ହୋଇଥିବା ଚାପକୁ ଛେଦ କରିବ । ଛେଦବିନ୍ଦୁର ନାମ B ଦିଅ ।

ଚତୁର୍ଥ ସୋପାନ :
\(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{BC}}\) ଅଙ୍କନ କର ।
∴ ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ।

ସୋପାନ – 1 : ଯେକୌଣସି ବାହୁର ଦୈର୍ଘ୍ୟକୁ ଭୂମିରୂପେ ନିଅ । ମନେକରୁ AB = 4 ସେ.ମି. ।
ସୋପାନ – 2 : A କୁ କେନ୍ଦ୍ରକରି 5 ସେ.ମି. ବ୍ୟାସାର୍କ ବିଶିଷ୍ଟ ଏକ ବୃହତ ଚାପ ଅଙ୍କନ କର ।
ସୋପାନ – 3 : ପୁନଶ୍ଚ ‘B’ କୁ କେନ୍ଦ୍ରକରି 6 ସେ.ମି. ବ୍ୟାସାର୍କ ବିଶିଷ୍ଟ ଏକ ଚାପ ଅଙ୍କନ କର ଯେପରିକି ଏହା ପୂର୍ବ ଚାପକୁ ଛେଦ କରିବ ।
ସୋପାନ – 4 : ଦୁଇଟି ଚାପର ଛେଦବିନ୍ଦୁ ହେଉଛି ତୃତୀୟ ବିନ୍ଦୁ ‘C’ । AC ଓ BC କୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ABC ପାଇବା

ଅଙ୍କନ କର :

Question 1.
ନିମ୍ନୋକ୍ତ ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ (ସେ.ମି.) ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ।
(a) 4, 4, 6
(b) 3, 4, 5
(c) 1, 5, 5
(d) 4, 6, 8
(e) 3.5, 3.5, 3.5
Solution:
(a) ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ 4 ସେ.ମି., 4 ସେ.ମି., ଓ 6 ସେ.ମି.
(i) ପ୍ରଥମ ସୋପାନ : 6 ସେ.ମି. ଦୈର୍ଘ୍ୟର \(\overline{\mathrm{AC}}\) ରେଖାଖଣ୍ଡ ଅଙ୍କନ କର ।

(ii) ଦ୍ଵିତୀୟ ସୋପାନ : A ଓ C କୁ ବ୍ୟାସାର୍ଦ୍ଧ କରି 4 ସେ.ମି. AB ଓ BC ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଚାପ ଅଙ୍କନ କରାଯାଉ ।
(iii) ତୃତୀୟ ସୋପାନ ; B ଠାରେ ଉତ୍ପନ୍ନ ବିନ୍ଦୁରୁ A ଓ C କୁ ଯୋଗକଲେ ABC ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ମିଳିବ ।

(b) ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ 3 ସେ.ମି., 4 ସେ.ମି., ଓ 5 ସେ.ମି.
(i) ପ୍ରଥମ ସୋପାନ : 5 ସେ.ମି. ଦୈର୍ଘ୍ୟର \(\overline{\mathrm{AC}}\) ରେଖାଖଣ୍ଡ ଅଙ୍କନ କର ।

(ii) ଦ୍ଵିତୀୟ ସୋପାନ : A କୁ ବ୍ୟାସାର୍ଦ୍ଧ କରି 3 ସେ.ମି. (AB) ବ୍ୟାସାର୍କ ବିଶିଷ୍ଟ ଏକ ଚାପ ଅଙ୍କନ କରାଯାଉ ।
(iii) ତୃତୀୟ ସୋପାନ : C କୁ କେନ୍ଦ୍ର କରି 4 ସେ.ମି. BC ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଚାପ ଅଙ୍କନ କର, ଯେପରି ତାହା ପୂର୍ବରୁ ଅଙ୍କା ହୋଇଥିବା ଚାପକୁ ଛେଦ କରିବ । ଛେଦବିନ୍ଦୁର ନାମ B ଦିଅ ।
(iv) ଚତୁର୍ଥ ସୋପାନ : \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{BC}}\) ଅଙ୍କନ କର ।
∴ ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ।

ଏହି ସୋପାନଗୁଡ଼ିକୁ ଅନୁସରଣ କରି ପିଲାମାନେ (c), (d) ଓ (e) ଅଙ୍କନ ନିଜେ କରିବେ ।
ଯେଉଁ ତ୍ରିଭୁଜର ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟ ସମାନ ତାହାକୁ ‘ସମବାହୁ’ ତ୍ରିଭୁଜ କୁହାଯାଏ । ଯେଉଁ ତ୍ରିଭୁଜର ଯେକୌଣସି ଦୁଇବାହୁର ଦୈର୍ଘ୍ୟ ସମାନ, ତାହାକୁ ‘ସମଦ୍ଵିବାହୁ’ ତ୍ରିଭୁଜ କୁହାଯାଏ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ଗୋଟିଏ ବୃତ୍ତର ଉପରିସ୍ଥ ବିନ୍ଦୁ କିମ୍ବା ବୃତ୍ତର କେନ୍ଦ୍ରବିନ୍ଦୁ ବ୍ୟବହାର କରି ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ।

Solution:
(i) ‘O’ କୁ କେନ୍ଦ୍ରବିନ୍ଦୁ ଭାବେ ନେଇ ଗୋଟିଏ ବୃତ୍ତ ଅଙ୍କନ କରାଯାଉ ।

(ii) ବୃତ୍ତ ଉପରେ ଦୁଇଟି ବିନ୍ଦୁ A ଓ B ନି ଆଯାଉ ।
(iii) OA, AB ଓ BO କୁ ଯୋଗ କର ।

(iv) Δ AOB ଏକ ଆବଶ୍ୟକୀୟ ସମଢ଼ିବାହୁ ତ୍ରିଭୁଜ ।

Question 2.
ସମ ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ବୃତ୍ତ ଉପରିସ୍ଥ ବିନ୍ଦୁ କିମ୍ବା ବୃତ୍ତର କେନ୍ଦ୍ରକୁ ନେଇ ସମବାହୁ ତ୍ରିଭୁଜ ଏବଂ ସମଜ୍ଜିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।

Solution:
ସମଦ୍ଵିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ :
(i) ପ୍ରଥମ ଚିତ୍ରରେ, ଗୋଟିଏ ବିନ୍ଦୁ C ନି ଆଯାଉ ।

(ii) AB, CB ଓ BA କୁ ଯୋଗକର ।

(ii) Δ ABC ଗୋଟିଏ ଆବଶ୍ୟକୀୟ ସମଙ୍ଗିବାହୁ ତ୍ରିଭୁଜ ।

ପ୍ରଥମ ଚିତ୍ରରେ ସମବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ :

(i) ସମବାହୁ ତ୍ରିଭୁଜ ପାଇବାକୁ ହେଲେ ଦୁଇଟି ବୃତ୍ତ ଛେଦବିନ୍ଦୁକୁ P ନାମରେ ନାମିତ କରାଯାଉ ।
(ii) AP, PB ଓ BA କୁ ଯୋଗ କର ।

(iii) Δ APB ଏକ ଆବଶ୍ୟକୀୟ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ।

ସମଦିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ :

(i) AB, BC ଓ CA କୁ ଯୋଗ କରାଯାଉ ।

(ii) Δ ABC ଏକ ଆବଶ୍ୟକୀୟ ସମଦ୍ଵିବାହୁ ତ୍ରିଭୁଜ ।

Question 3.
3 ସେ.ମି., 4 ସେ.ମି., 8 ସେ.ମି. ବାହୁର ଦୈର୍ଘ୍ୟ ନେଇ ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ କି ?
Solution:
ତ୍ରିଭୁଜର ଯେକୌଣସି ଦୁଇବାହୁର ସମଷ୍ଟି ତୃତୀୟ ବାହୁର ଦୈର୍ଘ୍ୟଠାରୁ ବୃହତ୍ତର । ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ । (3 + 4 = 7, ଯାହା 8 ଠାରୁ କ୍ଷୁଦ୍ରତର) ତେଣୁ ଏହା ସମ୍ଭବ ନୁହେଁ ।

Question 4.
2 ସେ.ମି., 3 ସେ.ମି., 6 ସେ.ମି. ବାହୁର ଦୈର୍ଘ୍ୟ ନେଇ ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ କି ?
Solution:
ତ୍ରିଭୁଜର ଯେକୌଣସି ଦୁଇବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି ତୃତୀୟ ବାହୁର ଦୈର୍ଘ୍ୟଠାରୁ ବୃହତ୍ତର । ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ । (2 + 3 = 5, ଯାହା 6 ଠାରୁ କ୍ଷୁଦ୍ରତର) ତେଣୁ ଏହା ସମ୍ଭବ ନୁହେଁ ।

Question 5.
ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ 10 ସେ.ମି., 15 ସେ.ମି. ଏବଂ 30 ସେ.ମି. ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ କରିବା ସମ୍ଭବପର କି ନୁହେଁ ପରୀକ୍ଷା କର ।
Solution:
ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ 10 ସେ.ମି., 15 ସେ.ମି. ଓ 30 ସେ.ମି.
ଏଠାରେ 10 + 15 < 30
ଏଠାରେ ତ୍ରିଭୁଜର ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ କ୍ଷୁଦ୍ରତ୍ତର ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

Question 6.
କୌଣସି ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ କରାଯାଇପାରିବ କି, ଯାହାର ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ 3 ସେ.ମି., 4 ସେ.ମି. ଏବଂ 5 ସେ.ମି. ? ଏହା ତୁମେ ଅଙ୍କନ ମାଧ୍ୟମରେ ପରୀକ୍ଷା କରି ଦେଖ ।
Solution:
ଏଥରେ ତ୍ରିଭୁଜର ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତର । ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ । ଏଠାରେ 3² + 4² = 5² ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ କଲେ ଏହା ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ ହେବ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ଆମେ ଅଙ୍କନ କରି ଜାଣିପାରିଲୁ ଯେ, 3 ସେ.ମି., 4 ସେ.ମି., 8 ସେ.ମି. ଏବଂ 2 ସେ.ମି., 3 ସେ.ମି., 6 ସେ.ମି.ର ଦୈର୍ଘ୍ୟ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ କରିବା ସମ୍ଭବ ନୁହେଁ । ଏହି ବିଷୟ ତ୍ରିଭୁଜ ଅଙ୍କନ ନକରି ତୁମେ ଜାଣିପାରିଛ କି ?
Solution:
ଯଦି ଯେକୌଣସି ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବଡ଼, ତେବେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।
ଏଠାରେ 3 + 4 = 7 । 8 ଠାରୁ 7 ବଡ଼ ନୁହେଁ । ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

Question 2.
ନିମ୍ନରେ ଦିଆଯାଇଥୁବା ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ଦୈର୍ଘ୍ୟଗୁଡ଼ିକ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ କି ନୁହେଁ, କହିପାରିବ କି ?
(a) 10 କି.ମି., 10 କି.ମି. ଏବଂ 25 କି.ମି.
(b) 5 ମି.ମି., 10 ମି.ମି. ଏବଂ 20 ମି.ମି.
(c) 12 ସେ.ମି., 20 ସେ.ମି. ଏବଂ 40 ସେ.ମି.
ତୁମେ ଅନୁଭବ କରିପାରୁଥିବେ ଯେ, ରଫ୍ ଚିତ୍ର ବ୍ୟବହାର କରି ସଳଖ ରାସ୍ତା ଓ ବୁଲାଣି ରାସ୍ତା ଓ ବୁଲାଣି ରାସ୍ତା ମାଧ୍ୟମରେ ପ୍ରତ୍ୟେକ ଦୈର୍ଘ୍ୟ ଓ ଅନ୍ୟ ଦୁଇଟି ଦୈର୍ଘ୍ୟର ଯୋଗଫଳ ସହିତ ସମ୍ପର୍କ ସ୍ଥାପନ କରିପାରିବା । ଏହିପରି ଭାବରେ ଆମକୁ ତିନୋଟି ପ୍ରକାରର ସମ୍ପର୍କ ସ୍ଥାପନ କରିବାକୁ ହେବ ।
Solution:
(a) 10 କି.ମି., 10 କି.ମି. ଏବଂ 25 କି.ମି. । ଏଠାରେ 10 + 10 < 25
ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ସାନ ହୋଇଥୁବାରୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

(b) 5 ମି.ମି., 10 ମି.ମି. ଏବଂ 20 ମି.ମି. । ଏଠାରେ 5 + 10 < 20
ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ସାନ ହୋଇଥିବାରୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

(c) 12 ସେ.ମି., 20 ସେ.ମି. ଏବଂ 40 ସେ.ମି.। ଏଠାରେ 12 + 20 < 40
ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁ ଠାରୁ ସାନ ହୋଇଥୁବାରୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

Question 3.
ପୂର୍ବରୁ ଆଲୋଚିତ ପ୍ରତ୍ୟେକ ଦୈର୍ଘ୍ୟମାନଙ୍କର ସେଟ୍‌କୁ ନେଇ ଅତିକମ୍ରେ ଦୁଇଥର ତୁଳନା କଲେ ଜାଣିପାରିବା ଯେ ସଳଖ ଦୈର୍ଘ୍ୟ ଅନ୍ୟ ଦୁଇଟିର ଦୈର୍ଘ୍ୟର ଯୋଗଫଳଠାରୁ କମ୍ । (ଯଦି ନୁହେଁ, ପୁନର୍ବାର ଯାଞ୍ଚ କର) ଉଦାହରଣ : 10 ସେ.ମି., 15 ସେ.ମି. ଏବଂ 30 ସେ.ମି.
ଏଠାରେ ଦୁଇଟି ତୁଳନା ହୋଇପାରିବ : 10 < 15 + 30, 15 < 10 + 30
କିନ୍ତୁ ଏହା ତୃତୀୟ ବାହୁର ଦୈର୍ଘ୍ୟ ପାଇଁ ସମ୍ଭବ ନୁହେଁ, ଯେହେତୁ : 30 > 10 + 15

(a) ଏହା କ’ଣ ସବୁବେଳେ ସମ୍ଭବ ? ବାହୁମାନଙ୍କର ଯେକୌଣସି ଦୈର୍ଘ୍ୟପାଇଁ ଅତିକମ୍‌ରେ ଦୁଇଟି ତୁଳନା ସର୍ବଦା ସମ୍ଭବ କି, ଯେଉଁଠାରେ ସଳଖ ଦୈର୍ଘ୍ୟ ଅନ୍ୟ ଦୁଇଟିର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟିଠାରୁ କମ୍ ହେବ ?
Solution:
ନା, ସବୁବେଳେ ତିନୋଟି ଦୈର୍ଘ୍ୟପାଇଁ ଏହା ସମ୍ଭବ ନୁହେଁ ।
ତିନୋଟି ଦୈର୍ଘ୍ୟ 5 ସେ.ମି., 6 ସେ.ମି. ଏବଂ 8 ସେ.ମି. ନେଲେ
5 + 6 > 8, 6 + 8 > 5 ଓ 5 + 8 > 6

(b) ପୁଣିଥରେ, ଦିଆଯାଇଥିବା ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟର ସେମାନଙ୍କ ମଧ୍ୟରୁ କେଉଁ ବାହୁର ଦୈର୍ଘ୍ୟ ଅନ୍ୟ ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟିଠାରୁ ସାନ ହେବ, ଏହା ବିନା ହିସାବରେ କହିହେବ କି ? (ସୂଚନା : ସଳଖ ଦୈର୍ଘ୍ୟକୁ ଉକ୍ରମରେ ବିଚାରକୁ ନେବା ।)
Solution:
ହଁ, ଏହା ବିନା ହିସାବରେ କହିବା ସମ୍ଭବ ହେବ । ଯଦି ଆମେ ତିନୋଟି ବାହୁର ଦୈର୍ଘ୍ୟକୁ ଉର୍ଦ୍ଧ୍ବକ୍ରମରେ ସଜାବ, ଏହା ସହଜ ହୋଇଯିବ ।
(i) ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ଯଥା – a < b < c (କ୍ଷୁଦ୍ରତରରୁ ବୃହତ୍ତର) ଭାବରେ ସଜାଯାଉ ।
(ii) ତା’ପରେ ଦେଖ‌ିବାକୁ ପଡ଼ିବ ଯେ, ସବୁଠାରୁ ବୃହତ୍ତର ବାହୁ ଅନ୍ୟ (a + b) ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟିଠାରୁ ସାନ ।
(iii) ଯଦି c < a + b ହୁଏ, ତେବେ ସମସ୍ତ ତ୍ରିଭୁଜଟି ଅସମାନତା ଧର୍ମ ଆପେଆପେ ସର୍ତ୍ତ ପୂରଣ କରିବ ।
(iv) ଯଦି c ≥ a + b, ତେବେ ତ୍ରିଭୁଜଟି ଗଠନ ସମ୍ଭବ ହେବ ନାହିଁ ।

(c) ଦିଆଯାଇଥବା ତ୍ରିଭୁଜର ତିନୋଟି ବାହୁର ଦୈର୍ଘ୍ୟ ପାଇଁ ତ୍ରିଭୁଜଟି ଅଙ୍କନ ସମ୍ଭବ କି ନୁହେଁ ଜାଣିବା ପାଇଁ ଆମେ କ’ଣ ତୁଳନା କରିବା ଆବଶ୍ୟକ ? Solution:
ଯଦି ତିନୋଟି ବାହୁର ଦୈର୍ଘ୍ୟ ଦତ୍ତ ଥିଲେ ତ୍ରିଭୁଜଟି ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ, ଜାଣିବା ପାଇଁ ଆମକୁ ତ୍ରିଭୁଜର ଅସମାନତା ଧର୍ମ ବ୍ୟବହାର କରିବା । ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟ ଅନ୍ୟ ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟିଠାରୁ ନିଶ୍ଚିତ ସାନ ହେବ । ଏଥିପାଇଁ a, b ଓ c ବାହୁଗୁଡ଼ିକ ପାଇଁ ଆମେ ନିଶ୍ଚୟ ଯାଞ୍ଚ କରିବା :
(1) a + b > c
(2) a + c > b ଏବଂ (3) b + c > a

ଯେତେବେଳେ ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟ ଅନ୍ୟ ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟିଠାରୁ ସାନ, ତାହା ତ୍ରିଭୁଜର ଅସମାନତା ଧର୍ମକୁ ସୂଚାଏ ।

ଉଦାହରଣ : ଦୈର୍ଘ୍ୟ ସେଟ୍ 3, 4, 5 ତ୍ରିଭୁଜ ଅସମାନତାକୁ ସୂଚାଏ ।
କିନ୍ତୁ 10, 15, 30 ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସୂଚାଏ ନାହିଁ ।
ଆମେ ଦେଖୁଲୁ ଯେ, ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ ଯଥା- 10, 15, 30 ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସୂଚାଏ ନାହିଁ । ତେଣୁ ଏଗୁଡ଼ିକର ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବ ନାହିଁ ।

Question 4.
4 ସେ.ମି., 5 ସେ.ମି. ଏବଂ 8 ସେ.ମି. ଦୈର୍ଘ୍ୟବିଶିଷ୍ଟ ବାହୁମାନଙ୍କୁ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବପର କି ?
Solution:
ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ 4 ସେ.ମି., 5 ସେ.ମି. ଏବଂ 8 ସେ.ମି. ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସୂଚାଇଅଛି । ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।
8 < 4 + 5 = 9

ବୃତ୍ତର ଅଙ୍କନ କଳ୍ପନା କରିବା :
• ଆମେ ସବୁଠାରୁ ବଡ଼ବାହୁ AB (8 ସେ.ମି.)କୁ ଭୂମି ରୂପେ ନେଇ ବୃତ୍ତ ଅଙ୍କନ ଆରମ୍ଭ କରିବା । ଏହାପରେ ଆମେ ପର୍ଯ୍ୟାୟକ୍ରମେ 4 ସେ.ମି., 5 ସେ.ମି. ଦୈର୍ଘ୍ୟ ପାଇଁ ଏକ ବୃହତ୍ ଚାପ ଅଙ୍କନ କରିବା । ଆମେ ଚାପ ପରିବର୍ତ୍ତେ ସମ୍ପୂର୍ଣ୍ଣ ବୃତ୍ତଟିଏ ଅଙ୍କନ କରିବା । ‘A’କୁ କେନ୍ଦ୍ର ଭାବେ ନେଇ 4 ସେ.ମି. ବ୍ୟାସାର୍କର ଏକ ବୃତ୍ତ ଅଙ୍କନ କରିବା ।

• ବର୍ତ୍ତମାନ ‘B’ କୁ କେନ୍ଦ୍ରକରି 5 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ନେଇ ଏକ ବୃତ୍ତ ଅଙ୍କନ କରିବା ।

Question 1.
ଯେଉଁ ଜ୍ୟାମିତିକ ଚିତ୍ରଟି ମିଳିବ ତାହାର ଏକ ବିଶ୍ଳେଷଣ ଚିତ୍ରଟିଏ (ରଫ ଚିତ୍ର) ଅଙ୍କନ କରିପାରିବ କି ?
Solution:
ନିମ୍ନରେ ଦିଆଯାଇଥୁବା ଚିତ୍ରରେ AX = 4 ସେ.ମି. ଏବଂ AB = 8 ସେ.ମି. ତେଣୁ BX = ? ଏହି ଦୈର୍ଘ୍ୟଟିରେ ଆମେ କ’ଣ ଚିତ୍ରଟି ଅନୁମାନ କରିପାରିବା ? ଚିତ୍ର ଅନୁଯାୟୀ ବିନ୍ଦୁ X, A ଏବଂ B ଏକ ସରଳରେଖାରେ ଅଛନ୍ତି ।
AB = AX + BX ⇒ BX = AB – AX = (8 – 4) ସେ.ମି. = 4 ସେ.ମି. ।

• ଯେହେତୁ BX = 4 ସେ.ମି. ଏବଂ ‘B’କୁ କେନ୍ଦ୍ରକରି ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 5 ସେ.ମି. ଅଟେ । ତେଣୁ ଏହା ନିର୍ଦ୍ଦିଷ୍ଟ ଯେ ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ ଦୁଇଟି ବିନ୍ଦୁରେ ଛେଦ କରିବେ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନଲିଖତ ଦୈର୍ଘ୍ୟସେଟ୍‌ରୁ କେଉଁଗୁଡ଼ିକ ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବ । ତୁମର ଉତ୍ତରକୁ ବର୍ଣନା କର । ମନେରଖ ଯେ, ପ୍ରତ୍ୟେକ ସେଟ୍‌ରେ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ମାପର ଏକକ ସମାନ ।
(a) 2, 2, 5
(b) 3, 4, 6
(c) 2, 4, 8
(d) 5, 5, 8
(e) 10, 20, 25
(f) 10, 20, 35
(g) 24, 26, 28
Solution:
(a) ଏଠାରେ ବାହୁତ୍ରୟ 2, 2, 5,
∴ 2 + 2 = 4 < 5 ( ଦୁଇ ବାହୁର ସମଷ୍ଟି ତୃତୀୟ ବାହୁଠାରୁ କ୍ଷୁଦ୍ରତ୍ତର) ତେଣୁ 2, 2, 5 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବ ନାହିଁ ।

(b) ଏଠାରେ ବାହୁତ୍ରୟ 3, 4, 6,
∴ 3 + 4 = 7 > 6, 3 + 6 = 9 > 4, 4 + 6 = 10 > 3
ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତମ ।
ତେଣୁ 3, 4, 6 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବେ ।

(c) ଏଠାରେ ବାହୁତ୍ରୟ 2, 4, 8
∴ 2 + 4 = 6 < 8 ତେଣୁ 2, 4, 8 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବ ନାହିଁ ।

(d) ଏଠାରେ ବାହୁତ୍ରୟ 5, 5, 8
∴ 5 + 5 = 10 > 8, 5 + 8 = 13 > 5, 5 + 8 = 13 > 5
ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତମ ।
ତେଣୁ 5, 5, 8 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବେ ।

(e) ଏଠାରେ ବାହୁତ୍ରୟ 10, 20, 25
∴ 10 + 20 = 30 > 25, 10 + 25 = 35 > 20, 20 + 25 = 45 > 10
ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତମ ।
ତେଣୁ 10, 20, 25 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବେ ।

(f) ଏଠାରେ ବାହୁତ୍ରୟ 10, 20, 35
∴ 10 + 20 = 30 < 35 ତେଣୁ 10, 20, 35 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବ ନାହିଁ ।

(g) ଏଠାରେ ବାହୁତ୍ରୟ 24, 26, 28
∴ 24 + 26 = 50 > 28, 24 + 28 = 52 > 26, 26 + 28 = 54 > 24
ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତମ ।
ତେଣୁ 24, 26, 28 ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ହୋଇପାରିବେ ।
ଯେତେବେଳେ ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସୂଚାଏ (ଦୈର୍ଘ୍ୟମାନଙ୍କର ସେଟ୍‌ଟିରେ ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟ ଅନ୍ୟ ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟିଠାରୁ ସାନ) ସେତେବେଳେ ସେହି ଦୈର୍ଘ୍ୟଗୁଡ଼ିକ ଏକ ତ୍ରିଭୁଜର ବାହୁ ହୋଇପାରିବେ ।

Question 2.
ଆମେ କିପରି ଭାବେ ନିଶ୍ଚିତ ହେବା ଯେ, ଯେଉଁ ଦୈର୍ଘ୍ୟମାନଙ୍କର ସେଟ୍ ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସୂଚାଏ ସେତେବେଳେ ହିଁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ?
Solution:
ହଁ, ତିନୋଟି ବାହୁର ଦୈର୍ଘ୍ୟର ସେଟ୍ ତ୍ରିଭୁଜର ଅସମାନତା ଧର୍ମକୁ ସୂଚାଏ, ତେବେ ତ୍ରିଭୁଜଗୁଡ଼ିକ ସର୍ବଦା ସମ୍ଭବପର ହେବ ।

Question 3.
ଦୁଇଟି ବୃତ୍ତ ଅଙ୍କନ କରି ନିମ୍ନୋକ୍ତ ପରିସ୍ଥିତିଗୁଡ଼ିକୁ ଆଲୋଚନା କର ।
Solution:

Question 4.
ଉପରୋକ୍ତ ତିନୋଟି ସମ୍ଭାବ୍ୟ ପରିସ୍ଥିତି ମଧ୍ୟରୁ କେଉଁଟି ତ୍ରିଭୁଜ ଅଙ୍କନ ପାଇଁ ସହାୟକ ହେବ ?
Solution:
ଯେତେବେଳେ ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ ଅନ୍ତର୍ଦେଶରେ ଛେଦ କରିବେ (ପରିସ୍ଥିତି 3) ସେତେବେଳେ ନିଶ୍ଚିତ ଭାବେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।

Question 5.
ଉପରେ ଦିଆଯାଇଥିବା ସମ୍ଭାବ୍ୟଗୁଡ଼ିକରୁ ଆମେ ବଡ଼ ବାହୁ ଏବଂ ଅନ୍ୟ ସାନ ବାହୁଦ୍ଵୟ ମଧ୍ଯରେ ସମ୍ପର୍କ ସ୍ଥିର କରିବା ।
Solution:
ପରିସ୍ଥିତି – 1 : ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ ଗୋଟିଏ ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରନ୍ତି ।
ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ଵୟର ସମଷ୍ଟି = AB
କିମ୍ବା, କ୍ଷୁଦ୍ରତମ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି = ବୃହତ୍ତମ ବାହୁର ଦୈର୍ଘ୍ୟ ।

ପରିସ୍ଥିତି – 2 : ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ ଅନ୍ତର୍ଦେଶରେ ଛେଦ କରନ୍ତି ନାହିଁ ।
ଏଠାରେ ବ୍ୟାସାର୍ଦ୍ଧଦ୍ଵୟର ସମଷ୍ଟି < AB
କିମ୍ବା କ୍ଷୁଦ୍ରତମ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି < ବୃହତ୍ତମ ବାହୁର ଦୈର୍ଘ୍ୟ ।

ପରିସ୍ଥିତି – 3 : ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ ଅନ୍ତର୍ଦେଶରେ ଛେଦ କରନ୍ତି । AB ଗୋଟିଏ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଏବଂ ଅନ୍ୟବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧର କିଛି ଅଂଶକୁ ନେଇ ଗଠିତ । ବ୍ୟାସାର୍ଷଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି > AB
କିମ୍ବା, କ୍ଷୁଦ୍ରତମ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି > ବୃହତ୍ତମ ବାହୁର ଦୈର୍ଘ୍ୟ ।

Question 6.
ଆମେ କ’ଣ ଏହି ବିଶ୍ଳେଷଣକୁ ତ୍ରିଭୁଜ ପାଇଁ ବ୍ୟବହାର କରିପାରିବା କି ? (ଯେତେବେଳେ ବାହୁର ଦୈର୍ଘ୍ୟଗୁଡ଼ିକ ତ୍ରିଭୁଜର ଅସମାନତା ପାଇଁ ସିଦ୍ଧ ହୁଏ ।)
Solution:
ଯେତେବେଳେ ଦିଆଯାଇଥିବା ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସିଦ୍ଧ କରେ, ସେତେବେଳେ କ୍ଷୁଦ୍ରତମ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି ବୃହତ୍ତମ ବାହୁର ଦୈର୍ଘ୍ୟଠାରୁ ବୃହତ୍ତମର ହେବ । ଅର୍ଥାତ୍‌ ଏହା ପରିସ୍ଥିତି 3 କୁ ସୂଚାଏ, ଯେଉଁଠାରେ ବୃତ୍ତଦ୍ଵୟ ପରସ୍ପରକୁ ଅନ୍ତର୍ଦେଶରେ ଛେଦ କରନ୍ତି । ତେଣୁ ତ୍ରିଭୁଜ ସମ୍ଭବ ଅଟେ।

Question 7.
କେଉଁ ପ୍ରକାରର ବୃତ୍ତଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ପାଇଁ ତ୍ରିଭୁଜ ଅସମାନତା ସିଦ୍ଧ ହେବ ନାହିଁ । ଏହି ପ୍ରକାରର 3ଟି ଉଦାହରଣ ଦିଅ ଯେଉଁଠାରେ ଦୈର୍ଘ୍ୟର ସେଗୁଡ଼ିକ ପାଇଁ ବୃତ୍ତଦ୍ଵୟ
(a) ପରସ୍ପରକୁ ଗୋଟିଏ ବିନ୍ଦୁରେ ସ୍ପର୍ଶ କରିବେ
(b) ପରସ୍ପରକୁ ଛେଦ କରିବେ ନାହିଁ ।
Solution:
(a) ଯେତେବେଳେ କ୍ଷୁଦ୍ରତମ ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି = ବୃହତ୍ତମ ବାହୁର ଦୈର୍ଘ୍ୟ ହେବ ।
ଉଦାହରଣ : (3, 4, 7); (5, 5, 10); (6, 8, 14)

(b) ପରସ୍ପରକୁ ଛେଦ କରିବେ ନାହିଁ : ଯେତେବେଳେ କ୍ଷୁଦ୍ରତମ ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି < ବୃହତ୍ତମ ବାହୁର ଦୈର୍ଘ୍ୟ ହେବ । ଉଦାହରଣ : (2, 3, 7); (4, 6, 11); (5, 7, 15)

Question 8.
ତ୍ରିଭୁଜଟିଏ ସମ୍ଭବପର ହେବାପାଇଁ ଏକ ସମ୍ପୂର୍ଣ୍ଣ ପ୍ରକ୍ରିୟା ନିର୍ଣ୍ଣୟ କର ।
Solution:
ଦିଆଯାଇଥିବା ତିନୋଟି ବାହୁର ଦୈର୍ଘ୍ୟ ମଧ୍ୟରୁ ଦୁଇଟି କ୍ଷୁଦ୍ରତମ ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି ଯଦି ବୃହତ୍ତର ବାହୁର ଦୈର୍ଘ୍ୟଠାରୁ ଅଧିକ ହୁଏ, ତେବେ ତ୍ରିଭୁଜଟି ସମ୍ଭବପର ହେବ ।
• ସିଦ୍ଧାନ୍ତ : ଯଦି କୌଣସି ଏକ ସେଟ୍‌ର ତିନୋଟି ଦୈର୍ଘ୍ୟ ତ୍ରିଭୁଜର ଅସମାନତା ପାଇଁ ସିଦ୍ଧ ହୁଏ, ତେବେ ତ୍ରିଭୁଜଟି ସମ୍ଭବପର ହେବ ଏବଂ ସେହି ଦୈର୍ଘ୍ୟଗୁଡ଼ିକ ତ୍ରିଭୁଜର ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ହେବ । ଯଦି ଏହି ଦୈର୍ଘ୍ୟଗୁଡ଼ିକ ତ୍ରିଭୁଜର ଅସମାନତାକୁ ସିଦ୍ଧ ନକରେ, ତେବେ ତାହା ତ୍ରିଭୁଜର ବାହୁ ହେବେ ନାହିଁ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ଦୈର୍ଘ୍ୟଗୁଡ଼ିକର ପ୍ରତ୍ୟେକ ସେଟ୍‌ରେ ତ୍ରିଭୂଜ ସମ୍ଭବ ହେବ କି ? ପରୀକ୍ଷା କର ।
(a) 1, 100, 100
(b) 3, 6, 9
(e) 1, 1, 5
(d) 5, 10, 12 ଆମେ ଜାଣିଛୁ ଯେ, ତ୍ରିଭୁଜର ଯେକୌଣସି ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତର ହେଲେ, ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ । (a) ଏଠାରେ 1, 100, 100 1 + 100 = 101 > 100, 1 + 100 = 101 > 100, 100 + 100 = 200 > 1
ଯେହେତୁ ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତର, ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।

(b) ଏଠାରେ 3, 6, 9
3 + 6 = 9 = 9, 6 + 9 = 15 > 3, 3 + 9 = 12 > 6
3 + 6 = 9 ଠାରୁ 9 ବଡ଼ ନୁହେଁ, ସମାନ ।
∴ ତ୍ରିଭୁଜ ସୂତ୍ର ଅନୁସାରେ ଏଠାରେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

(c) ଏଠାରେ 1, 1, 5
1 + 1 = 2 < 5
∴ ତ୍ରିଭୁଜ ସୂତ୍ର ଅନୁସାରେ ଏଠାରେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

(d) ଏଠାରେ 5, 10, 12 5 + 10 = 15 > 12, 5 + 12 = 17 > 10, 10 + 12 = 22 > 5
ଯେହେତୁ ଦୁଇ ବାହୁର ଦୈର୍ଘ୍ୟ ତୃତୀୟ ବାହୁଠାରୁ ବୃହତ୍ତର, ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।

Question 2.
50, 50, 50 କୁ ନେଇ ସମବାହୁ ତ୍ରିଭୁଜ ସମ୍ଭବ ହେବ କି ? ସାଧାରଣତଃ, ଯେକୌଣସି ବାହୁର ଦୈର୍ଘ୍ୟ ପାଇଁ ସମବାହୁ ତ୍ରିଭୁଜ ସମ୍ଭବ ହେବ କି ? ତାହା ବର୍ଣ୍ଣନା କର ।
Solution:
ସମବାହୁ ତ୍ରିଭୁଜର ସମସ୍ତ ବାହୁଗୁଡ଼ିକ ସମାନ ।
ଏଠାରେ ବାହୁଗୁଡ଼ିକ 50, 50, 50
50 + 50 = 100 > 50
ତ୍ରିଭୁଜ ଅଙ୍କନ ସୂତ୍ର ଅନୁସାରେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।

Question 3.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ମୂଲ୍ୟ ସହ ତୃତୀୟ ବାହୁର ସମ୍ଭାବ୍ୟ ଅତି କମ୍‌ରେ 5ଟି ମୂଲ୍ୟ ନିର୍ଦ୍ଧାରଣ କର, ଯେପରିକି ଏହା ତ୍ରିଭୁଜର ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ହେବ । (ଦଶମିକ ମୂଲ୍ୟ ମଧ୍ଯ ନିର୍ଦ୍ଧାରଣ କରାଯାଇପାରେ)
(a) 1, 100
(b) 5, 5
(c) 3, 7
Solution:
(a) ଆମେ ଜାଣିଛେ ତ୍ରିଭୁଜ ଅଙ୍କନ ବେଳେ ତୃତୀୟ ବାହୁଟି ଦୁଇଟି ବାହୁର ଦୈର୍ଘ୍ୟର ସମଷ୍ଟି ଠାରୁ ସାନ ହେବା ଦରକାର । ତୃତୀୟ ବାହୁଟି ଦୁଇଟି ବାହୁର ବିୟୋଗଫଳଠାରୁ ବୃହତ୍ତର ହେବା ଦରକାର ।
ତେଣୁ ସମ୍ଭାବ୍ୟ ତୃତୀୟ ବାହୁଟି = 99.1, 99.2, 99.3, 99.4, 99.5, 99.6, ….. 100.9

(b) ତୃତୀୟ ବାହୁଟି 0 (5 – 5) ଠାରୁ ବଡ଼ ଓ 10 (5 + 5) ଠାରୁ ସାନ ହେବା ଦରକାର ।
ତେଣୁ ସମ୍ଭାବ୍ୟ ତୃତୀୟ ବାହୁଟି = 1, 2, 3, 4, 5, 6, 7, 8, 9

(c) ତୃତୀୟ ବାହୁଟି 4 (7 – 3) ଠାରୁ ବଡ଼ ଓ 10 (3 + 7) ଠାରୁ ସାନ ହେବା ଦରକାର । ତେଣୁ ସମ୍ଭାବ୍ୟ ତୃତୀୟ ବାହୁଟି = 5, 6, 7, 8, 9

7.3 ତ୍ରିଭୁଜର କେତେକ ବାହୁ ଓ କୋଣ ଦତ୍ତ ଥିଲେ, ତ୍ରିଭୁଜ ଅଙ୍କନ :

Page No. (163 to 169)

ଦୁଇଟି ବାହୁ ଓ ସେମାନଙ୍କ ଅନ୍ତର୍ଗତ କୋଣ :
ଦୁଇଟି ବାହୁ ଓ ସେମାନଙ୍କ ଅନ୍ତର୍ଗତ କୋଣକୁ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ ।
(i) କମ୍ପାସ ବ୍ୟବହାର କରି ଗୋଟିଏ ବାହୁର ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କରାଯାଉ ।
(ii) ଏହି ବାହୁର ଗୋଟିଏ ପାର୍ଶ୍ଵ ବିନ୍ଦୁରେ 60 କୋଣ ଅଙ୍କନ କରାଯାଉ ।
(iii) ସେହି ବିନ୍ଦୁରୁ ଅନ୍ୟ ବାହୁର ଦୈର୍ଘ୍ୟ ନେଇ ଚାପ ଅଙ୍କନ କରାଯାଉ ।
(iv) ସମସ୍ତ ବିନ୍ଦୁଗୁଡ଼ିକୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ମିଳିବ ।

Question 1.
ତ୍ରିଭୁଜ ABC ଅଙ୍କନ କର ଯାହାର AB = 5 ସେ.ମି., AC = 4 ସେ.ମି., ∠A = 45°



Solution:
ଦିଆଯାଇଥିବା ବାହୁଗୁଡ଼ିକ ମଧ୍ୟରୁ AB କୁ ତ୍ରିଭୁଜର ଭୂମିରୂପେ ନିଅ ।

ସୋପାନ – 1 : କମ୍ପାସ ସାହାଯ୍ୟରେ ଗୋଟିଏ ବାହୁ AB = 5 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କରାଯାଉ ।
ସୋପାନ – 2 : ପ୍ରୋଟ୍ରାକ୍ଟରରେ ∠A ଠାରେ 45 କୋଣ ଅଙ୍କନ କରାଯାଉ ।
ସୋପାନ – 3 : A ବିନ୍ଦୁକୁ କେନ୍ଦ୍ର କରି 4ସେ.ମି. ଦୈର୍ଘ୍ୟର ଏକ ଚାପ କାଟି ଏହାର ନାମ C ଦିଆଯାଉ
ସୋପାନ – 4 : BC କୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ABC ମିଳିବ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ମାପକୁ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯେଉଁଠାରେ କୋଣଟି ବାହୁଦ୍ୱୟର ଅନ୍ତର୍ଗତ ଅଟେ ।
(a) 3 ସେ.ମି., 75°, 7 ସେ.ମି.
(b) 6 ସେ.ମି., 25°, 3 ସେ.ମି.
(c) 3 ସେ.ମି., 120°, 8 ସେ.ମି.
Solution:
(a) 3 ସେ.ମି., 75°, 7 ସେ.ମି.

(i) ଗୋଟିଏ ବାହୁ AB = 3 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କର ।
(ii) ∠A = 75° ଅଙ୍କନ କର ଏବଂ କୋଣର ଅନ୍ୟ ବାହୁଟିକୁ ଅଙ୍କନ ପାଇଁ B ବିନ୍ଦୁକୁ କେନ୍ଦ୍ରକରି 7 ସେ.ମି. ଦୈର୍ଘ୍ୟର ଏକ ଚାପ ଅଙ୍କନ କର, ଯାହା C ବିନ୍ଦୁରେ ଛେଦ କରିବ ।
(iii) ଅନ୍ୟ ବାହୁ AB କୁ ଛାଡ଼ି ୯ ବିନ୍ଦୁ ଚିହ୍ନଟ କର ଯେପରି AC = 7 ସେ.ମି. ହେବ ।
(iv) BC କୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ମିଳିବ ।

(b) 6 ସେ.ମି., 25°, 3 ସେ.ମି.

(i) ଗୋଟିଏ ବାହୁ AB = 6 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କରାଯାଉ ।
(ii) ∠A = 250 ଅଙ୍କନ କର ଏବଂ କୋଣର ଅନ୍ୟ ବାହୁଟିକୁ ଅଙ୍କନ କର ।
(iii) ଅନ୍ୟ ବାହୁ AB କୁ ଛାଡ଼ି C ବିନ୍ଦୁ ଚିହ୍ନଟ କର ଯେପରି AC = 3 ସେ.ମି. ହେବ ।
(iv) BC କୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ମିଳିବ ।

(c) 3 ସେ.ମି., 120°, 8 ସେ.ମି.

(i) ଗୋଟିଏ ବାହୁ AB = 8 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କର ।
(ii) ∠A = 120° ଅଙ୍କନ କର ଏବଂ କୋଣର ଅନ୍ୟ ବାହୁଟିକୁ ଅଙ୍କନ କର ।
(iii) ଅନ୍ୟ ବାହୁ AB କୁ ଛାଡ଼ି C ବିନ୍ଦୁ ଚିହ୍ନଟ କର ଯେପରି AC = 3 ସେ.ମି. ହେବ ।
(iv) BC କୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ମିଳିବ ।

Question 2.
ଆମେ ଜାଣୁ ଯେ ବାହୁମାନଙ୍କର ଯେକୌଣସି ଦୈର୍ଘ୍ୟପାଇଁ ତ୍ରିଭୂଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ । ସେହିପରି କିଛି ମାପ ଅଛି କି ଯାହା ଦୁଇବାହୁ ଏବଂ ଏହାର ଅନ୍ତର୍ଗତ କୋଣକୁ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ? ତ୍ରିଭୁଜ ଅଙ୍କନ ସମୟରେ ତୁମେ ଯାହା ଲକ୍ଷ୍ୟ କଲ ତାହାକୁ ଆଧାରିତ କରି ଦର୍ଶାଅ ।
Solution:
ଯେତେବେଳେ ଅନ୍ତର୍ଗତ କୋଣର ପରିମାଣ 0° ବା 180° ଥ‌ିବ, ସେହି ସମୟରେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବପର ନୁହେଁ ।

ଦୁଇଟି କୋଣ ଓ ଏହାର ଅନ୍ତର୍ଗତ ବାହୁ :

ଏଠାରେ ଦୁଇଟି କୋଣ ଓ ଗୋଟିଏ ବାହୁର ମାପ ଦିଆଯାଇଛି । ସେହି ବାହୁଟିକୁ ଆମ କୋଣଦ୍ଵୟର ଅନ୍ତର୍ଗତ ବାହୁ କହିଥାଉ । ଯଥା –

Question 3.
ABC ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯାହାର AB = 5 ସେ.ମି., ∠A = 45°, ∠B = 80°

Solution:
ସୋପାନ – 1 : ଗୋଟିଏ ବାହୁ AB = 5 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କରାଯାଉ ।
ସୋପାନ – 2 : ∠A = 45° ଓ ∠B = 80° ପରିମାଣ ନେଇ A ଓ B ବିନ୍ଦୁରେ କୋଣ ଅଙ୍କନ କରାଯାଉ ।
ସୋପାନ – 3 : ଦୁଇଟି ରେଖାଖଣ୍ଡର ଛେଦବିନ୍ଦୁଟି ତୃତୀୟ ଶୀର୍ଷ ବିନ୍ଦୁ C ହେବ ।
ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ଅଟେ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନଲିଖୂତ ମାପକୁ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ।
(a) 75°, 5 ସେ.ମି., 75°
(b) 25°, 3 ସେ.ମି., 60°
(c) 125°, 6 ସେ.ମି., 30°
Solution:
(a) 75°, 5 ସେ.ମି., 75°

(i) ଗୋଟିଏ ବାହୁ AB = 5 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କରାଯାଉ ।
(ii) ∠A = 75° ଓ ∠B = 75° ପରିମାଣ ନେଇ କୋଣ ଅଙ୍କନ କରାଯାଉ ।
(iii) ଦୁଇଟି ରେଖାଖଣ୍ଡର ଛେଦବିନ୍ଦୁଟି ତୃତୀୟ ଶୀର୍ଷ ବିନ୍ଦୁ C ନିଅ ।
ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ଅଟେ ।

(b) 25° 3 ସେ.ମି., 60°

(i) ଗୋଟିଏ ବାହୁ AB = 3 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କର ।
(ii) ∠A = 25° ଓ ∠B = 60° ପରିମାଣ ନେଇ କୋଣ ଅଙ୍କନ କର ।
(iii) ଦୁଇଟି ରେଖାଖଣ୍ଡର ଛେଦବିନ୍ଦୁଟି ତୃତୀୟ ଶୀର୍ଷ ବିନ୍ଦୁ C ହେବ । ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ଅଟେ ।

(c) 125°, 6 ସେ.ମି., 30°

(i) ଗୋଟିଏ ବାହୁ AB = 6 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କର ।
(ii) ∠A = 125° ଓ ∠B = 30° ପରିମାଣ ନେଇ କୋଣ ଅଙ୍କନ କର ।
(iii) ଦୁଇଟି ରେଖାଖଣ୍ଡର ଛେଦବିଦୁଟି ତୃତୀୟ ଶୀର୍ଷ ବିନ୍ଦୁ C ହେବ ।
ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ଅଟେ ।

ତ୍ରିଭୁଜଗୁଡ଼ିକର ଅଙ୍କନ ସବୁବେଳେ ସମ୍ଭବ କି ? :

Question 1.
ବିଭିନ୍ନ ପ୍ରକାରର ଦୁଇଟି କୋଣ ଓ ସଲଗ୍ନ ବାହୁ ନେଇ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ କି ? ବର୍ଣ୍ଣନା କର ।
Solution:
ଯଦି ତ୍ରିଭୁଜର ଦୁଇକୋଣର ସମଷ୍ଟି 180° କିମ୍ବା 180° ରୁ ଅଧିକ ହେଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ନାହିଁ । କାରଣ ଦୁଇଟି ରେଖା ପରସ୍ପରକୁ ଛେଦ କରିବେ ନାହିଁ ।

Question 2.
ଏମିତି କିଛି ଉଦାହରଣ ଦିଅ ଯାହାକି ଦୁଇଟି କୋଣର ପରିମାଣ ଓ ଏହାର ଅନ୍ତର୍ଗତ ବାହୁନେଇ ତ୍ରିଭୁଜ ସମ୍ଭବପର ହେବ ନାହିଁ ।
Solution:
ଉଦାହରଣ :
(a) ଯଦି ଦୁଇଟି କୋଣର ପରିମାଣ 90° ଲେଖାଏଁ ହେବ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବପର ହେବ ନାହିଁ ।
(b) ଯଦି ଗୋଟିଏ କୋଣର ପରିମାଣ 100° ଓ ଅନ୍ୟ କୋଣଟିର ପରିମାଣ 90° ହେବ ତ୍ରିଭୁଜ ଅଙ୍କନ
ସମ୍ଭବପର ହେବ ନାହିଁ ।
ଆମେ ସେହି ପ୍ରକାରର ପରିସ୍ଥିତିକୁ କଳ୍ପନା କରିବା । ଭୂମିଟିଏ ଅଙ୍କନ କଲାପରେ ଅନ୍ୟ ପାର୍ଶ୍ଵ ଗୁଡ଼ିକ ଏପରି ଭାବରେ ଅଙ୍କନ କରିବା ଯେପରି ସେମାନେ ପରସ୍ପରକୁ ଛେଦ କରିବେ ନାହିଁ । ସେହିପରି କେତେକ ଉଦାହରଣ ନିମ୍ନରେ ଦିଆଗଲା ।

ଯଦି କୋଣଦ୍ଵୟ 90° ରୁ ଅଧିକ କିମ୍ବା 90° ସହ ସମାନ ହୁଏ, ତେବେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ । ବର୍ତ୍ତମାନ ଭୂମି ଉପରିସ୍ଥ ବାହୁର କୌଣସି ଏକ କୋଣ ସୂକ୍ଷ୍ମକୋଣ ହେଉ । ମନେକର 40 । ଅନ୍ୟ କୋଣଟି ସମ୍ଭାବ୍ୟ କେତେ ହେବା ଦରକାର ଯାହାଦ୍ଵାରା ବାହୁଦ୍ୱୟ ପରସ୍ପରକୁ ଛେଦ କରିବେ ନାହିଁ ।

ଏହା ସ୍ପଷ୍ଟ ଯେ, B ଠାରୁ ରେଖାଟି ଯଥେଷ୍ଟ ଡାହାଣପଟକୁ ଅଙ୍କନ ହେଲେ, AC ରେଖା ଛେଦ କରିବ ନାହିଁ ।

ଉପରୋକ୍ତ ଚିତ୍ରରୁ ଏହା ସ୍ପଷ୍ଟ ଯେ ଯେଉଁରେଖା ସର୍ବନିମ୍ନ କୋଣ B ଠାରେ ସୃଷ୍ଟି କରୁଛି ତାହା ‘?’ ର ସମାନ୍ତରାଳ ହେବା ଆବଶ୍ୟକ । ମନେକର ସେହି ସମାନ୍ତରାଳ ରେଖାଟି ‘m’ ।

Question 3.
∠B ର ପ୍ରକୃତ ମୂଲ୍ୟ କୁହାଯାଇପାରିବ କି ?
Solution:
ଆମେ ଜାଣୁ ଯେ ଦୁଇଟି ସରଳରେଖାକୁ ଗୋଟିଏ ଛେଦକ ଛେଦକଲେ ଛେଦକର ଏକ ପାର୍ଶ୍ୱସ୍ଥ ଅନ୍ତସ୍ଥ କୌଣଦ୍ଵୟର ସମଷ୍ଟି 180° । ତେଣୁ ∠B = 140° ।

Question 4.
∠Bର କେଉଁ ମୂଲ୍ୟ ପାଇଁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବପର ନୁହେଁ । AB ରେଖାଖଣ୍ଡ ଏଠାରେ କୌଣସି ଭୂମିକା ଗ୍ରହଣ କରେ କି ?
Solution:
ଉପରୋକ୍ତ ଆଲୋଚନାରୁ ଆମେ ଜାଣିପାରିଲୁ ଯେ, ତ୍ରିଭୁଜ ଅଙ୍କନ ପାଇଁ AB କୌଣସି ଭୂମିକା ଗ୍ରହଣ କରେ ନାହିଁ । ଆମେ କହିପାରିବା ଯେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବପର ନୁହେଁ ଯେତେବେଳେ ∠B ପରିମାଣ 140 ସହ ସମାନ କିମ୍ବା ତା’ଠାରୁ ଅଧିକ ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନଲିଖୁତ ପ୍ରତ୍ୟେକ କୋଣପାଇଁ ଅନ୍ୟ ଏକ କୋଣ ନିର୍ଣ୍ଣୟ କର, ଯାହା ପାଇଁ ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ (i) ସମ୍ଭବ, (ii) ସମ୍ଭବ ନୁହେଁ । ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ଅତିକମ୍‌ରେ ଦୁଇଟି କୋଣ ନିଶ୍ଚୟ କର ।
(a) 30°
(b) 70°
(c) 54°
(d) 144°
Solution:
ସମ୍ଭବ ହେବ : 60° : 30° + 60° = 90° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
80° : 80° + 30° = 110° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180 ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
ସମ୍ଭବ ହେବ ନାହିଁ : 160° : 30° + 160° = 190°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180ରୁ ଅଧ୍ବକ, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ)
150° : 30° + 150° = 180°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ ।)

(b) ସମ୍ଭବ ହେବ : 50° : 50° + 70° = 120° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
80° : 80° + 70° = 150° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180 ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
ସମ୍ଭବ ହେବ ନାହିଁ : 120° : 120° + 70° = 190°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180ରୁ ଅଧ୍ବକ, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ)
110° : 110° + 70° = 180°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ ।)

(c) ସମ୍ଭବ ହେବ : 60° : 54° + 60° = 114° (ଯେହେତୁ ଦୁଇଟି କୌଣର ସମଷ୍ଟି 180° ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
80° : 54° + 80° = 134° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
ସମ୍ଭବ ହେବ ନାହିଁ : 130° : 130° + 54° = 184°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180ରୁ ଅଧ୍ବକ, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ) 126° : 126° + 54° = 180°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ ।)

(d) ସମ୍ଭବ ହେବ : 30° : 144° + 30° = 174° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
20° : 144° + 20° = 164° (ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍, ତୃତୀୟ କୋଣ ଯୋଗ କରିଲେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।)
ସମ୍ଭବ ହେବ ନାହିଁ : 50° : 144° + 50° = 194°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180°ରୁ ଅଧ‌ିକ, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନୁହେଁ)
40°: 40° + 144° = 184°
(ଯେହେତୁ ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180°, ତୃତୀୟ କୋଣ ଯୋଗ ହୋଇପାରିବ ନାହିଁ ।)

Question 2.
ନିମ୍ନଲିଖତ ଯୋଡ଼ିଗୁଡ଼ିକ ମଧ୍ୟରୁ କେଉଁଟି ତ୍ରିଭୁଜର କୋଣ ହୋଇପାରିବ ଏବଂ କେଉଁଗୁଡ଼ିକ ନୁହେଁ ?
(a) 35°, 150°
(b) 70°, 30°
(c) 90°, 85°
(d) 50°, 150°
Solution:
(a) ଦୁଇଟି କୋଣକୁ ଯୋଗ କଲେ = 35° + 150° = 185°
ଯେହେତୁ ଦୁଇଟି କୋଣର ପରିମାଣ 180° ରୁ ଅଧୁକ, ତେବେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

(b) ଦୁଇଟି କୋଣକୁ ଯୋଗ କଲେ = 70° + 30° = 100°
ଯେହେତୁ ଦୁଇଟି କୋଣର ପରିମାଣ 180° ରୁ କମ୍, ତେବେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ।

(c) ଦୁଇଟି କୋଣକୁ ଯୋଗ କଲେ = 90° + 85° = 175°
ଯେହେତୁ ଦୁଇଟି କୋଣର ପରିମାଣ 180 ରୁ କମ୍, ତେବେ ତ୍ରିଭୁଜ ଅଙ୍କନ, ସମ୍ଭବ ।

(d) ଦୁଇଟି କୋଣକୁ ଯୋଗ କଲେ = 50° + 150° = 200°
ଯେହେତୁ ଦୁଇଟି କୋଣର ପରିମାଣ 180° ରୁ ଅଧୂକ, ତେବେ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।

ଯେତେବେଳେ ` ପ୍ରଦତ୍ତ କୋଣଦ୍ଵୟର ସମଷ୍ଟି 180° ରୁ କମ୍ ସେତେବେଳେ ଏହି କୋଣଗୁଡ଼ିକ ସହ ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ । ଯଦି ଯୋଗଫଳ 180° ସହ ସମାନ କିମ୍ବା ବଡ଼ ହୁଏ ସେତେବେଳେ ଏହି କୋଣଗୁଡ଼ିକ ସହ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହୋଇ ନଥାଏ ।

Question 3.
ଦୁଇଟି କୋଣ ନେବା ଯଥା- 60° ଏବଂ 70°, ଯାହାର ସମଷ୍ଟି 180° ରୁ କମ୍ । ଏହାର ଅନ୍ତର୍ଗତ ବାହୁ 5 ସେ.ମି. ହେଉ । ତୃତୀୟ କୋଣର ମାପ କେତେ ହୋଇପାରିବ ? ଯଦି ବାହୁର ଦୈର୍ଘ୍ୟ ପରିବର୍ତ୍ତନ ହୁଏ ତେବେ ତୃତୀୟ କୋଣର ମାପ ପରିବର୍ତ୍ତିତ ହେବ କି ? ମନେକର ବାହୁଟି 7 ସେ.ମି. ହୁଏ, ତ୍ରିଭୁଜଟି ଅଙ୍କନ କର ।
Solution:

ଆମେ ଜାଣିଛେ ଯଦି ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍ ହେବ, ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ହେବ ।
ଏଠାରେ ଦୁଇଟି କୋଣ 60° ଓ 70°
ଦୁଇଟି କୋଣର ସମଷ୍ଟି = 60° + 70° = 130°
∴ ତୃତୀୟ କୋଣ = 180° – 130° = 50°
ଯଦି ବାହୁର ଦୈର୍ଘ୍ୟ ପରିବର୍ତ୍ତନ ହୁଏ ତେବେ ତୃତୀୟ କୋଣର ମାପରେ କୌଣସି ପରିବର୍ତ୍ତନ ହେବ ନାହିଁ ।
ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜଟି ନିମ୍ନ ପ୍ରକାରର ହେବ ।

Question 4.
ସାଧାରଣତଃ ଯଦି ଦୁଇଟି କୋଣର ପରିମାଣ ସ୍ଥିର ତେବେ ତୃତୀୟ କୋଣର ପରିମାଣ ଅନ୍ତର୍ଗତ ବାହୁର ମାପ ଉପରେ ନିର୍ଭର କରେ କି ? ବିଭିନ୍ନ ପ୍ରକାର କୋଣ ଦୁଇଟି ଏବଂ ଦୈର୍ଘ୍ୟ ସହ ଚେଷ୍ଟା କର ।
Solution:
ମାପଗୁଡ଼ିକରୁ ଜାଣିପାରିବା ଯେ ତୃତୀୟ କୋଣ ପାଇଁ ବାହୁର କୌଣସି ପ୍ରଭାବ ନାହିଁ (କିମ୍ବା ଅତି ନଗଣ୍ୟ) ।

Question 5.
ABC ତ୍ରିଭୁଜଟି ଅଙ୍କନ କର ଯାହାର ∠B = 50° ଏବଂ ∠C = 70° । ଆସ ଦେଖୁବା କିପରି କୌଣସି ଅଙ୍କନ ନକରି ଆମେ ∠A ର ମାପକୁ ଲେଖିବା ?
Solution:

∠A + ∠B + ∠C = 180° ⇒∠A + 50° + 70° = 180°
⇒∠A + 120° = 180° ⇒∠A = 180° – 120° = 60°
ଆମେ ଦେଖୁଲୁ ଯେ ସମାନ୍ତର ସରଳରେଖାର ଧାରଣା ମାଧ୍ୟମରେ ଏକ ତ୍ରିଭୁଜର ଯେକୌଣସି ଦୁଇଟି କୋଣର ସମଷ୍ଟି 180° ରୁ କମ୍ ନିର୍ଣ୍ଣୟ କରିବା ଉପଯୋଗୀ ଥିଲା । ତୃତୀୟ କୋଣ ∠BACର ପରିମାଣ ଖୋଜି ବାହାର କରିବାରେ ମଧ୍ୟ ସମାନ୍ତର ସରଳରେଖା ସହାୟକ ହେବ ।

ମନେକର ‘A’ ଶୀର୍ଷବିନ୍ଦୁ ମଧ୍ୟଦେଇ BC ସହ ସମାନ୍ତର XY ସରଳରେଖା ଅଙ୍କନ କରାଯାଇଛି ।

Question 6.
ଆମେ ଏଠାରେ ∠XAB ଏବଂ ∠YAC ସୃଷ୍ଟି ହେବାର ଦେଖିପାରୁଛୁ । ସେମାନଙ୍କର ପରିମାଣ କେତେ ?
Solution:
ଯେହେତୁ, XY ସରଳରେଖା BC ସହ ସମାନ୍ତର, ∠XAB = ∠B ଏବଂ ∠YAC = ∠C ।
କାରଣ ସେମାନେ AB ଓ AC ଛେଦକଦ୍ୱୟର ଏକାନ୍ତର କୋଣ ।
ତେଣୁ ∠XAB = 50° ଏବଂ ∠YAC = 70°
∠XAB, ∠YAC ଓ ∠BAC ଏକତ୍ର 180° କୋଣ ସୃଷ୍ଟି କରନ୍ତି ।
ଆମେ ଜାଣୁ, ∠XAB + ∠YAC + ∠BAC = 180°
⇒ 50° + 70° + ∠BAC = 180°
⇒ 120° + ∠BAC = 180°
⇒ ∠BAC = 180° – 120° = 60°
∴ ତ୍ରିଭୁଜର ∠A = 60°, ∠B = 50° ଓ ∠C = 70°

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ଦୁଇଟି କୋଣକୁ ନେଇ ତୃତୀୟ କୋଣ ନିର୍ଣ୍ଣୟ କର । (ସମାନ୍ତର ସରଳରେଖା ବ୍ୟବହାର କର)
(a) 36°, 72°
(b) 150°, 15°
(c) 90°, 30°
(d) 75°, 45°
Solution:
(a) ମନେକର ‘A’ ଶୀର୍ଷବିନ୍ଦୁ ମଧ୍ୟଦେଇ BC ସହ ସମାନ୍ତର XY ସରଳରେଖା ଅଙ୍କନ କରାଯାଇଛି । ଏହାଦ୍ଵାରା ଏଠାରେ ∠XAB ଏବଂ ∠YAC ସୃଷ୍ଟି ହୋଇଛି । ଯେହେତୁ, XY ସରଳରେଖା BC
ସହ ସମାନ୍ତର, ∠XAB = ∠B ଏବଂ ∠YAC = ∠C । କାରଣ ସେମାନେ AB ଓ AC ଛେଦକଦ୍ୱୟର ଏକାନ୍ତର କୋଣ ।
ତେଣୁ ∠XAB = 36° ଏବଂ ∠YAC = 72°
ଆମେ ଜାଣୁ, ∠XAB + ∠YAC + ∠BAC = 180°
⇒ 36° + 72° + ∠BAC = 180°
⇒ 108° + ∠BAC = 180°
⇒ ∠BAC = 180° – 108° = 72°
∴ ତ୍ରିଭୁଜର ∠A = 72°, ∠B = 36° ଓ ∠C = 72°

(b) ମନେକର ‘A’ ଶୀର୍ଷବିନ୍ଦୁ ମଧ୍ୟଦେଇ BC ସହ ସମାନ୍ତର XY ସରଳରେଖା ଅଙ୍କନ କରାଯାଇଛି । ଏହାଦ୍ଵାରା ∠XAB ଏବଂ ∠YAC ସୃଷ୍ଟି ହୋଇଛି । ଯେହେତୁ, XY ସରଳରେଖା BC ସହ ସମାନ୍ତର, ∠XAB = ∠B ଏବଂ ∠YAC = ∠C । କାରଣ ସେମାନେ AB ଓ AC ଛେଦକଦ୍ଵୟର ଏକାନ୍ତର କୋଣ ।
ତେଣୁ ∠XAB = 150°
ଏବଂ ∠YAC = 15°
ଆମେ ଜାଣୁ,
∠XAB + ∠YAC + ∠BAC = 180°
⇒ 150° + 15° + ∠BAC = 180°
⇒ 165° + ∠BAC = 180° ⇒ ∠BAC = 180° – 165° = 15°
∴ ତ୍ରିଭୁଜର ∠A = 15°, ∠B = 150° ଓ ∠C = 15°

(c) ମନେକର ‘A’ ଶୀର୍ଷବିନ୍ଦୁ ମଧ୍ୟଦେଇ BC ସହ ସମାନ୍ତର XY ସରଳରେଖା ଅଙ୍କନ କରାଯାଇଛି । ଏଠାରେ ∠XAB ଏବଂ ∠YAC ସୃଷ୍ଟି ହୋଇଛି । ଯେହେତୁ, XY ସରଳରେଖା BC ସହ ସମାନ୍ତର,
∠XAB = ∠B ଏବଂ ∠YAC = ∠C I କାରଣ ସେମାନେ AB ଓ AC ଛେଦକଦ୍ୱୟର ଏକାନ୍ତର କୋଣ ।
ତେଣୁ ∠XAB = 90° ଏବଂ ∠YAC = 30°
ଆମେ ଜାଣୁ, ∠XAB + ∠YAC + ∠BAC = 180°
= 90° + 30° + ∠BAC = 180°
⇒ 120° + ∠BAC = 180°
⇒ ∠BAC = 180° – 120° = 60°
∴ ତ୍ରିଭୁଜର ∠A = 60, ∠B = 90° ଓ ∠C = 30°

(d) ମନେକର ‘A’ ଶୀର୍ଷବିନ୍ଦୁ ମଧ୍ୟଦେଇ BC ସହ ସମାନ୍ତର XY ସରଳରେଖା ଅଙ୍କନ କରାଯାଇଛି ।
ଏଠାରେ ∠XAB ଏବଂ ∠YAC ସୃଷ୍ଟି ହୋଇଛି । ଯେହେତୁ, XY ସରଳରେଖା BC ସହ ସମାନ୍ତର, ∠XAB = ∠B ଏବଂ ∠YAC = ∠C । କାରଣ ସେମାନେ AB ଓ AC ଛେଦକଦ୍ଵୟର ଏକାନ୍ତର କୋଣ ।
ତେଣୁ ∠XAB = 75° ଏବଂ ∠YAC = 45°
ଆମେ ଜାଣୁ, ∠XAB + ∠YAC + ∠BAC = 1800
= 75° + 45 + ∠BAC = 180
⇒ 120 + ∠BAC = 180
⇒ BAC = 180 – 120 = 60
∴ ତ୍ରିଭୁଜର ∠A = 60°, ∠B = 75° ଓ ∠C = 45°

Question 2.
ପ୍ରତ୍ୟେକ କୋଣର ପରିମାଣ 70° ଥାଇ ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ କି ? ଯଦି ଦୁଇଟି କୋଣର ପରିମାଣ 70° ଲେଖାଏଁ ହୁଏ, ତେବେ ତୃତୀୟ କୋଣର ପରିମାଣ କେତେ ? ଯଦି ପ୍ରତ୍ୟେକ କୋଣର ପରିମାଣ ସମାନ ହୁଏ, ତେବେ ଏହି କୋଣର ପରିମାଣ କେତେ ହେବ ?
Solution:

ନା, ପ୍ରତ୍ୟେକ କୋଣର ପରିମାଣ 70° ହେଲେ, 70° + 70° + 70° = 210° ଯାହା 180° ରୁ ବଡ଼ ।
ତେଣୁ ତ୍ରିଭୁଜ ଅଙ୍କନ ସମ୍ଭବ ନୁହେଁ ।
ଯଦି ଦୁଇଟି କୋଣର ପରିମାଣ 70° ହୁଏ, ତେବେ ∠A = ∠B = 70°
ABC ତ୍ରିଭୁଜରେ ∠A + ∠B + ∠C = 180°
= 70° + 70° + ∠C = 180°
= 140° + ∠C = 180°
= ∠C = 180° – 140° = 40°
ଯଦି ପ୍ରତ୍ୟେକ କୋଣ ସମାନ ହୁଏ ତେବେ A ABC ରେ ∠A = ∠B = ∠C
⇒ ∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠A + ∠A = 180° ⇒ 3 × ∠A = 180° ⇒ ∠A = 60°
∴ ତ୍ରିଭୁଜର ପ୍ରତ୍ୟେକ କୋଣ 60

Question 3.
ଏଠାରେ ଏକ ତ୍ରିଭୁଜ ଅଛି ଯେଉଁଥ‌ିରେ ∠B = ∠C ଏବଂ ∠A = 50° ।
ତୁମେ ∠B ଏବଂ ∠Cର ପରିମାଣ ନିର୍ଣ୍ଣୟ କରିପାରିବ କି ?

Solution:
ଦଉଅଛି ∠B = ∠C, ∠A = 50°
ଆମେ ଜାଣୁ ତ୍ରିଭୁଜର ତିନିକୋଣର ସମଷ୍ଟି = 180°
∴ ∠A + ∠B + ∠C = 180°
⇒ 50° + ∠B + ∠B = 180° ⇒ 50° + 2∠B = 180°
⇒ 2∠B = 180° – 50° = 130° ⇒ ∠B = 130° ÷ 2 = 65°
∴ ତ୍ରିଭୁଜର ∠A = 50°, ∠B = 65° ଓ ∠C = 65°

ତ୍ରିଭୁଜର କୋଣ ପରିମାଣ ସମ୍ପୃକ୍ତ ଧର୍ମ :
Question 1.
ପ୍ରମାଣ କର ଯେ, ତ୍ରିଭୁଜର ତିନିକୋଣର ପରିମାଣର ସମଷ୍ଟି 180° ।

Solution:

ABC ଏକ ତ୍ରିଭୁଜ ନିଆଯାଉ । ‘A’ ଶୀର୍ଷବିନ୍ଦୁ ମଧ୍ୟଦେଇ BC ସହ ସମାନ୍ତର ସରଳରେଖାଟିଏ ଅଙ୍କନ କର ।
ଆମକୁ ∠A + ∠B + ∠C ନିର୍ଣ୍ଣୟ କରିବାକୁ ହେବ ।
ଆମେ ଜାଣୁ ଯେ ∠B = ∠XAB, ∠C = ∠YAC
ତେଣୁ ∠A + ∠B + ∠C
= ∠A + ∠XAB + ∠YAC = 180°
ତେଣୁ ଆମେ ପ୍ରମାଣ କଲୁ ଯେ ଯେକୌଣସି ତ୍ରିଭୁଜର ତିନିକୋଣର ପରିମାଣର ସମଷ୍ଟି 180° ।
ଏହା ତ୍ରିଭୁଜର କୋଣ ପରିମାଣ ସମ୍ପୃକ୍ତ ଧର୍ମ ଅଟେ ।

ଆରମ୍ଭରେ ତୃତୀୟ କୋଣ ଏବଂ ଅନ୍ୟ ଦୁଇଟି କୋଣ ମଧ୍ୟରେ ସମ୍ପର୍କ ସ୍ପଷ୍ଟ ନଥିଲା । ତ୍ରିଭୁଜର ଉପର ଶୀର୍ଷ ବିନ୍ଦୁ ମଧ୍ୟଦେଇ ଅଙ୍କନ କରାଯାଇଥିବା ସମାନ୍ତର ସରଳରେଖା ଧାରଣା ଯୋଗୁଁ କୋଣ ମଧ୍ଯରେ ସମ୍ପର୍କ ସ୍ପଷ୍ଟ ହୋଇପାରିଲା । ଏହି ଚତୁର ଚିନ୍ତାଧାରା ଗଣିତ ଇତିହାସରେ “The Elements” ନାମକ ଏକ ପ୍ରଭାବଶାଳୀ ପୁସ୍ତକରୁ ମିଳିପାରିବ । ଏହି ପୁସ୍ତକ ଗ୍ରୀକ୍ ଗଣିତଜ୍ଞ “Euclid”ଙ୍କୁ ସମର୍ପଣ ଅଟେ ।

କୋଣର ପରିମାଣ ସମ୍ପୃକ୍ତ ଧର୍ମ ପରୀକ୍ଷାର ଆଉ ଏକ ସରଳ ଉପାୟ ରହିଅଛି । ଏକ କାଗଜରେ ତ୍ରିକୋଣୀୟ ଭାଙ୍ଗ ନେଇ ନିମ୍ନରେ ଦର୍ଶାଯାଇଥିବା ଉପାୟ
ଅବଲମ୍ବନ କରାଯାଇପାରିବ । ଏହି କାଗଜ ଭାଙ୍ଗଦ୍ଵାରା ତୁମେ ଦେଖିପାରୁଛ କି ତ୍ରିଭୁଜର ତିନିକୋଣର ସମଷ୍ଟି 180 ° ।

ବହିଃସ୍ଥ କୋଣ :

କୌଣସି ଏକ ତ୍ରିଭୁଜର ଅନ୍ତଃସ୍ଥ କୋଣର ସନ୍ନିହିତ ପରିପୂରକ କୋଣକୁ ତ୍ରିଭୁଜର ବହିଃସ୍ଥ କୋଣ କୁହାଯାଏ । ଚିତ୍ରରେ ∠ACD ତ୍ରିଭୁଜର ଏକ ବହିଃସ୍ଥ କୋଣ ଅଟେ ।

Question 1.
∠ACD ନିଶ୍ଚୟ କର, ଯଦି ∠A = 50 ଏବଂ ∠B = 60° ।
Solution:
∠A + ∠B + ∠ACB = 180°
⇒ 50° + 60° + ∠ACB = 180° ⇒∠ACB = 180° – 110° = 70°
ତେଣୁ ∠ACD = 180° – 70° = 110°
(ଯେହେତୁ ∠ACB ଓ ∠ACD ପରସ୍ପର ସନ୍ନିହିତ ପରିପୂରକ କୋଣ ଅଟେ ।)

Question 2.
∠A ଓ ∠B ବିଭିନ୍ନ କୋଣର ମାପ ପାଇଁ ବହିଃସ୍ଥ କୋଣ ନିର୍ଣ୍ଣୟ କର । ବହିଃସ୍ଥ କୌଣ ଓ ଅନ୍ୟ ଦୁଇଟି ଅନ୍ତଃସ୍ଥ କୋଣ ମଧ୍ଯରେ କ’ଣ ସମ୍ପର୍କ ଲକ୍ଷ୍ୟ କରୁଛ କି ?
ସୂଚନା : କୋଣ ପରିମାଣ ସମ୍ପୃକ୍ତ ଧର୍ମରୁ ∠A + ∠B + ∠ACB = 180°
ପୁନଶ୍ଚ ∠ACD + ∠ACB = 180° (ସନ୍ନିହିତ ପରିପୂରକ)
Solution:
∠A + ∠B + ∠ACB = 180°
⇒ 50° + 60° + ∠ACB = 180° ⇒∠ACB = 180° – 110° = 70°
ତେଣୁ ∠ACD = 180° – 70° = 110° ଅର୍ଥାତ୍ ∠ACD = ∠A + ∠B
(ଯେହେତୁ ∠ACB ଓ ∠ACD ପରସ୍ପର ସନ୍ନିହିତ ପରିପୂରକ କୋଣ ଅଟେ ।)
∴ ଏକ ତ୍ରିଭୁଜର ବହିଃସ୍ଥ କୋଣର ପରିମାଣ ଏହାର ଅନ୍ତଃସ୍ଥ ଦୂରବର୍ତ୍ତୀ କୋଣର ସମଷ୍ଟି ସଙ୍ଗେ ସମାନ ।

7.4 ତ୍ରିଭୁଜର ଉଚ୍ଚତା ସମ୍ପର୍କିତ ଅଙ୍କନ :

Page No. 170 

Question 1.
ABC ତ୍ରିଭୁଜ ବିଚାରକୁ ନିଆଯାଉ । ଶୀର୍ଷବିନ୍ଦୁ ‘A’ ରୁ ବିପରୀତ ବାହୁ ‘BC’ ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ ବା ଉଚ୍ଚତାର ଦୈର୍ଘ୍ୟ କେତେ ଏବଂ ଏହାକୁ କିପରି ମାପିପାରିବା ?

Solution:
ABC ତ୍ରିଭୁଜର ଶୀର୍ଷ ବିନ୍ଦୁ ‘A’ ରୁ ବିପରୀତ ବାହୁ ‘BC’ ପ୍ରତି ଲମ୍ବ ଅଙ୍କନ କରାଯାଉ । ‘BC’ ବାହୁକୁ ବର୍ଜିତ କରି B ଦିଗରେ D ପର୍ଯ୍ୟନ୍ତ ନିଆଯାଉ । AD କୁ ମାପ କରି ଏହାର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କରାଯାଉ ।

କାଗଜ ଭାଙ୍ଗ ମାଧ୍ୟମରେ ଉଚ୍ଚତା :
ତ୍ରିଭୁଜାକୃତି ଖଣ୍ଡିଏ କାଗଜ ନିଅ । ଏହାର ଗୋଟିଏ ପାର୍ଶ୍ଵକୁ ଭୂମି ଭାବରେ ଗ୍ରହଣ କରି ଏହାକୁ ଏପରି ଭାବରେ ଭାଙ୍ଗି ଦିଅ ଯେପରିକି ଭାଙ୍ଗ ପଡ଼ିଥିବା ଦାଗଟି ଉପର ଶୀର୍ଷ ବିନ୍ଦୁରୁ ଭୂମି ପ୍ରତି ଲମ୍ବ ହେବ ।

ତ୍ରିଭୁଜର ଉଚ୍ଚତା ଅଙ୍କନ :

କୌଣସି ଏକ ତ୍ରିଭୁଜ ଅଙ୍କନ କର । BC କୁ ଭୂମିରୂପେ ନେଇ A, B, C ଶୀର୍ଷବିନ୍ଦୁ ନାମିତ କର । A ଠାରୁ BC ପ୍ରତି ଲମ୍ବ (ଉଚ୍ଚତା) ଅଙ୍କନ କରାଯାଉ । ଏହା ହିଁ ଏକ ତ୍ରିଭୁଜ ସମ୍ଭବ କି ଯାହାର ଗୋଟିଏ ବାହୁ ଏକ ଉଚ୍ଚତା ହେବ ? ସମ୍ଭବ ହେବ, ଯେତେବେଳେ ତ୍ରିଭୁଜର ଗୋଟିଏ କୋଣର ପରିମାଣ ସମକୋଣ ।

ଯେଉଁ ତ୍ରିଭୁଜର ଗୋଟିଏ କୋଣର ପରିମାଣ 90° (ସମକୋଣ), ସେହି ତ୍ରିଭୁଜକୁ ସମକୋଣୀ ତ୍ରିଭୁଜ କୁହାଯାଏ ।
A ଠାରୁ BC ପ୍ରତି ଲମ୍ବ (ଉଚ୍ଚତା) = AB

7.5 ତ୍ରିଭୁଜର ପ୍ରକାରଭେଦ :

Page No. (173 to 174)

ନିଜେ କରି ଦେଖ :

Question 1.
ABC ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯାହାର BC = 5 ସେ.ମି., AB = 6 ସେ.ମି., CA = 5 ସେ.ମି. ଏବଂ A ବିନ୍ଦୁଠାରୁ BC ପ୍ରତି ଲମ୍ବ ଅଙ୍କନ କର ।
Solution:

(i) BC = 5 ସେ.ମି.ଦୈର୍ଘ୍ୟର ଏକ ରେଖା ଅଙ୍କନ କର ।
(ii) କମ୍ପାସ ସାହାଯ୍ୟରେ B ବିନ୍ଦୁରୁ 6 ସେ.ମି. ଦୈର୍ଘ୍ୟର ଏକ ଚାପ ଅଙ୍କନ କର ।
(iii) C ବିନ୍ଦୁରୁ 5 ସେ.ମି. ଦୈର୍ଘ୍ୟ ନେଇ ଏକ ଚାପ ଅଙ୍କନ କର ।
ଦୁଇଟି ଚାପ ଯେଉଁ ବିନ୍ଦୁରେ ଛେଦ କରିବେ ତାହାକୁ A ଦିଅ ।
AB ଓ AC କୁ ଯୋଗ କର ।
(iv) ABC ଗୋଟିଏ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ।
(v) ସେୟାର ସାହାଯ୍ୟରେ A ବିନ୍ଦୁରୁ 90° ମାପ କରି ତାହାକୁ D ନାମ ଦିଅ । AD କୁ ଯୋଗ କର । AD, BC ପ୍ରତି ଲମ୍ବ ।

Question 2.
POR ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯାହାର QR = 4 ସେ.ମି., PQ = 7 ସେ.ମି, ∠Q = 140° । P ବିନ୍ଦୁଠାରୁ QR ପ୍ରତି ଲମ୍ବ ଅଙ୍କନ କର ।
Solution:

(i) ଗୋଟିଏ ବାହୁ QR = 4 ସେ.ମି. ଦୈର୍ଘ୍ୟ ଅଙ୍କନ କର ।
(ii) ∠Q = 140, ଅଙ୍କନ କର ଏବଂ କୋଣର ଅନ୍ୟ ବାହୁଟିକୁ ଅଙ୍କନ କର ।
(iii) ଅନ୍ୟ ବାହୁ QR କୁ ଛାଡ଼ି P ବିନ୍ଦୁ ଚିହ୍ନଟ କର ଯେପରି PQ = 7 ସେ.ମି. ହେବ ।
(iv) PR କୁ ଯୋଗକଲେ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ମିଳିବ ।
(v) ସେଟ୍‌ୟାର ସାହାଯ୍ୟରେ P ବିନ୍ଦୁରୁ 90° ମାପ କରି ତାହାକୁ S ନାମ ଦିଅ । PS କୁ ଯୋଗ କର । PS, QR ପ୍ରତି ଲମ୍ବ ।

Question 3.
ABC ସମକୋଣୀ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯାହାର ∠B = 90°, AC = 5 ସେ.ମି. । ଏହି ମାପନେଇ ଆଉ କେତୋଟି ତ୍ରିଭୁଜ ଅଙ୍କନ କରାଯାଇପାରିବ ?
(ସୂଚନା : ଧ୍ୟାନ ଦେବା ଯେ, ଅନ୍ୟ ମାପଗୁଡ଼ିକ ଯେକୌଣସି ମୂଲ୍ୟ ହୋଇପାରେ । AC କୁ ଭୂମି ଭାବରେ ନିଆଯାଉ । ∠A ଓ ∠C କେଉଁ ପରିମାଣ ହୋଇପାରିବ ଯାହାଫଳରେ ଅନ୍ୟକୋଣଟି 90° ହେବ ?)
Solution:

ଯେହେତୁ AC = 5 ସେ.ମି. ଓ ∠B = 90
ମନେକର ∠A = 40°
∴ ∠A + ∠B + ∠C = 180°
⇒ 40° + 90° + ∠C = 180°
⇒ ∠C = 180° – 130° ⇒ ∠C = 50°
(i) AC = 5 ସେ.ମି. ଦୈର୍ଘ୍ୟର ଏକ ରେଖା ଅଙ୍କନ କର ।
(ii) A ବିନ୍ଦୁରୁ 40° କୋଣ ଅଙ୍କନ କର ଓ B ବିନ୍ଦୁରୁ 50° କୋଣ ଅଙ୍କନ କର । ଦୁଇଟି ରେଖା ଯେଉଁ ବିନ୍ଦୁରେ ମିଳିତ ହେଉଛନ୍ତି ତାହାକୁ B ନାମ ଦିଅ ।
(iii) ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ହେବ ।
(iv) A ର ମୂଲ୍ୟ ଯେକୌଣସି ମୂଲ୍ୟ ନେଇପାରିବେ ଏବଂ ∠C ର ମୂଲ୍ୟ ପାଇପାରିବେ, ଯେପରି ଉଭୟ କୋଣର ଯୋଗଫଳ ସର୍ବଦା 90° ହେବ ।

Question 4.
ଅଙ୍କନ ମାଧ୍ୟମରେ ପରୀକ୍ଷଣ କରି ଦେଖାଅ ଯେ, ଏପରି ସମବାହୁ ତ୍ରିଭୁଜ ସମ୍ଭବପର କି ? ଯାହାର
(i) ଗୋଟିଏ କୋଣ ସମକୋଣ
(ii) ଗୋଟିଏ ସ୍ଥୂଳକୋଣଏବଂ ଏକ ସମଦିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ କର ଯାହାର
(i) ଗୋଟିଏ କୋଣ ସମକୋଣ
(ii) ଗୋଟିଏ ସ୍ଥୂଳକୋଣ ।
Solution:
(i) ସମବାହୁ ତ୍ରିଭୁଜର ତିନି କୋଣର ପରିମାଣ ସମାନ । ଯଦି ଏହାର ଗୋଟିଏ କୋଣ ସମକୋଣ ଅର୍ଥାତ୍ 90°, ତେବେ ଅନ୍ୟ କୋଣ ଦୁଇଟି 90° ଲେଖାଏଁ ହେବ ।
ଅର୍ଥାତ୍ କୋଣତ୍ରୟର ସମଷ୍ଟି 90° + 90° + 90° = 270° ଯାହା 180° ଠାରୁ ବଡ଼ । ଏହା ଅସମ୍ଭବ ।
(ii) ସମବାହୁ ତ୍ରିଭୁଜର ଗୋଟିଏ କୋଣ ସ୍ଥୂଳକୋଣ ଅର୍ଥାତ୍ 90° ରୁ ବେଶୀ ହେଲେ ଅନ୍ୟ ସମସ୍ତ କୋଣ ମଧ୍ୟ 90° ରୁ ବେଶୀ ହେବ । ତେଣୁ ଏହା ଅସମ୍ଭବ ।

ସମକୋଣୀ ସମଦ୍ବିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ :

(i) BC = 5 ସେ.ମି. ଅଙ୍କନ କର ।
(ii) ∠A + ∠B + ∠C = 180° ⇒ ∠A + ∠B + ∠A = 180°
⇒ 2∠A + 90° = 180° = 2∠A = 180° – 90° = 90°
∴ ∠A = 45°
(iii) B ବିନ୍ଦୁରେ 90° ଓ C ବିନ୍ଦୁରେ 45° ଅଙ୍କନ କର ।
(iv) ଦୁଇଟି ରେଖା ଯେଉଁ ବିନ୍ଦୁରେ ମିଳିତ ହେବେ ତାହାକୁ A ନାମ ଦିଅ ।
(v) ବର୍ତ୍ତମାନ ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ।

ସ୍ଥଳକୋଣୀ ସମଦ୍ଵିବାହୁ ତ୍ରିଭୁଜ ଅଙ୍କନ :

(i) BC = 6 ସେ.ମି. ଅଙ୍କନ କର ।
(ii) ମନେକର ∠B = 120°
(iii) ଯେହେତୁ ଏହା ସମଙ୍ଗିବାହୁ ତ୍ରିଭୁଜ, ∠A = ∠C
⇒ ∠A + ∠B + ∠C = 180° ⇒ ∠A + 120 + ∠A = 180°
⇒ 2∠A + 120° = 180° ⇒ 2∠A = 180° – 120° = 60°
∴ ∠A = 30° = ∠C
(iv) B ବିନ୍ଦୁରୁ 120° ଓ ୯ ବିନ୍ଦୁରୁ 30° ଅଙ୍କନ କର । ଯେଉଁଠି ସେମାନେ ମିଳିତ ହେବେ ତାହାର ନାମ A ଦିଅ ।
(v) ABC ଏକ ଆବଶ୍ୟକୀୟ ତ୍ରିଭୁଜ ।

Students can include Class 7 Curiosity Extra Questions and Class 7 Science Chapter 10 Life Processes in Plants Extra Question Answer in their daily self-study routine.

Class 7 Science Chapter 10 Life Processes in Plants Extra Questions

Class 7 Science Chapter 10 Extra Questions on Life Processes in Plants

Life Processes in Plants Class 7 Very Short Question Answer

Question 1.
For the synthesis of glucose, what are the raw materials required by a plant?
Answer:
For synthesis of glucose plants require carbon dioxide and water along with the presence of sunlight and chlorophyll in leaves.

Question 2.
How do plants store the energy captured from sunlight during photosynthesis, and how is this stored energy released during respiration?
Answer:
The energy captured from sunlight during photosynthesis is stored in the form of glucose by the plant. During respiration, this glucose is broken down in the presence of oxygen to release energy, which is then used for various life processes.

Question 3.
Sunlight, chlorophyll, carbon dioxide, water and minerals are raw materials essential for photosynthesis. Do you know where they are available? Fill in the blanks with the appropriate raw materials.

(i) Available in the plant: …………
Answer:

(ii) Available in the soil: …………
Answer:
Water and minerals

(iii) Available in the air: …………
Answer:
Carbon dioxide

(iv) Available during day: …………
Answer:
Sunlight

Question 4.
Which part/parts of a plant are involved in respiration?
Answer:
The process of respiration is performed by all living parts of a plant, as energy is required for various metabolic activities. Growing parts like floral buds, seeds of plants perform more respiration compared to other parts. Hence, all plant parts – green or non-green carry out respiration.

Question 5.
Study the given equation and answer the questions that follow.

(i) What does the equation depict?
Answer:
The equation depicts the process of photosynthesis.

(ii) Where does it occur?
Answer:
It occurs in the leaves.

(iii) What happens to the products?
Answer:
Glucose is broken down to release energy through respiration, while oxygen released is breathed in by other living organisms.

(iv) What are the requirements for this process?
Answer:
Water, chlorophyll, carbon dioxide, and sunlight.

Question 6.
During an experiment, Aman observed that lime water in a test tube connected to a flask containing germinating bean seeds turned milky. Explain his observation. Would the same result occur with boiled beans?
Answer:

• Germinating seeds were actively respiring and releasing carbon dioxide, which turned the lime water milky.
• No, if boiled beans are used, the lime water will not turn milky because boiled seeds are dead and do not respire (no carbon dioxide is released).

Question 7.
A goat eats away all the leaves of a small plant (balsam). However, in a few days, new leaves could be seen sprouting in the plant again. How did the plant survive without leaves?
Answer:
Plants store the food they produce by photosynthesis in the stems and roots. So, when the goat eats all the leaves, the plant fulfils its requirement by utilising the food stored in stems and roots.

Question 8.
Radhika wants to find out whether a purple plant can make its own food. She keeps a purple plant in sunlight for four days. Which other type of coloured plant should Radhika keep as part of her study to compare with the purple plant?

Answer:
Radhika should also keep a green plant as part of her study to compare with the purple plant. This comparison would make the experiment more reliable and accurate.

Question 9.
Potato and ginger are both underground parts that store food. Where is the food prepared in these plants?
Answer:
Food in potato and ginger plants is prepared in the leaves through photosynthesis and then transported via phloem to the underground parts for storage.

Question 10.
Shiv while changing ink for his pen, dropped the ink refill in a small flower vase kept on his study table. He noticed red colour in the stem of the flower kept in the vase after a few hours. What is the reason behind his observation?
Answer:
The red ink moved upwards through the stem due to the presence of xylem, which is responsible for the transport of water and minerals in plants. The xylem draws the coloured liquid (in this case, the ink mixed with water) from the vase up through the stem to the flower.

Question 11.
In our body, blood is the medium that transports nutrients from one place to another. What are the counterparts of blood, which transport nutrients in plants?
Answer:
Xylem and phloem are the tissues in plants for the transport of water and nutrients (food), respectively.

Life Processes in Plants Class 7 Short Question Answer

Question 1.
The following flowchart was given by Ashu’s teacher in homework. Help Ashu complete the flowchart.

Answer:
(i) Carbon dioxide
(ii) Present in (green) leaves
(iii) Oxygen

Question 2.
Renu wanted to perform an experiment to show her brother Sunil the raw materials required for photosynthesis. She took two similar green plants- one of them she placed under shade and the other under sunlight.

(i) Which plant will show a positive result for the iodine test?
Answer:
The plant kept under sunlight will give a positive result for the iodine test.

(ii) Why will the plant you selected give a positive result for the iodine test, and the other not?
Answer:
Under sunlight, the green plant will perform photosynthesis, resulting in the production of starch. The other plant kept under shade will be unable to perform photosynthesis and produce starch without sunlight.

(iii) Will there be any differences in the appearance of both the potted plants? State your reason.
Answer:
Yes, the plant kept in sunlight will be healthy, green, with more number of leaves. The plant kept under shade will be weak, have paleyellowish leaves and show poor growth.

Question 3.
Name the two structures involved in the transportation of substances in plants, and compare them based on the following features:

Answer:
The two structures involved in the transportation of substances in plants are xylem and phloem.

Feature	A: Xylem	B: Phloem
Material transported	Transports water and minerals	Transports food
Direction of transport	Upward (from roots to leaves)	Both upward and downward

Question 4.
Sunita saw the picture given alongside in an encyclopedia and wondered what it is. Study the given diagram and answer the following questions:

(i) Which structure is represented by the diagram?
(ii) Where can you find these structures in a plant?
(iii) What is the role of this structure?
Answer:
(i) The diagram represents stomata.
(ii) Stomata can be found on leaves, especially on the lower leaf surface.
(iii) Stomata facilitates exchange of gases during photosynthesis and respiration.

Question 5.
A student uprooted a rose plant from the soil. Most of the root tips, with root hair were left behind in the soil. She planted it in a pot with new soil and watered it regularly. Will the plant grow or die? Give reason for your answer.
Answer:
The following possibilities exist:

• Without the root hairs, the roots will not be able to absorb water and nutrients and the plant may die.
• The rose plant may grow new roots and the plant will live.
• The rose plant may not be able to survive in a different type of soil.

Question 6.
Observe the figures carefully and answer the following questions.

(i) In which jar, will the amount of carbon dioxide be the highest and why?
(ii) In which jar, will the amount of carbon dioxide be the lowest and why?
Answer:
(i) Jar (C) will have the highest amount of carbon dioxide because the mice inside the jar are continuously respiring, using up oxygen and releasing carbon dioxide. Since there is no plant inside the jar to absorb the carbon dioxide, its amount keeps increasing and becomes maximum.
(ii) Jar (A) will have the lowest amount of carbon dioxide because there are no mice inside the jar. The plant continuously uses carbon dioxide for photosynthesis, reducing its amount, while releasing oxygen.

Question 7.
Water transportation is said to form a oneway traffic, while the transport of food forms a two-way traffic in plants. Justify.
Answer:
Water transport is one-way because xylem carries water and minerals from the roots upwards to the leaves and other parts of the plant. In contrast, food transport is two-way because phloem carries food from the leaves, where it is produced, to all the parts that require it, such as the roots, developing flowers, or fruits, allowing movement in both upward and downward directions.

Question 8.
Mention the importance of respiration.
Answer:
Respiration involves the break down of glucose (food) and the release of energy; required for the growth and repair of various plant parts. Supporting growth and other life processes like preparing food, transportation, etc., require a lot of energy. Thus, respiration is essential for survival as it provides the energy required to carry out all such functions.

Question 9.
Two identical plants were kept under different conditions: [Scientific Temper and Critical Thinking Skill]
Plant A: Sunlight + water
Plant B: Sunlight + water + leaves coated with grease
After a few days, Plant B showed stunted growth.
(i) Which plant structure’s function do you think was affected?
(ii) Why did the greased plant show poor growth?
(iii) Mention any one life process that was impacted in Plant B.
Answer:
(i) The affected structure is stomata.
(ii) Grease blocked the stomata on the leaves, preventing the exchange of gases, thereby affecting food preparation and energy release.
(iii) Photosynthesis or respiration

Life Processes in Plants Class 7 Long Question Answer

Question 1.
Just as in human beings, the transport of water, minerals and food is also crucial for the growth and development of plants. Explain.
Answer:
Transport of water, minerals and food are essential for growth to take place. All activities utilise water and minerals. Synthesis of substances inside the body is a high energy consuming process. This energy is obtained by oxidation of food, i.e., glucose during respiration. This is why the transportation of food, water and minerals, all are essential for growth.

• Water and minerals are transported through xylem, and food is transported through phloem.
• Transport of water and minerals dissolved in it occurs only in one direction from the roots to upper plant parts; transport is unidirectional.
• Transport of food occurs from leaves to all growing parts of a plant; transport is bidirectional.

Question 2.
Plants have different mechanisms for synthesising, transporting, and utilising food to obtain energy. Discuss.
Answer:
Plants prepare food, distribute it, and use it to meet their energy needs through coordination between the following life processes:

• Synthesis of food: Plants prepare their own food by photosynthesis. Green leaves containing chlorophyll use sunlight, carbon dioxide, and water to make food in the form of glucose, which is stored as starch.
• Transport of food: The food prepared in the leaves is transported to all parts of the plant through the phloem. This process helps supply food to growing and storage parts like roots, stems, and fruits. Xylem transports water and minerals from the roots to the leaves, which are essential raw materials for photosynthesis. Thus, both xylem and phloem ensure that water, minerals, and food reach all parts of the plant for healthy growth.
• Utilisation of food: Plants use the transported food for obtaining energy through respiration. During respiration, food (glucose) is broken down using oxygen to release energy needed for growth, repair, and other life activities.

Question 3.
Discuss how photosynthesis and respiration are essential for the survival of various living organisms.
Answer:
For the survival of all living organisms, energy is required. Animals are incapable of capturing light energy. Hence, they depend upon plants for this energy. Plants capture light energy from the Sun during photosynthesis and convert it into chemical energy. This chemical energy is available in the form of glucose. During respiration, glucose is oxidised to liberate energy. This liberated energy is used for performing various metabolic activities. This is why both photosynthesis and respiration are essential for the survival of all living organisms.

Question 4.
Priya read that organisms breathe in oxygen and breathe out carbon dioxide, in a continuous manner. She wonders then there should have been less oxygen and more carbon dioxide in the atmosphere. Based on your understanding, help Priya clear her confusion. Also, compare the life processes involved.
Answer:
This balance is maintained by two life processes: photosynthesis and respiration. Green plants take in carbon dioxide from the atmosphere and release oxygen during photosynthesis, while plants, animals, and humans take in oxygen and release carbon dioxide during respiration. This natural cycle ensures that the gases in the atmosphere are continuously exchanged and maintained.

These two processes can be compared in the following manner:

Feature	Photosynthesis	Respiration
Purpose	Food (glucose) is prepared by plants.	Food is broken down to release energy.
Location	Occurs primarily in the leaves and other green parts of the plant.	Occurs in all parts of the plant.
Raw materials	Requires carbon dioxide and water, along with chlorophyll and the presence of sunlight.	Requires oxygen and glucose
Products	Produces glucose and oxygen.	Produces carbon dioxide, water, and energy.
Time	Occurs mainly during the daytime.	Occurs day and; night (does not require sunlight).
Importance	Helps plants produce food.	Provides energy for growth and life activities through the breakdown of the food produced.

Life Processes in Plants Class 7 Case Based Questions

Read the following passages and answer the questions that follow:

Question 1.
Ved believes that plants also give out something during the process of making their food. He puts up an experimental set-up, as shown in the figure given below to collect evidence for the release of a particular gas during photosynthesis. Based upon his observation, answer the following questions:

(i) Ved observed the formation of air bubbles. Explain his observation.
Answer:
The oxygen gas produced in this set-up caused bubbles to emerge and get accumulated in the inverted test tube.

(ii) What is the source of the gas released?
Answer:
The source of the oxygen gas released is photosynthesis performed by the aquatic plant in the experimental set-up.

(iii) How will the set-up behave if kept in a dark room?
Answer:
No air bubble formation will be observed in the set-up when kept in dark as no photosynthesis will occur in the absence of sunlight.

OR

(iii) Will the leaves of the aquatic plant form starch when kept in dark? How will Ved find out?
Answer:
No. Ved can confirm this with iodine test on a leaf plucked from the plant kept in the dark room. The leaf will not turn blue-black, indicating no starch formation.

Question 2.
Sara went on a vacation with her family for a few weeks and forgot to tell the caretaker to water the plant pots. When they returned, she noticed that the plants in their balcony appeared pale and unhealthy with wilted leaves, even though the plants were kept in sunlight. Sara wondered what may have caused this.

(i) Which life process in plants is mainly affected when water is not supplied?
Answer:
Photosynthesis

(ii) Name the structure through which water reaches the leaves.
Answer:
Xylem

(iii) Roots absorb minerals from the soil along with water. Can you tell why these minerals are important for the plant?
Answer:
Minerals are nutrients needed for healthy growth and proper functioning of the plant.

OR

(iii) Even though sunlight was available, why did the plants show poor growth?
Answer:
Plants showed poor growth even in the presence of sunlight because water is an essential raw material required for photosynthesis; food preparation cannot take place in its absence, thereby impacting the overall growth of the plant.

Life Processes in Plants Class 7 Picture Based Questions

Observe the given pictures and answer the questions that follow:

Question 1.
The given picture represents a microscopic view of a leaf peel with visible tiny pores on its surface.

(i) What are these tiny pores called?
Answer:
Stomata

(ii) Name the life process(es) for which these structures are essential.
Answer:
Photosynthesis and Respiration

(iii) Which atmospheric gases enter and exit the plant through them? Explain how the opening and closing of these pores help the plant survive.
Answer:
Carbon dioxide and oxygen. The opening and closing of these pores help the plant exchange gases with their surroundings and lose water for excretion, another important life process. They are thus essential for a plant’s survival.

OR

(iii) Predict what would happen if all these pores remained permanently closed.
Answer:
If all stomata remained permanently closed, the exchange of gases would stop. As a result, the plant would neither be able to make food nor respire, and would eventually die.

Question 2.
Sarla, a Class 7 science teacher, showed a diagram to explain plant transport systems to her students. Arrows in the image are indicating the direction of the movement of two different substances by two different pathways. Some arrows are labelled as ‘A’ and some are labelled as ‘B’.

(i) The grey-coloured type A arrows in the picture depict the movement of which substance?
Answer:
Grey arrows, labelled as A depict the movement of water and minerals.

(ii) The black-coloured type B arrows in the picture depict the movement of which substance?
Answer:
Black arrows, labelled as B depict the movement of food.

(iii) Give correct names of the structures denoted by arrows A and B respectively.
Answer:
A – Xylem
B – Phloem

OR

(iii) How does the movement of substances through these two pathways differ?
Answer:
Movement of water, through xylem, is unidirectional, while movement of food, through phloem, is bidirectional.

Life Processes in Plants Class 7 Activity Based Questions

Read the following activity and answer the questions that follow:

1. Sarita, a Grade 7 science teacher, kept a healthy potted plant inside a dark room for thirty-six hours. On the third day, she plucked a leaf from the plant and demonstrated the starch test to her students during their practical class. Although the leaf was still green, it did not turn blue-black when tested with iodine solution.

(i) Why did the leaf not turn blue-black in this case?
Answer:
The leaf did not turn blue-black because no photosynthesis occurred in the absence of sunlight, and all the stored starch was utilised. As a result, no starch was left or formed to react with iodine.

(ii) Name one condition necessary for food preparation that was missing.
Answer:
Sunlight

(iii) Even in the absence of food preparation, which life process must have continued in the plant? Explain your answer.
Answer:
Respiration, because it is independent of sunlight.

OR

(iii) How will a leaf from this potted plant, exposed to 24 hours of sunlight, behave if half of it is inserted in a bottle containing caustic soda and the other half is exposed to air, before performing the iodine test?
Answer:
The part of the leaf inside the bottle will not turn blue-black, as no food will be made in that part, in the absence of carbon dioxide (absorbed by caustic soda), while the part exposed to air will turn blue-black because photosynthesis can occur there.

Life Processes in Plants Extra Questions for Practice

Question 1.
As part of a revision test, Shubham was asked to identify the correct statement regarding life processes in plants from the options given below. Which option should he choose?
(a) During photosynthesis and respiration, exchange of water takes place.
(b) Glucose, synthesised during photosynthesis serves as an instant source of energy.
(c) Stomata help in capturing sunlight during photosynthesis.
(d) Phloem transports water and minerals from roots to all parts of the plant.
Answer:
(b) Glucose, synthesised during photosynthesis serves as an instant source of energy.

Question 2.
Growth in plants is indicated by:
(a) Increase in number of leaves and emergence of new branches
(b) Increase in length and thickness of its stem
(c) Increase in the number of flowers only
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 3.
Ashi was instructed by her teacher to select the plant part that participates in both photosynthesis and respiration. Which part should she choose from the ones listed below?
(a) Xylem
(b) Stomata
(c) Flower
(d) Seed
Answer:
(b) Xylem does not participate in the exchange of gases; flower is the reproductive organ of the plant; seed is a dormant structure that respires at a low rate but is not involved in photosynthesis as it lacks chlorophyll.

Question 4.
If the xylem of a plant stem is damaged, which process(es) will be most directly affected?
(a) Synthesis of food
(b) Transport of food to roots
(c) Storage of food in plant parts
(d) Both (a) and (b)
Answer:
If the xylem is damaged, water supply to leaves is reduced, directly affecting the synthesis of food (photosynthesis), not the transport of food to roots (done by phloem).

Questions 5 and 6 consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 5.
Assertion (A): Food reaches from developing buds to leaves through the phloem.
Reason (R): Phloem can transport food in different directions depending on the needs of the plant.
Answer:
Food reaches from leaves to all plant parts, such as buds, through phloem.

Question 6.
Assertion (A): During the day, plants usually show a net release of oxygen.
Reason (R): The rate of photosynthesis is higher than the rate of respiration in the presence of sunlight.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Question 7.
Suman claims only water and sunlight are the raw materials used for photosynthesis. Is the claim made by her right?
Answer:
No, Suman is incorrect: the process of photosynthesis also requires carbon dioxide from air and chlorophyll pigment present in leaves, along with water and sunlight.

Question 8.
Mention the visible changes that occur in a plant when it is kept in dark for a long period of time.
Answer:
When a plant is kept in the dark for a long period of time, the leaves will gradually lose chlorophyll and become yellowish, and no new leaves will be formed. Overall growth of the plant will be stunted (no food production in absence of sunlight).

Question 9.
Why do non-green patches of leaves fail to provide a positive result for the iodine test? Also, state why broad leaves are more efficient in photosynthesis.
Answer:
The non-green leaf patches usually lack chlorophyll or may not have sufficient chlorophyll to prepare enough starch to be detected using the iodine test.
Broad leaves are more efficient in photosynthesis because they contain more chlorophyll and have a larger surface area to capture or absorb more sunlight.

Question 10.
Manoj, a farmer, observed that even when crops received enough sunlight, reduced rainfall during the growing season resulted in low crop yield compared to the previous year. He wondered what the reason could be. Do you think the internal processes of plants could have affected crop production under such conditions?
Answer:
Yes, the internal processes of plants like photosynthesis, respiration, and the transportation of water, minerals, and food are interdependent and together determine successful crop production. In the given situation:

• Reduced rainfall limited the water supply, which is an essential raw material needed for photosynthesis, so less food must have been prepared.
• Poor transport of water and food through xylem and phloem would have affected the supply of nutrients to all other parts of the plant.
• Respiration depends on the food prepared to release energy for growth and development.
  With less food prepared. respiration released less energy, resulting in poor growth, weaker plants, and thus, a poor crop yield.

Question 11.
Joseph Priestley was a remarkable scientist of the 18th century. Priestley conducted an experiment in which he placed a mouse and a burning candle inside a sealed bell jar, but the mouse died. However, when he placed a mint plant along with the mouse in the bell jar with a burning candle, the mouse survived and the candle continued to burn. Based on this observation answer the following questions:

(i) Which feature of plants did Priestley interpret from his observations?
Answer:
Priestley interpreted that plants restore to the air whatever breathing animals and the burning candle remove.

(ii) Why did the mouse die in the first case, in the absence of the mint plant?
Answer:
The oxygen gas, within the bell jar, got consumed by the burning candle (burning requires oxygen) and the respiring mouse. In the absence of oxygen, the mouse could not respire and therefore died.

(iii) Identify the process responsible for the survival of the mouse in the second case. What are the products of this process?
Answer:
The mouse survived due to the availability of oxygen released during photosynthesis, by the mint plant. During photosynthesis, plants consume carbon dioxide, produce glucose, and release oxygen gas.
OR
(iii) Name the other life process that is occurring inside the bell jar. Out of scenarios A and B , in which jar will be the amount of carbon dioxide high and low respectively. Give reason.
Answer:
The process is respiration.

• In scenario A, the mouse will breathe out carbon dioxide, thereby increasing its amount inside the bell jar.
• In scenario B, the carbon dioxide released by the respiring mouse is used up by the mint plant in performing photosynthesis, hence the carbon dioxide concentration will be low.

Question 12.
Explore the consequences of the disappearance of photosynthetic organisms on the Earth.
Answer:

The disappearance of photosynthetic organisms would make the Earth uninhabitable for most living beings.

• Plants produce oxygen through photosynthesis, which is essential for animals to breathe and respire. Without plants, the oxygen level in the atmosphere would drop, causing difficulty in breathing and suffocation.
• Photosynthesis utilises and maintains the balance of carbon dioxide in the atmosphere. Without this process, carbon dioxide levels would increase, trapping more heat, which may lead to global warming and climate change.
• Plant roots help to hold soil in place. Without plants, soil would be easily washed or blown away, leading to soil erosion.

Students can include Class 7 Curiosity Extra Questions and Class 7 Science Chapter 9 Life Processes in Animals Extra Question Answer in their daily self-study routine.

Class 7 Science Chapter 9 Life Processes in Animals Extra Questions

Class 7 Science Chapter 9 Extra Questions on Life Processes in Animals

Life Processes in Animals Class 7 Very Short Question Answer

Question 1.
Name the parts of the alimentary canal where
(i) water gets absorbed from undigested food.
(ii) digested food gets absorbed.
(iii) taste of the food is perceived.
(iv) bile juice is produced.
Answer:
(i) Large intestine
(ii) Small intestine
(iii) Tongue
(iv) Liver

Question 2.
If breathing is a part of respiration, then what is the difference between them? Give your answer in a tabular form.
Answer:

Breathing	Respiration
It is the process of inhaling and exhaling air.	It is the process of breaking down food to release energy in the presence of oxygen.
It is a physical process.	It is a chemical process.

Question 3.
As Ajay climbed higher up the mountain, he realised that his breathing had become faster and his chest was moving rapidly. This made him wonder how inhalation and exhalation occurs during breathing, and if the diaphragm plays any role during inhalation. Provide an answer for the same.
Answer:

• Inhalation is taking in oxygen-rich air, and exhalation is breathing out air rich in carbon dioxide.
• During inhalation, the diaphragm contracts and moves downwards.

Question 4.
During a class discussion, Prerna said that respiration and breathing are interdependent. To which, Manav countered that respiration and breathing are different processes and can occur independently of each other. Based on your understanding, whose claim is correct and why?
Answer:
Prerna’s claim is right. Breathing supplies oxygen needed for respiration, and respiration releases energy that helps in breathing. Thus, although they are different processes, respiration and breathing are interdependent.

Question 5.
After the doctor advised Harsh’s uncle to avoid eating food rich in fats due to a liver infection, Harsh asked his teacher the following questions:
(i) Why did the doctor recommend avoiding food rich in fats?
Answer:
The doctor told Harsh’s uncle to avoid eating food rich in fats because the liver helps with fat digestion. When the liver is affected, it cannot function properly, thereby affecting the digestion of fats.

(ii) What does the liver secrete that helps in the digestion of fats?
Answer:
Liver secretes bile which breaks down fats into small droplets making it easier to digest and also neutralises acid present in the food (coming from stomach) in the small intestine.

Question 6.
Following a football match in his school, Anish noticed that he was breathing heavily. His teacher explained that it is because his body requires more energy. What is this energy-releasing process called? Give its word equation.
Answer:
Glucose + Oxygen → Carbon dioxide + Water + Energy

Question 7.
While visiting a dairy farm, Preeti noticed that cows and buffalo swallow grass quickly without chewing much. Later, she saw them sitting calmly and chewing the same food again. Can you explain why ruminants show this type of feeding behaviour?
Answer:
Ruminants like cows and buffalo first swallow their food quickly and store it in a specialised structure called the rumen. Later, this partially digested food is brought back to the mouth and chewed again through a process called rumination. This helps them digest tough plant materials like grass properly.

Question 8.
Name the main components of the circulatory system. What is its primary function in the human body?
Answer:

• Circulatory system includes the heart, blood, and blood vessels.
• It helps in the transport of nutrients, oxygen, and other substances throughout the body.

Question 9.
As Aarav was eating bread, it felt soft and sweet after chewing. Curious, he asked his mother why this happened. She explained that it is due to a substance found in the mouth. What is this substance called, and what is its function in the mouth?
Answer:
Saliva is the watery fluid produced in the mouth. It contains a digestive juice that helps break down starch into sugar.

Question 10.
Give the pathway of air flow in the human respiratory system. How are lungs protected?
Answer:
Nostrils → Nasal passages → Windpipe → Bronchi → Smaller and finer branches of bronchi → Alveoli (balloon-like sacs)
– Lungs are protected by the rib cage.

Life Processes in Animals Class 7 Short Question Answer

Question 1.
Write down the function of the following parts:
(i) Oesophagus
(ii) Anus
(iii) Liver
Answer:
(i) Oesophagus: It is a long, flexible tube that contracts and relaxes in a wave-like motion to push the food down into the stomach.
(ii) Anus: It is the opening at the end of the digestive tract that expels the solid waste (stool) out of the body.
(iii) Liver: It secretes bile, which helps in the breakdown of fats.

Question 2.
During a biology field trip, Ved came across the following animals:
(i) Tadpole
(ii) Snake
(iii) Fish
(iv) Lizard
(v) Earthworm
(vi) Goat
Can you help Ved name the respiratory organs of the given animals?
Answer:
(i) Gills
(ii) Lungs
(iii) Gills
(iv) Lungs
(v) Moist skin
(vi) Lungs

Question 3.
Apart from digestive and respiratory systems, what other systems are present in the human body? Write down their functions.
Answer:
The other two systems that work in coordination with digestive and respiratory systems are circulatory and excretory systems.

• Circulatory system: It helps in the transport of nutrients, oxygen, carbon dioxide, and other substances via blood.
• Excretory system: It removes waste products such as excess water and salts from the body.

Question 4.
During a science class, Mr. Joshi displayed an image of a balloon-shaped structure, essential in the process of respiration and asked the students to identify it and explain its function. What is the name of this structure, and what does it do?
Answer:
The balloon-like structures are called alveoli. These are small air sacs in the lungs where exchange of oxygen and carbon dioxide takes place. They have thin walls surrounded by fine tubes containing blood.

As blood flows through these fine tubes, it carries carbon dioxide from all parts of the body. which passes into the alveoli and is removed from the body when we breathe out. At the same time, oxygen from the air inside the alveoli enters the blood and is then carried to different parts of the body. These structures provide a larger surface area for the exchange of gases during respiration.

Question 5.
On an average, an adult human being at rest breathes 15-18 times per minute. The breathing rate, however, may differ under different conditions. Arrange the following activities given in the box in order of increasing breathing rates and give reason for your answer.
Sleeping, Cycling, Brisk walk, Watching T.V.
Answer:

• Order of increasing breathing rate: Sleeping < Watching TV < Brisk walk < Cycling
• Breathing rate increases with the energy demand of the body. While sleeping, the body’s energy requirement is lowest, so breathing is slowest. Watching TV requires slightly more energy, so breathing is comparatively faster. Brisk walking is a physical activity that requires more oxygen, increasing the breathing rate, and cycling is the most physically demanding among these activities, so the breathing rate is highest.

Question 6.
‘Different animals have different mechanism for the digestion of food’. Comment on this statement taking into consideration the digestion in ruminants and birds.
Answer:
The differences in the mechanism of digestion in ruminants and birds are listed below:

Feature	Ruminants (cows and buffalo)	Birds
Type of food	Plant material (grass)	Grains, seeds, insects, etc.
Chewing	Chew partially digested food	Do not chew; swallow food as a whole
Grinding food	Mouth (chewing); teeth present	Teeth absent, food is broken down in a chamber called the gizzard

Life Processes in Animals Class 7 Long Question Answer

Question 1.
Shweta was studying the human digestive system and learnt that the pancreas is a vital organ located in the abdomen. It produces a secretion ‘X’ that is released into the small intestine, which is crucial for breaking down the food we eat into simpler substances that can be absorbed by the body.
Based on this information, answer the following questions.
(i) Name the secretion X .
(ii) What is the nature of X ?
(iii) What are the functions of X ?
(iv) Can digestion occur properly without X? Give reason for your answer.
Answer:
(i) Pancreatic juice
(ii) Basic in nature
(iii)

• Neutralises acid present in the food.
• Helps in the breakdown of carbohydrates, proteins, and fats.

(iv) No, digestion cannot occur properly without pancreatic juice because it breaks down carbohydrates, proteins, and fats. Without it, these nutrients cannot be digested and absorbed efficiently by the body.

Question 2.
(i) How do fishes and earthworms breathe?
(ii) What are the different adaptations for respiration used by amphibians like frogs during their life cycle? How is it helpful for them?
Answer:
(i) Fishes have specialised structures called gills for breathing, which are richly supplied with blood vessels. The exchange of oxygen and carbon dioxide between the blood and the gases dissolved in surrounding water takes place across them. Whereas, earthworms use their moist skin for the exchange of gases from the environment.

(ii) Amphibians, like frogs, use different body parts for breathing during different stages of their life:

• Tadpoles breathe through gills, similar to fish.
• Adult frogs breathe using lungs when on land and through their moist skin when underwater.

These adaptations help them to survive both in water and on land.

Question 3.
(i) Rahul was learning about how humans breathe and wanted to know what respiration means and which parts of the body help us breathe.
How would you explain the human respiratory system to Rahul?
Answer:
The process of breaking down food to release energy is called respiration.
Main parts of the human respiratory system are:

• Nostrils
• Nasal passages
• Windpipe
• Lungs
• Alveoli
• Diaphragm

(ii) Give the word equation for respiration.
Answer:
Glucose + Oxygen → Carbon + Water + Energy dioxide

(iii) Write down the function(s) of:
(a) Nasal passages
(b) Alveoli
Answer:
(a) Nasal passages: It filters dust particles and moistens the air we breathe in.
(b) Alveoli: Alveoli are small balloon-like air sacs where the exchange of gases (oxygen and carbon dioxide) takes place.

Question 4.
One afternoon, while having lunch with her mother, Isha wondered how the food she eats is broken down inside the body. As she chewed her food, she started asking her mother a few questions about digestion. Based on their conversation, answer the following questions:
(i) Where does digestion begins in the body, and what is the role of saliva in digestion?
(ii) What does the stomach secrete to aid the process of digestion?
(iii) Define: (a) Mechanical digestion (b) Egestion
Answer:
(i) Digestion of food starts in the mouth. Saliva plays an important role in digestion by moistening the food, making it easier to chew and swallow. It contains a digestive juice that helps break down starch into sugar.
(ii) Stomach secretes digestive juice, acid, and mucus that aids in the process of digestion.
(iii)
(a) Mechanical digestion: The process of initial breakdown of food into fine pieces is called mechanical digestion. This process begins in the mouth as the teeth break down the food we eat into smaller pieces through crushing and chewing.
(b) Egestion: The process of removing undigested and unabsorbed food materials from the body in the form of stool, through the anus, is called egestion.

Life Processes in Animals Class 7 Case Based Questions

Read the following passages and answer the questions that follow:

Question 1.
Mohan was fascinated while watching his pet parrot, Peeku, eat grains quickly. He wondered how birds eat and digest food since they do not have teeth. Suddenly, he remembered his science class, where Mrs. Rita explained about a special structure, that helps birds in the process of digestion.

(i) Name the structure.
Answer:
Gizzard

(ii) Do ruminants also have this structure?
Answer:
No, ruminants such as cows and buffalo do not have a gizzard. Instead, they have a rumen, which helps in the digestion of food.

(iii) How does this structure help in the digestion of food?
Answer:
The gizzard has muscular walls and often contains small stones (grit) swallowed by the bird. The contraction and relaxation of the walls help to grind and crush the food into smaller pieces, making digestion easier.
OR
(iii) Is the process of digestion in birds different from that in humans? Give reason.
Answer:
Birds do not have teeth and they use a gizzard for grinding food, whereas humans chew food with their teeth before swallowing. This special structure called gizzard is absent in humans.

Question 2.
During the annual sports day at school, Rohan and Ravi participated in the 100 metre sprint.
After finishing the race, Rohan was gasping for breath. His face turned red, and his heart was beating rapidly. They counted their breaths per minute and observed that Rohan was breathing faster than Ravi. [Critical Thinking and Observation Skills]

(i) Why was Rohan breathing faster than Ravi after the race?
Answer:
Rohan might not be physically fit and healthy, or he might have run faster than Ravi, which required a higher oxygen supply to his muscles and resulted in faster breathing.

(ii) What are the end products formed during respiration in our body?
Answer:
Carbon dioxide, water, and energy.

(iii) What happens during respiration? Name the primary respiratory organ in humans.
Answer:

• During respiration, food in the form of glucose is broken down to release energy.
• Lungs are the primary respiratory organs in humans.

OR

(iii) Does blood play any role during the process of respiration?
Answer:
Yes, blood has a vital role in respiration. It functions as a transport medium for oxygen and carbon dioxide, facilitating gaseous exchange between the lungs and the body parts.

Life Processes in Animals Class 7 Picture Based Questions

Observe the given picture and answer the questions that follow:

1. (i) Where are these structures found?
Answer:
Inner lining of the small intestine

(ii) Mention a structural characteristic of these structures.
Answer:
These are finger-like projections surrounded by tiny blood vessels.

(iii) What are their functions?
Answer:
These finger-like projections increase the surface area for absorption and allow digested nutrients to pass into the blood, which is carried to different parts of the body to provide energy.
OR
(iii) Mention any two other roles of the location where these structures are found.
Answer:

• The small intestine receives bile from the liver to break down fats into tiny droplets.
• It also receives pancreatic juice to neutralise the acid in food.
• Its inner lining secretes digestive juices that further break down fats, proteins, and carbohydrates.

Life Processes in Animals Class 7 Activity Based Questions

Read the following activity and answer the questions that follow:

Question 1.
Objective: To demonstrate the mechanism of breathing using a simple lung model.
Materials required: A transparent plastic bottle, two small balloons, one large rubber sheet, Y-shaped hollow tube, rubber bands and tape

Procedure:

• Take a clean transparent plastic bottle and cut off its bottom.
• Attach two small balloons to the ends of a Y-shaped tube using rubber bands.
• Insert the tube into the mouth of the bottle and seal with tape to make it airtight.
• Stretch the rubber sheet tightly over the open bottom of the bottle.
• Pull the rubber sheet downward and then push it upward gently, while observing the balloons.

Observation: When the rubber sheet is pulled downwards, the balloons inside the bottle inflate.
When the rubber sheet is pushed upwards, the balloons deflate.
(i) Which process of breathing is represented when the balloons inflate?
(ii) What changes occur inside the bottle when the rubber sheet is pulled downwards?
(iii) What does the deflation of balloons indicate about the movement of air during exhalation?
Answer:
(i) Inhalation
(ii) The space inside the bottle increases, allowing air to enter into the balloons and they inflate.
(iii) Deflation of balloons indicates that air moves out of the lungs during exhalation due to reduced volume of the chest cavity.
OR
(iii) What should be kept the same in this activity for a fair demonstration? Mention ‘Yes’ or ‘No’ for each and give reason.

S. No.	Should this be kept the same?	Yes or No?	Reason
(a)	Size of the bottle
(b)	Number of balloons
(c)	Airtight sealing of the bottle

Answer:
(a) Yes; the bottle represents the chest cavity, changing its size affects the space for air.
(b) Yes; the balloons represent the lungs; changing their number is not possible, since humans have two lungs.
(c) Yes; air leakage will disturb air pressure and affect inflation and deflation of balloons.

Life Processes in Animals Extra Questions for Practice

Question 1.
As part of a revision test, Mohan was asked to identify ‘A’ and ‘B’ in the diagram given alongside. Which among the following options should he choose?

(a) A – Carbon dioxide, B – Oxygen
(b) A – Oxygen, B – Carbon dioxide
(c) A – Carbon dioxide, B – Hydrogen
(d) A – Hydrogen, B – Oxygen
Answer:
(a) During respiration carbon dioxide is removed from the body when we breathe out. At the same time, oxygen from the air inside the alveoli enters the blood and is then carried to different parts of the body.

Question 2.
While on a walk through the neighbourhood park, Ananya noticed some tadpoles swimming together in a pond. She wondered how these organisms breathe underwater. Which of the following structures help tadpoles in gaseous exchange?
(a) Gills
(b) Lungs
(c) Alveoli
(d) Moist skin
Answer:
(a) Gills

Question 3.
The image alongside shows a child who is blowing foggy air out of his mouth. Which terminology would be most appropriate for defining the act of blowing air out?

(a) Breathing
(b) Exhalation
(c) Inhalation
(d) Respiration
Answer:
(b) Exhalation

Question 4.
While helping his father plant sunflower seedlings, Mihir noticed an earthworm with moist skin, as given alongside. Curious, he asked his father why earthworms have such skin. What could be the most appropriate explanation Mihir’s father can give?

(a) It helps them escape from a predator’s grip.
(b) It helps them reflect sunlight during summer.
(c) It helps them absorb oxygen from the surroundings.
(d) It helps them absorb nutrients from the soil.
Answer:
(c) Earthworms breathe through their moist skin, which allows gaseous exchange.

Questions 5 and 6 consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the correct option:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 5
Assertion (A): Breathing is the process through which fresh air from outside enters the lungs and fills the alveoli.
Reason (R): Alveoli are balloon-like structures that help in increasing the surface area for exchange of gases.
Answer:
(b) Both A and R are true but R is not the correct explanation of A.

Question 6.
Assertion (A): The large intestine absorbs water and some salts from the undigested food, thus making the waste semi-solid, and it is called stool.
Reason (R): The stool is stored in the upper part of the large intestine, called the rectum.
Answer:
(c) Stool is stored in the lower part of the large intestine, called the rectum.

Question 7.
While teaching her younger brother, Arna explained that the small intestine is in fact the longest part of the digestive system and plays a major role in digestion of food. Her brother, Arsh wondered what exactly it does to help the body obtain energy. What functions are performed by the small intestine?
Answer:

• The small intestine secretes digestive juice that further breaks down fats, proteins, and carbohydrates.
• It has multiple finger-like projections that help in the absorption of nutrients into the blood.

Question 8.
Do different organisms have different breathing organs? Explain.
Answer:
Yes, different organisms have different breathing organs, as this enables them to adapt better to their environment.
For example, most aquatic organisms like fish breathe through gills. On the other hand, terrestrial animals such as birds, elephants, cows, and humans breathe through their lungs.

Question 9.
Ms. Rekha gave her Class 7 students an assignment to make a poster on alveoli. The poster should highlight their function and specific characteristics. What points should the students include in their posters?
Answer:

• Function: Alveoli increase the surface area for efficient exchange of oxygen and carbon dioxide.
• Characteristics: Alveoli are tiny, balloon-like air sacs in the lungs where exchange of gases take place. They have thin walls and are surrounded by blood vessels. They allow oxygen to enter the blood and carbon dioxide to exit the body.

Question 10.
Observe the given figures and answer the following questions.
(i) Name the processes depicted by figures A and B.
Answer:
Figure A indicates the process of inhalation and figure B indicates the process of exhalation.

(ii) In the given figures indicate the direction of

(a) movement of air
(b) movement of diaphragm
(c) movement of ribs
Answer:

Question 11.
Sana, a Class 7 student, was very curious to learn about nutrients present in food. One day at school, her teacher conducted an experiment by adding a few drops of iodine solution to some boiled rice. To her surprise, the rice turned blue-black. Sana later repeated the experiment at her home, with slices of potato and chapati and saw similar results.

(i) What is starch?
Answer:
Starch is a type of complex carbohydrate present in our food.

(ii) Why did the food items show a blue-black colour when iodine was added?
Answer:
Rice, potatoes, and chapati turn blue-black when iodine solution is added to them because these food items contain starch.

(iii) In which part of the digestive system is starch first acted upon, and into which simpler substance is it converted?
Answer:
Digestion of starch begins in the mouth by the action of saliva which contains a digestive juice that helps break down starch into simple sugars.
OR
(iii) Can you think of any two food items that do not contain starch? How can you prove this using the iodine test?
Answer:
Examples of food items that do not have starch include lemon, cucumber, etc. These food items will not turn blue-black when tested with iodine solution.

Question 12.
During a school field trip to a science museum, Aisha saw a model of the human digestive system and was curious about how it works. She asked her teacher to explain the different parts involved in digestion and how they function. How would the teacher explain the different parts of the digestive system to Aisha?
Answer:
Digestion in humans is a step-by-step process that takes place in the alimentary canal. The main organs and their respective roles are:

• Mouth: Site where the food is chewed (mechanical digestion, aided by teeth), and mixed with saliva. Saliva contains digestive juice that begin the digestion of starch.
• Oesophagus: Pushes food to the stomach using muscular movements (contraction and relaxation in a wave-like motion).
• Stomach: Secretes acid and digestive juice that helps in the breakdown of protein, and the mucus secreted protects the stomach lining from the acid.
• Small Intestine: Receives digestive secretions from three sources – the inner lining of the small intestine itself, liver (bile) and pancreas (pancreatic juice) that aids in the complete digestion of food. The nutrients are absorbed into the blood through finger-like projections present on their inner walls.
• Large Intestine: Absorbs water and salts, and then forms solid waste (stool) from the undigested food that is temporarily stored in the rectum before expulsion.
• Anus: Removes undigested waste from the body through the process of egestion.