CHSE Odisha Class 12 Alternative English Grammar Phrasal Verbs

Odisha State Board CHSE Odisha Class 12 Alternative English Solutions Grammar Phrasal Verbs Exercise Questions and Answers.

CHSE Odisha Class 12 Alternative English Grammar Phrasal Verbs

Phrasal Verbs:
We have to remember the phrasal verbs with their meaning perfectly. Let us discuss some important phrasal verbs.

Intransitive:

1. break down: stop working.
Ex:-. My car broke down twice during our journey.
2. break out: start suddenly
Ex:- Cholera has broken out in our locality.
3. burst out: begin to do something suddenly
Ex:- The children burst out laughing.
4. come about: happen
Ex:- How did the accident come about?
5. come out: published, become known
Ex:- This magazine comes out once in a week. Our results came out yesterday.
6. come off: happen, take place
Ex:- My sister’s wedding came off in a grand way.
7. come on: say to encourage
Ex:- Come on, let’s try again.
8. come round: regain consciousness, cure, recover
Ex:- He is still unconscious. He has not come around.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

9. die away: become weak, disappear gradually
Ex:- The noise gradually died away.
10. draw up: approach and stop
Ex:- A car drew up beside me.
11. drop in: call on somebody
Ex:- Why don’t you drop in and see me sometime?
12. drop out: withdraw
Ex:- Ajit has dropped out (of the team)
13. fall out: quarrel
Ex:- You should not fall out for such a trivial problem.
14. getaway: escape
Ex:- Two of the thieves got away.
15. get along: image
Ex:- It is very difficult to get along without money.
16. get on: make progress
Ex:- How are you getting on at college?
17. go off: explode
Ex:- Many people died when a bomb went off in the busy market area.
18. goon: continue
Ex:- The two friends, went on talking for hours.
19. go out: extinguish
Ex:- The lamp went out in the wind.
20. get up: the rise
Ex:- I get up early in the morning.
21. give in: surrender, accept defeat
Ex:- The tired soldiers finally gave in to their enemy.
22. give out: come to an end
Ex:- Our food supply gave out after a week.
23. give up: cease, stop, abandon
Ex:- I have given up smoking
24. hold on: maintain one’s position
Ex:- Out troops held on resolutely refusing to yield an inch.
25. look out: pay attention, be careful
Ex:- Look out: there is a heavy truck coming very fast behind us.
26. lookup: become better
Ex:- The weather is looking up now.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

27. makeup: replace a loss
Ex: It will take a long time to make up the loss.
28. pull up: come to a stop
Ex:- The driver pulled up at the traffic lights.
29. put up: stay, live
Ex:- We are putting up in a small house.
30. run down: lose a store of energy
Ex:- The battery has run down.
31. run out: to come to an end
Ex:- All our food has run out.
32. set out / off: begin a journey
Ex:- We set out / off our journey early in the morning.
33. set in: begin
Ex:- The trains have set in early this year.
34. shut up: be quiet
Ex:- Shut up and leave me alone.
35. take off: leave the ground
Ex:- The airplane took off at 70’ clock.
36. turn up: come usually to a meeting
Ex:- The meeting was postponed, as only Haifa do-can people turned up.
37. wear out: became unfit for use
Ex:- This part of the machine has worn out. Cheap shoes wear out easily.

Transitive:

1. account for: give a reason for
Ex:- You must account for your absence at college yesterday.
2. agree with: be good for health.
Ex:- Egg does not agree with me.
3. break into: enter by force
Ex:- Thieves broke into our house last night.
4. (i) burst into: a sudden show of emotion.
Ex:- She burst into tears on getting the bad news.
(ii) to come in suddenly
Ex:- The angry men burst into my room and shouted at me.
5. call on: visit
Ex:- We called on our new neighbor yesterday.
6. come across: meet, find by chance.
Ex:- I come across some friends in the marketplace yesterday.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

7. come over: influence
Ex:- He looks very sad, what has come over him?
8. count on: rely on
Ex:- Can I count on your help during my difficulties?
9. do without: manage without
Ex:- He can’t do without tea.
10. get on: progress, conduct
Ex:- How are you getting on with your study?
11. get over: to recover from an illness/loss/difficulty?
Ex:- Leena has not got over the shock of her husband’s death.
12. go into: investigate
Ex:- The auditors have gone into our accounts.
13. go through: read
Ex:- I have gone through this novel.
14. jump at: accept eagerly
Ex: The children jumped at the proposal of visiting Nandakanan.
15. keep off: remain at a distance
Ex:- Keep off the grass
16. live on: have as food
Ex:- A baby lives on milk only.
17. look after: take care of
Ex:- The old man has nobody to look after him.
18. look after: to consider
Ex:- I look up to Rajesh as my own brother.
19. look into: examine carefully
Ex:- The police are looking into the theft at present.
20. look for: try to find
Ex:- I looked for my lost pen but found it nowhere
21. standby: help
Ex:- If they trouble you we’ll stand by you.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

22. stand for: represent
Ex:- M.O. stands for a money order.
23. take after: resemble, look alike
Ex:- The baby takes after its mother.
24. take to: start a habit
Ex:- Rahul has taken to drinking after his wife’s death.

Verb + Object + Particle

1. answer back: reply rudely
Ex:- It is not good to answer your parent’s back.
2. count in: include
Ex:- If you are going to the circus, count me in.
3. order about: call to do something
Ex:- Don’t try to order me about, I am not your servant.
4. take for: mistake
Ex:- My aunt took me forAnil.
5. fell apart: consider separate
Ex:- I have never been able to fall the two brothers apart.
6. try on: put on a garment to see whether it fits
Ex:- You must try these shoes on before you buy them.

Verb + Particle + Object
(Or)
Verb + Object + Practical

1. blow up: break into pieces by an explosion
Ex:- The bridge was blown up by enemy soldiers.
2. bring about: cause to happen
Ex:- The new principal brought about several changes in the college.
3. bring out: Publish
Ex:- My father will bring out a new book next week.
4. bring up: rear, educate
Ex:- The mother worked hard to bring up her children.
5. call off: cancel
Ex:- We called off the strike after an agreement with the government.
6. carry on: Continue
Ex:- Smita carried on singing for a long time.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

7. carry out: obey, do successfully
Ex:- You should carry out the order of your parents.
8. cut down: diminish, reduce
Ex:-You should cut down your expenses.
9. give up: stop
Ex:- You should give up smoking.
10. keep up: maintain, retain
Ex:- You should keep up the glory of your motherland.
11. keep away: remain at a distance Ex:- Keep away the children from fire.
12. lay by: keep for future use
Ex:- You should lay by something for old age.
13. leave out: omit
Ex:- You can leave out the questions you can not answer.
14. let down: opposite of back up.
Ex:- You have promised to stand by me. You won’t let me down, will you?
15. let off: not punish
Ex:- I’ll let you off this time, but I’ll punish you if you do it again.
16. lookup: search for a word in a dictionary.
Ex:- Look up a word in a dictionary, if you do not know its meaning.
17. make out: understand
Ex:- Can you make out the meaning of this sentence?
18. makeup: replace a loss
Ex:- It will take a long time to make up for the loss.
19. make over: to hand over charges
Ex:- The outgoing Principal made over the charge to the new Principal
20. pack up: stop working
Ex:- It is time to pack up and go home.
21. pick out: choose
Ex:- She picked out a frock that she liked most.
22. pulldown: Destroy
Ex:- The old building was pulled down.
23. put on: begin to wear, and dress oneself.
Ex:- Don’t forget to put your coat on.
24. put down: suppress by force
Ex:- The violent agitation was put down in no time.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

25. put off: postpone, keep for a later time.
Ex:- Don’t put off today’s work for tomorrow.
26. put out: extinguish
Ex:- Put out the light before you sleep.
27. run over: knockdown – by traffic.
Ex:- Hundreds of pedestrians are run over in the streets every year.
28. setup: establish
Ex:- The government has set up a hospital in our village.
29. take in: cheat
Ex:- He was taken in by the shopkeeper.
30. takeoff: remove clothes, hat, etc.
Ex:- Take off your shoes before entering a temple.
31. take over: accept the duty
Ex:- Ramesh, took over the business from his father.
32. turndown: reject
Ex:- He turned down my request.
33. turn on: start the flow of
Ex:- I turned on the tap.
34. turnoff: stop the flow of
Ex:- Please turn off all lights before going to bed.
35. windup: bring to an end
Ex:- It is time for him to wind up his speech.
36. workout: calculate correctly
Ex:- An intelligent child can work out this sum.

Verb + Adverb Particle + Preposition + Object

1. catch up with: come from behind and reach someone in front by going faster.
Ex:- Drive fester, they are catching up with us.
2. do away with: abolish, get rid of
Ex:- You can’t do away with violence by using violence.
3. get up with: make progress in something you are doing.
Ex:- How are you getting on with your business?
4. go back on: fail to keep.
Ex:- I can’t go back on my word.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

5. go in for: choose something as your job or interest.
Ex:- I thought of going in for teaching.
6. grow out of: became too big for
Ex:- He has grown out of that shirt.
7. look forward to: to be excited and pleased about something that is going to happen.
Ex:- We are looking forward to our uncle’s visit.
8. look down upon/on: hate, despise
Ex:- We should not look down upon the poor.
9. make up for: compensate for
Ex: – Hard work can often make up for lack of intelligence.
10. put up with: tolerate, bear
Ex:- I can’t put up with your rudeness any.
11. run out of: use all of something
Ex:-We have run out of sugar
12. watch out for: Keep looking and waiting for something/someone
13. keep up with: manage to go or learn as far as someone.
Ex:- The new boy can’t keep up with the class.

Exercise For Practice
1. Use appropriate phrasal verbs for the underlined verbs in the following sentences:

1. My brother has read this novel.
2. I can’t tolerate his insulting words.
3. His grandfather died yesterday.
4. He has postponed the meeting.
5. I can’t understand his speech.
6. We would not hate uncivilized people.
7. You should not try to cheat me.
8. Cholera has began in our locality.
9. The boy resembles his father.
10. That book has been published.
11. You should obey the words of the elders.
12. Pramila belongs to a royal family.
13. You should rise early in the morning.
14. Our Principal distributed the prizes.
15. You should maintain the prestige of your parents.
16. Stop the computer.
17. He has solved all the sums.
18. He has established a factory.
19. The police followed the thief.
20. The Pakistan army had to yield.

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

Answer:
1. My brother has gone through this novel.
2. I can’t put up with his insulting words.
3. His grandfather passed away yesterday.
4. He has called off the meeting.
5. I can’t make out his speech.
6. We would not look down upon uncivilized people.
7. You should not try to take me in.
8. Cholera has broken out in our locality.
9. The boy takes after his father.
10. That book has been brought out.
11. You should carry out the words of the elders.
12. Pramila comes of a royal family.
13. You should get up early in the morning.
14. Our Principal gave away the prizes.
15. You should keep the prestige of your parents.
16. Turn off the computer.
17. He has worked out all the sums.
18. He has set up a factory.
19. The police ran after the thief.
20. The Pakistan army had to give in.

Exercise For Practice
2. Supply a phrasal verb of the same meaning as indicated in the brackets.

1. All the lights __________ when the power supply was cut off. (stopped giving light).
2. Please check if you have __________ any name. (omitted).
3. The workers __________ working for a long time. (continued)
4. You should __________ something for your children’s education. (save for future use).
5. Priya has __________ her illness only recently. (recovered from)
6. It is dishonest to __________ one’s words. (fail to keep a promise)
7. Ranjit was __________ by a car. (hit).
8. The tires of my cycle have __________. (become unfit for use)
9. Mr. Patra has __________ a school in his village. (established)
10. You should __________ wild animals in a jungle. (be careful of)
11. This clock has __________. (stopped working)
12. You must __________ your misbehavior. (give a reason for)
13. I __________ and an old friend at a shop. (met by chance)
14. The new boy __________ with almost everybody in the class. (quarrel)

CHSE Odisha Class 12 Alternative English Solutions Phrasal Verbs

Exercise For Practice
3. Choose the correct particles to make the sentences meaningful.

1. I called __________ my friend, (off / on)
2. We took a long time to work __________ the problem, (out / at)
3. The minister has promised to think the __________ matter. (about / over)
4. It is hard to make __________ their purpose, (out / upon)
5. We are looking __________ the problem carefully, (at/into/for)
6. Always keep __________ from danger, (out/away)
7. The thief got __________ with my car. (out / away)
8. An accident brought __________ a change in his life, (about / out)
9. It is bad manners to answer __________. (to, at, back)
10. The robbers broke __________ the bank at night, (down / into)
11. Take __________ the dirty clothes, (of / off / out)
12. The child is looking __________ the birds, (to/at/ for)
13. I rang __________ Aju in the morning, (to / up / for)
14. They are bringing __________ a new book, (about / out / to)
15. Would you care __________ a tea? (for/to / on)
16. They pulled __________ the old house, (off / top/down)
17. The reporters took __________ the speech, (down/off/to)
18. I am looking __________ your problem (to with/up with/in with)
19. I can’t put __________ rude people (to with / up with / in with)
20. The soldiers blew __________ the bridge, (down / off / up)
21. I can’t make __________ the meaning of this word, (to/of / out)
22. She carried __________ singing for a long time, (into/of / on)

CHSE Odisha Class 12 Alternative English Grammar Tense and Aspect

Odisha State Board CHSE Odisha Class 12 Alternative English Solutions Grammar Tense and Aspect Exercise Questions and Answers.

CHSE Odisha Class 12 Alternative English Grammar Tense and Aspect

We have already discussed about ”Tense” in the 1 st year. Let us do some exercises now.
Exercise For Practice (Solved):

1. Fill in the blanks with the correct very forms (Present Tense) from those in brackets.
1. My brother _________ (read) a play by Kalidas.
2. The students _________ (play) much attention to their studies.
3. Who _________ (say), I am the wrong.
4. _________ the birds not _________ (chirp) early in the morning?
5 _________ the students _________ (swim) in the river?
6 _________ your mother not _________ (keep) fit these days?
7 _________ they _________ (refuse) to help you?
8. Puspa _________ not _________ I (iron) her clothes.
9. _________ your sister know how to swim?
10. Rakesh _________ not _________ (take) coffee without sugar.
11. _________ we not _________ (see) many ups and downs in the life?
12. I _________ drop a five rupee note somewhere.
13. _________ I not _________ (invite) him to dinner?
14. Pinki _________ not _________ (keep) awake till midnight these days.
15. Rajeswari _________ (travel) round the world.
16 _________ it _________ (drizzle) since room?
17. She _________ (withdraw) her name from the debate.
18. She _________ (wait) for you for an hour.
19 _________ the maidservant _________ (wash) the floor?
20. It _________ no _________ (rain) her for the last two days.

CHSE Odisha Class 12 Alternative English Solutions Tense and Aspect

Answer:
1. My brother is reading a play by Kalidas.
2. The students are playing much attention to their studies.
3. Who says. I am the wrong.
4. Do the birds not chirp early in the morning?
5. Are the students swimming in the river?
6. Does your mother not keep fit these days?
7. Have they refused to help you?
8. Puspa is not ironing her clothes.
9. Does your sister know how to swim?
10. Rakesh does not take coffee without sugar.
11. Have we not seen many ups and downs in the life?
12. I have dropped a five rupee note somewhere.
13. Am I not inviting him to dinner?
14. Pinki does not keep awake till midnight these days.
15. Rajeswari has traveled round the world.
16. Has it been drizzling since the room?
17. She has withdrawn her name from the debate.
18. She has been waiting for you for an hour.
19. Has the maidservant washed the floor?
20. It has not raining her for the last two days.

Exercise For Practice:
2. Fill in the blanks with correct verb forms (Present Tense) from those in brackets.

1. _________ God not _________ (protect) us all?
2. _________ you sister _________ (pass) the examination?
3. Hari _________ recently _________ (sell) his house.
4. I _________ (read) English for eight years.
5 _________ you _________ graze the cattle since morning?
6. Malaria _________ (rage) in the city for two years.
7. Vegetables and fruits _________ not _________ (harm) us in any way.
8 she not _________ (visit) her home every year?
9. She _________ not _________ (bathe) in hot water during summer.
10. Whom _________ you _________ (like) the most?
11. I _________ (learn) the verses from the Gita.
12. _________ they _________ (travel) by train?
13. Seema _________ not _________ (wash) her clothes.
14. _________ the police not _________ (chose) the thieves?
15. _________ those forests, not _________ (look) green?
16. How _________ you _________ (pull) on with your brother?
17. Who _________ (teach) you since morning?
18. _________ they been _________ (boil) since for ten minutes?
19. Tap _________ not _________ (run) for an hour.
20. Whose umbrella _________ you _________ (use) since last two days?

CHSE Odisha Class 12 Alternative English Solutions Tense and Aspect

Exercise For Practice:
3. Fill in the blanks with correct verb forms (Past Tense) from those in brackets.

1. My father _________ (give) me this present on my birthday.
2. When I _________ (visit) her house, she _________(sleep).
3 _________ Suraj _________ (write) a romantic novel?
4. We _________ (reach) the station before the train _________ (leave).
5. She _________ (sleep) since 8 p.m.
6 _________ Gandhi always _________ (speak) the truth?
7. It _________ (drizzle) since 4 o’ clock.
8. It _________ (rain) heavily at 10 o’ clock.
9. Hari _________ (try) to grind his own axe.
10. The teacher _________ (not) _________ (punish) the naughty boys.
11. I _________ not _________ (talk) to Rahim the other day.
12. He _________ (go) to the post office after the rain _________ (stop).
13. I _________ (wait) for you when the bell _________(ring).
14. _________ the old man _________ (cross) the road very carefully.
15. Mother _________ (prepare) tea for five minutes.
16. Shakil _________ not _________(entertain) the guests with her titbits.
17. The train _________ (run) continuously for four hours.
18. Which God _________ you _________ (worship) in the temple?
19. Ranjana _________ not _________(call) on me last night.
20. In whose house _________ Sheela _________ (stay)?

Answer:
1. My father gave me this present on my birthday.
2. When I visited her house, she was sleeping.
3. Was Suraj writing a romantic novel?
4. We had reached the station before the train left.
5. She had been sleeping since 8 p.m.
6. Did Gandhi always speak the truth?
7. It had been drizzling since 4 o’clock.
8. It had been raining heavily at 10 o’clock.
9. Hari was trying to grind his own axe.
10. The teacher did not punish the naughty boys.
11. I was not talking to Rahim the other day.
12. He want to go to the post office after the rain has stopped.
13. I was waiting for you when the bell rang.
14. Did the old man cross the road very carefully?
15. Mother had been preparing tea for five minutes.
16. Shakil was not entertaining the guests with her titbits.
17. The train had been running continuously for four hours.
18. Which God had you worshipping in the temple?
19. Ranjana did not call on me last night.
20. In whose house was Sheela staying?

CHSE Odisha Class 12 Alternative English Solutions Tense and Aspect

Exercise For Practice:
4. Fill in the blanks with correct verb forms (Past Tense) from those in brackets.

1. You _________ (listen) to Radio for half an hour.
2. Whose clothes _________ you _________ (fold)?
3. Whom _________ you _________ (teach) Grammar?
4. When I _________ (teach), he _________ (doze).
5. It _________ not _________ (rain) when we _________ (leave) for.
6 _________ it _________ heavily at 10 o ’ clock yesterday (rain)?
7. I _________ (read) a novel the whole day long.
8. When you _________ (send) her a telegram?
9 _________ an accident not _________ (take) place here yesterday?
10. The police _________ not _________ (arrest) the thieves knowingly?
11. _________ I _________ (lend) her some money yesterday?
12. He _________ (solve) the difficult sum at once.
13. Mohan _________ not _________ (work) in the worship for several days.
14. _________ he not _________ (knock) at the door for five minutes?
15. Where _________ he _________ (hide) for two days?
16. Which book _________ you _________ (land) _________ to me ?
17. _________ the sun not _________ (set) when the farmers _________ (return) home ?
18. I _________ not _________ (receive) any letter from my uncle.
19. Whose like _________ (fly) high?
20. Who _________ (shatter) this glass into pieces?

BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ

Odisha State Board BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ Textbook Exercise Questions and Answers.

BSE Odisha Class 7 Science Solutions Chapter 11 ଗତି ଓ ସମୟ

Question 1.
ତଳେ ଦିଆଯାଇଥ‌ିବା ପ୍ରତ୍ୟେକ ଗତି କେଉଁ ପ୍ରକାର ଗତି (ସରଳରେଖକ, ବୃତ୍ତାକାର, ଦୋଳନ, ଆବୃତ୍ତି) ବା ଏକାଧ୍ଵ ଗତିର ସମ୍ମିଶ୍ରଣ ଲେଖ ।
(କ) ତୁମେ ସିଧା ରାସ୍ତାରେ ଦୌଡ଼ିଲା ବେଳେ ତୁମ ହାତର ଗତି ।
(ଖ) ଗୋଟିଏ ବିଦ୍ୟୁତ୍ ଘଣ୍ଟିରେ ହାତୁଡ଼ିର ଗତି (ଘଣ୍ଟି ବାଜିଲାବେଳେ) ।
(ଗ) ବର୍ଷା ହେଉଥ‌ିବା ବେଳେ ସିଧା ରାସ୍ତାରେ ଯାଉଥ‌ିବା ଗୋଟିଏ କାର୍‌ର ସାମନା କାଚ ୱାଇପର୍‌ର ଗତି
(ଘ) ପବନ ସାମ୍‌ନାରେ ଥିବା କାଗଜ ତିଆରି ଚକ୍ରିର ଗତି ।
(ଙ) ଗୋଟିଏ ବିଦୁପତ୍ ତାଲିତ ଖେଲନା ଖାହ୍ ଦକାଲିଲା ବେଳେ ହେଲନାର ହାତର ଉଚି |
(ଚ) ଗୋଟିଏ ସିଧା ପୋଲ ଉପରେ ଯାଉଥିବା ଗୋଟିଏ ଟ୍ରେନ୍‌ର ଗତି ।
Solution:
(କ) ଦୋଳନ ଓ ସରଳରେଖ
(ଗ) ଦୋଳନ ଗତି
(ଙ) ଦୋଳନ ଗତି/ଆବୃତ୍ତି
(ଖ) ଦୋଳନ |ଆବୃତ୍ତି
(ଚ) ସରଳରେଖ୍ୟ ଗତି

BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ

Question 2.
ତୁମେ ଦେଖିବା ତଥା ସରଳରେଖ୍କ ଗତି କରୁଥିବା ଗୋଟିଏ ବସ୍ତୁର ନାମ ଲେଖ ।
Solution:

  • ସଳଖ ଓ ସମତଳ ରାସ୍ତାରେ ଗଡ଼ିଯାଉଥ‌ିବା ପେଣ୍ଡୁର ଗତି, ସଳଖ ରାସ୍ତାରେ ଏକ ନିଦ୍ଦିଷ୍ଟ ଦିଗରେ ଗତି କରୁଥ‌ିବା ଯାନ ।
  • ସଳଖ ରାସ୍ତାରେ ଏକ ନିଦିକୁ ବିଗରେ ଗତି କରୁଥିବା ଯାନ |

Question 3.
ଚିତ୍ର ୧୧.୧୧ ବ୍ୟବହାର କରି କାର୍‌ର ବେଗ ନିର୍ଣ୍ଣୟ କର ।
Solution:
BSE Odisha 7th Class Science Solutions Chapter 11 Img 1
୩୦ ମିନିଟ୍‌ରେ କାର୍‌ର ଅତିକ୍ରାନ୍ତ ଦୂରତା ୨୦ କି.ମି. ।
BSE Odisha 7th Class Science Solutions Chapter 11 Img 2

BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ

Question 4.
ସମୟ ଅନୁସାରେ ଗୋଟିଏ କାର୍ ଗତି କରିଥିବା ଦୂରତାର ସାରଣୀ ତଳେ ଦିଆଯାଇଛି ।
BSE Odisha 7th Class Science Solutions Chapter 11 Img 3
ତୁମେ ଆଗରୁ ଆଙ୍କିଥିବା ସମୟ-ଦୂରତା ଗ୍ରାଫ୍‌ରେ ଏଇ ସାରଣୀ ବ୍ୟବହାର କରି ଅନ୍ୟ ଏକ ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଅଙ୍କନ କର ।
(କ) ଦୁଇଟି ଗ୍ରାଫ୍ ମଧ୍ୟରେ କ’ଣ ପ୍ରଭେଦ ଲକ୍ଷ୍ୟ କରୁଛ ?
(ଖ) ପ୍ରତ୍ୟେକ ଗ୍ରାଫ୍‌ X-ଅକ୍ଷ ସହିତ ଆନତ କୋଣ କ’ଣ ସମାନ ?
(ଗ) ଏହି ଆନତ କୋଣ ସହିତ ବସ୍ତୁର ବେଗର କେଉଁ ସଂପର୍କ ଲକ୍ଷ୍ୟ କରୁଛ ଲେଖ ।
Solution:
BSE Odisha 7th Class Science Solutions Chapter 11 Img 4
BSE Odisha 7th Class Science Solutions Chapter 11 Img 5
(i) ପ୍ରଥମ ଗ୍ରାଫ୍‌ଟି ଏକ ସରଳରେଖା, ଏବଂ ଏହା ମୂଳ ବିନ୍ଦୁ O ରୁ ବାହାରିଛି । ଏଥୁରୁ ବସ୍ତୁର ସମବେଗର ସୂଚନା ମିଳିଥାଏ ।
(ii) ଦ୍ବିତୀୟ ଗ୍ରାଫ୍‌ଟି ଏକ ସରଳରେଖା ନୁହେଁ ଏବଂ ଏହା ମୂଳ ବିନ୍ଦୁରୁ ନ ବାହାରି X- ଅକ୍ଷରୁ ବାହାରିଥାଏ । ଏଥୁରୁ ବସ୍ତୁର ଅସମ ବେଗର ସୂଚନା ମିଳିଥାଏ ।
(ଖ) ପ୍ରତ୍ୟେକ ଗ୍ରାଫ୍‌ X- ଅକ୍ଷ ସହିତ ଆନତ କୋଣ ସମାନ ନୁହେଁ ।
(ଗ) ଆନତ କୋଣ କମ୍ ହେଲେ ବେଗ କମ୍ ହୋଇଥାଏ । ଆନତ କୋଣ ଅଧିକ ହେଲେ ବେଗ ଅଧ‌ିକ ହୋଇଥାଏ ।

Question 5.
ଗୋଟିଏ ବସ୍ତୁର ସମୟ -ଦୂରତା ଗ୍ରାଫ୍ ତଳେ ଦିଆଗଲା । ବସ୍ତୁଟି କେଉଁ ପ୍ରକାର ଗତି କରୁଛି ଲେଖ ।
BSE Odisha 7th Class Science Solutions Chapter 11 Img 6
Solution:
ଏଠାରେ ବସ୍ତୁଟିର ଦୂରତା ସମୟ ଅନୁସାରେ ପରିବର୍ତ୍ତନ ହେଉନାହିଁ । ତେଣୁ ବସ୍ତୁଟି ସ୍ଥିର ରହିଥ‌ିବାର ସୂଚନା ମିଳୁଛି ।

Question 6.
ତୁମେ ସାଇକେଲ୍ ଚଳାଇ ଗଲାବେଳେ ତୁମର ବେଗ ୧୨ କି.ମି./ଘଣ୍ଟା । ଗୋଟିଏ ମହୁମାଛି ଉଡ଼ିଲାବେଳେ ତା’ର ବେଗ ୫ ମି./ସେ. । ତୁମର ଓ ମହୁମାଛିର ବେଗ ଭିତରେ କାହାର ବେଗ ଅଧୁକ ଅଟେ ?
Solution:
BSE Odisha 7th Class Science Solutions Chapter 11 Img 7

Question 7.
ଗୋଟିଏ କାର୍ ୧୫ ମିନିଟ୍ କାଳ ୪୦ କି.ମି. |ଘଣ୍ଟା ବେଗରେ ଗତି କଲା । ତା’ପରେ ୨୦ ମିନିଟ୍ କାଳ ୬୦ କି.ମି. | ଘଣ୍ଟା ବେଗରେ ଗତି କଲା । ଏଇ ଯାତ୍ରାରେ କାର୍‌ର ହାରାହାରି ବେଗ କେତେ ?
Solution:
BSE Odisha 7th Class Science Solutions Chapter 11 Img 8

BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ

Question 8.
ନିମ୍ନୋକ୍ତ ମଧ୍ୟରୁ ଭୁଲ ଉକ୍ତିଗୁଡ଼ିକ ସଂଶୋଧନ କରି ତୁମ ଖାତାରେ ଲେଖ ।
(କ) ସମୟର ମୌଳିକ ଏକକ ଘଣ୍ଟା ଅଟେ ।
(ଖ) ଦୁଇଟି ସହର ମଧ୍ୟରେ ଦୂରତା କିଲୋମିଟରରେ ମପାଯାଏ ।
(ଗ) ପ୍ରତ୍ୟେକ ବସ୍ତୁ ଗତି କଲାବେଳେ ସାଧାରଣତଃ ଏକ ସମାନ ବେଗରେ ଗତି କରିଥାନ୍ତି ।
(ଘ) ଗୋଟିଏ ସରଳ ଦୋଳକର ଦୋଳନ ସମୟ ଧ୍ରୁବ ନୁହେ ।
(ଙ) ଟ୍ରେନ୍‌ର ବେଗ ମି./ ଘଣ୍ଟାରେ ପ୍ରକାଶ କରାଯାଏ ।
Solution:
(କ) ସମୟର ମୌଳିକ ଏକକ ସେକେଣ୍ଡ ଅଟେ ।
(ଖ) ଦୁଇଟି ସହର ମଧ୍ୟରେ ଦୂରତା କିଲୋମିଟରରେ ମପାଯାଏ ।
(ଗ) ପ୍ରତ୍ୟେକ ବସ୍ତୁ ଗତି କଲାବେଳେ ସାଧାରଣତଃ ଏକ ଅସମ ବେଗରେ ଗତି କରିଥାନ୍ତି ।
(ଘ) ଗୋଟିଏ ସରଳ ଦୋଳକର ଦୋଳନ ସମୟ ଧ୍ରୁବ ଅଟେ ।
(ଙ) ଟ୍ରେନର ବେଗ କି.ମି.|ଘ.ରେ ପ୍ରକାଶ କରାଯାଏ ।

Question 9.
ବେଗର ପୌଲିକ ଏକକ
BSE Odisha 7th Class Science Solutions Chapter 11 Img 9

Question 10.
ତଳେ ଚାରୋଟି ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଦିଆଯାଇଛି । ସେଥ୍ମଧ୍ୟରୁ କେଉଁଟି ଗୋଟିଏ କାର୍‌ର ଅସମ ଗତିର ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଅଟେ ?
BSE Odisha 7th Class Science Solutions Chapter 11 Img 10
Solution:
(ଗ) ଗ୍ରାଫ୍‌ଟି କାରର ଅସମ ଗତିର ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ।

Question 11.
ଗୋଟିଏ ଗତିଶୀଳ କାରର ଓଡ଼ୋମିଟରର ମାପାଙ୍କ ପୂର୍ବାହ୍ନ ୦୮.୦୦ରେ ୬୩୨୧୯.୦ କି.ମି. ଓ ପୂର୍ବାହ୍ନ ୦୮.୩୫ରେ ୬୩୩୦୯.୦ କି.ମି. ଅଟେ । ଏଇ ସମୟ ଅନ୍ତରାଳରେ କାର୍‌ର ବେଗ କି.ମି. | ମିନିଟ୍ ତଥା କି.ମି. | ଘଣ୍ଟାରେ ପ୍ରକାଶ କର ।
Solution:
ଗତିଶୀଳ କାର୍‌ର ଓଡ଼ୋମିଟରର ମାପାଙ୍କ ପୂର୍ବାହ୍ନ ୮.୦୦ରେ ୬୩୨୧୯.୦ କି.ମି. ଥିଲା ଏବଂ ପୂର୍ବାହ୍ନ ୮.୩୫ରେ ୬୩୩୦୯.୦ କି.ମି. ହେଲା ।
BSE Odisha 7th Class Science Solutions Chapter 11 Img 11

Question 12.
ମନେକର ଯେ ଚିତ୍ର ୧୧.୧ ଓ ୧୧.୨ରେ ଦର୍ଶାଯାଇଥିବା ଫଟୋ ଦୁଇଟି ୧୦ସେ. ଅନ୍ତରାଳରେ ନିଆଯାଇଛି । ଯଦି ଫଟୋ ଗୁଡ଼ିକରେ ୧୦୦ମି. ଦୂରତା ୧ ସେ.ମି. ରୂପରେ ଦର୍ଶାଯାଇଛି ତେବେ ନୀଳକାର୍‌ର ବେଗ ନିର୍ଣ୍ଣୟ କର ।
BSE Odisha 7th Class Science Solutions Chapter 11 Img 12

ବିପଯୁବସ୍ତୁ ସପୂଜାପ ପୂଚନା ଓ ବିଶେଷଣ :

→ ଗତି :

  • ଦିଗ ଓ ଦୂରତା ସମୟ ଅନୁସାରେ ନ ବଦଳିଲେ ବସ୍ତୁଟି ସ୍ଥିର ଏବଂ ଏଥୁରୁ କୌଣସି ଗୋଟିଏ ବି ବଦଳିଲେ ତାହା ଗତିଶୀଳ ।
  • ଗତି ସାଧାରଣତଃ ଚାରି ପ୍ରକାରର ଅଟେ ; ଯଥା – ସରଳରୈଖ୍ୟକ ଗତି, ବୃତ୍ତୀୟ ଗତି, ଆବର୍ତୀ ଗତି ଏବଂ ହୋଇନ ଗତି |

→ ସରଳରୈଖ୍ୟକ ଗତି :
ବସ୍ତୁଟି ଗୋଟିଏ ସରଳରେଖାରେ ଗତି କଲେ ତା’ର ଗତିକୁ ସରଳରେଖ୍ୟ ଗତି କୁହାଯାଏ ।
ଉଦାହରଣ ସ୍ଵରୂପ, ପେଣ୍ଡୁଟିଏ ଗଡ଼େଇଲେ ସେ ସରଳରେଖାରେ ଗତି କରିଥାଏ, ଆଲୋକ ଏକ ସରଳରେଖାରେ ଗତି କରିଥାଏ ।

→ ବୃତ୍ତୀୟ ଗତି :
ଯଦି କୌଣସି ବସ୍ତୁ ବୃତ୍ତାକାର ପଥରେ ଗତି କରୁଥାଏ, ତେବେ ତା’ର ଗତିକୁ ବୃତ୍ତୀୟ ଗତି କୁହାଯାଏ । ଉଦାହରଣ ସ୍ୱରୂପ – ଚକ୍ତିବାଣର ଗତି, ସାଇକେଲ୍ ଚକର ଗତି, ଚକ୍ରିଦୋଳିର ଗତି ଇତ୍ୟାଦି ।

→ ଆବର୍ତ୍ତୀ ଗତି :
ସୂର୍ଯ୍ୟ ଚାରିପାଖରେ ପୃଥ‌ିବୀର ଗତି, ନିଜର ଅକ୍ଷ ଚାରିପାଖରେ ବୁଲୁଥବା ପୃଥ‌ିବୀର ଗତି ଇତ୍ୟାଦି ଆବର୍ତୀ ଗଢିର ଉଦାହରଣ |

→ ଦୋଳନ ଗତି :
ତୁମେ ଦୋଳି ଖେଳୁଥ‌ିବାବେଳେ ଦୋଳିଟି ଏପଟ ସେପଟ ହୋଇ ଅର୍ଥାତ୍‌ ଥରେ ଆଗକୁ ଓ ପଛକୁ ଗତି କରିଥାଏ । ସେହିପରି ଭାବରେ ଘଣ୍ଟାରେ ଥ‌ିବା ଦୋଳକ ଓ ପେଣ୍ଡୁଲମ୍ ଏପଟ ଓ ସେପଟ ହୋଇ ଗତି କରିଥାଏ । ଏହି ପ୍ରକାର ଗତିକୁ ଦୋଳନ ଗତି କୁହାଯାଏ ।

BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ

→ ଧୀର ଅଥବା ଦୃତ ଗତି :

(i) କେତେକ ବସ୍ତୁ ଅନ୍ୟ କିଛି ବସ୍ତୁ ତୁଳନାରେ ଦ୍ରୁତତର ଗତିରେ ଗତି କରିଥାନ୍ତି ।
(ii) ଉଦାହରଣ ସ୍ୱରୂପ ମଟରଗାଡ଼ି, ଶଗଡ଼ଠାରୁ ଦ୍ରୁତତର ଗତିରେ ଗତି କରିଥାଏ । ଏଠାରେ ମଟରଗାଡ଼ି ଦ୍ରୁତ ଗତିରେ ଗତି କରିଥାଏ ଏବଂ ଶଗଡ଼ଗାଡ଼ି ଧୀର ଗତିରେ ଗତି କରିଥାଏ ।
(iii) ଗୋଟିଏ ବସ୍ତୁ ବିଭିନ୍ନ ସମୟରେ ଧୀର ଅଥବା ଦ୍ରୁତ ଗତିରେ ଗତି କରିଥାଏ ।
(iv) ଏକା ସମୟ ଅନ୍ତରାଳରେ ଯେଉଁ ଯାନଟି ଅଧ୍ୟା ଦୂରତା ଅତିକ୍ରମ କରେ ତାହା ଅପେକ୍ଷାକୃତ ଦ୍ରୁତତର ଗତିରେ ଯାଉଛି ବୋଲି କୁହାଯାଏ ।

→ ବେଗ :

  • ଏକକ ସମୟରେ ଗୋଟିଏ ବସ୍ତୁ ଯେତିକି ଦୂରତା ଅତିକ୍ରମ କରେ, ତାହାକୁ ସେ ବସ୍ତୁର ବେଗ କୁହାଯାଏ ।
    BSE Odisha Class 7 Science Solutions Chapter 11 Img 1
  • ଯେକୌଣସି ଯାନ ଗତିଶୀଳ ଥିଲାବେଳେ ଏକ ସମାନ ବେଗରେ ଗତିକରେ ନାହିଁ । ଗତିଶୀଳ ଅବସ୍ଥାରେ ସର୍ବଦା ବେଗ କମ୍ ବା ବେଶି ହୋଇଥାଏ । ତେଣୁ ଗତିଶୀଳ ଯାନର ବେଗ ହେଉଛି ପ୍ରକୃତରେ ତାହାର ହାରାହାରି ବେଗ ।
    BSE Odisha Class 7 Science Solutions Chapter 11 Img 2
  • ଗୋଟିଏ ବସ୍ତୁ ଗତି କଲାବେଳେ ଯଦି ତାର ବେଗ ବାରମ୍ବାର ପରିବର୍ତ୍ତିତ ହୁଏ, ଏପରି ଗତିକୁ ନୈକସମାନ ବା ଅସମଗତି କୁହାଯାଏ ।
  • ଯଦି କୌଣସି କ୍ଷେତ୍ରରେ ଏକ ସରଳରେଖାରେ ଗତି କରୁଥିବା ଗୋଟିଏ ବସ୍ତୁର ବେଗ ଅପରିବର୍ତ୍ତିତ ରହେ, ତେବେ ସେପରି ଗତିକୁ ସମଗତି କୁହାଯାଏ ।
  • ଗୋଟିଏ ବସ୍ତୁର ବେଗ ଗଣନା କରିବା ପାଇଁ ବସ୍ତୁଟି ଗତି କରିଥିବା ମୋଟ ଦୂରତ୍ୱ ଏବଂ ସେହି ଗତି ପାଇଁ ବସ୍ତୁଟି ନେଇଥିବା ମୋଟ ସମୟ ନିର୍ଣ୍ଣୟ କରିବା ଦରକାର । `

→ ସମୟର ମାପ :

  • ଗୋଟିଏ ସୂର୍ଯ୍ୟୋଦୟଠାରୁ ତା ଗୋଟିଏ ଦିନ କୁହାଯାଏ । ଗୋଟିଏ ଦିନ ଠାରୁ କମ୍ ଘଣ୍ଟା ବ୍ୟବହାର କରାଯାଏ ।
  • ପୂର୍ବକାଳରେ ସମୟ ମାପିବା ଯନ୍ତ୍ର ବା ଘଡ଼ି ବ୍ୟବହାର କରାଯାଉଥିଲା ; ଯଥା – ବାଲୁକାଘଡ଼ି, ଜଳଘଡ଼ି, ସୂର୍ଯ୍ୟଘଡ଼ି ଇତ୍ୟାଦି ।
    BSE Odisha Class 7 Science Solutions Chapter 11 Img 3

→ ସରଳ ଦୋଳକ :

  • ସରଳ ଦୋଳକର୍ ସୂତାର ଦୈର୍ଘ୍ୟ ପ୍ରାୟ 1 ମିଟର । ସୂତାରୁ ଝୁଲିଥିବା ଛୋଟ, ଭାରୀ (ସାଧାରଣତଃ ଗୋଲାକାର) ବସ୍ତୁକୁ ଗୋଲକ (Bob) କୁହାଯାଏ ।
  • ଦୋଳକର ବକୁ A ପାର୍ଶ୍ଵକୁ ଟାଣି ଛାଡ଼ିଦେଲେ ବର୍‌ଟି ଦୋଳନ କରି ମାଧ୍ୟ ଅବସ୍ଥାନ ‘O’କୁ ଅତିକ୍ରମ କରି ‘B’ ପାର୍ଶ୍ଵକୁ ଯିବ ।
  • B ପାର୍ଶ୍ଵରୁ ବକ୍‌ଟି ଦୋଳନ କରି ମାଧ ଅବସ୍ଥାନ ଠ ଦେଇ À ପାର୍ଶ୍ଵକୁ ଫେରିବ ।
  • ଏହିପରି ବବ୍ ନିରନ୍ତର ଗତିକରି ଚାଲିବ । ବତ୍‌ର ଏହି ଗତିକୁ ଦୋଳନ ଗତି କୁହାଯାଏ ।
    BSE Odisha Class 7 Science Solutions Chapter 11 Img 4
  • ବଟି ସ୍ଥିତି Aରୁ ସ୍ଥିତି Bକୁ ଯାଇ ପୁଣି ସ୍ଥିତି Aକୁ ଫେରିଆସିବାକୁ ଯେତିକି ସମୟ ନିଏ ତାହା ଏହି ନିର୍ଦ୍ଦିଷ୍ଟ ସରଳ ଦୋଳକର ଦୋଳନ ସମୟ (Period of Oscillation) କୁହାଯାଏ ।

BSE Odisha 7th Class Science Solutions Chapter 11 ଗତି ଓ ସମୟ

→ ସମୟର ଏକକ :

  • ସମୟର ମୌଳିକ ଏକକ ସେକେଣ୍ଡ । ସମୟର ବୃହତ୍ତର ଏକକ ମିନିଟ୍ ବା ଘଣ୍ଟା ।
  • ବୟସ ପ୍ରକାଶ କରିବା ପାଇଁ ବର୍ଷକୁ ଏକକ ରୂପେ ବ୍ୟବହାର କରାଯାଏ ।

→ ବେଗର ଏକକ :
BSE Odisha Class 7 Science Solutions Chapter 11 Img 5

  • ଗୋଟିଏ ମିଟର ଯାନର ବେଗ ଦର୍ଶାଇଥାଏ କାରଣ ସେଥ‌ିରେ କି.ମି.|ଘଣ୍ଟା ଲେଖା ହୋଇଥାଏ । ଏହି ମିଟରକୁ ବେଗ ମିଟର (Speedometer) କୁହାଯାଏ ।
  • ଯାନମାନଙ୍କରେ ମିଟର ମଧ୍ୟ ଲାଗିଥାଏ ଯାହା ଯାନଟି ଯାତ୍ରା କରିଥିବା ଦୂରତ୍ୱ କିଲୋମିଟରରେ ଦର୍ଶାଇଥାଏ । ଅନେକ ସମୟରେ ବେଗ ମିଟରରେ ହିଁ ଏକ ଆୟତାକାର ଜାଗାରେ ଏହି ମିଟର ଥାଏ ଓ ତା’ ଉପରେ କି.ମି. ଲେଖା ହୋଇଥାଏ । ଏହି ମିଟରକୁ ଓଡୋମିଟର କହନ୍ତି ।

→ ଗତିଶୀଳ ବସ୍ତୁର ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ :

  • ଦୁଇଟି ସରଳରେଖାର ଛେଦ ବିନ୍ଦୁକୁ O ମୂଳବିନ୍ଦୁ (Origin) କୁହାଯାଏ । X’OX ଓ Y’OY କୁ ଅକ୍ଷ କୁହାଯାଏ । ମୂଳ ବିନ୍ଦୁର ଡାହାଣକୁ X- ର ମୂଲ୍ୟ ଯୁକ୍ତାତ୍ମକ ଓ ବାମକୁ ବିଯୁକ୍ତାତ୍ମକ ହୋଇଥାଏ । ସେହିପରି ମୂଳବିନ୍ଦୁରୁ ଉପରକୁ Y- ର ମୂଲ୍ୟ ଯୁକ୍ତାତ୍ମକ ଓ ତଳକୁ ବିଯୁକ୍ତାତ୍ମକ ହୋଇଥାଏ । ଗ୍ରାଫ୍ Y-ଅକ୍ଷରେ ନିଆଯାଇଥାଏ ।
    BSE Odisha Class 7 Science Solutions Chapter 11 Img 6
  • ଗୋଟିଏ ବସ୍ତୁର ଗତିକୁ ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଦ୍ଵାରା ଉପସ୍ଥାପନ କରାଯାଏ ।
  • ଏକ ସମାନ ବେଗରେ ଗତି କରୁଥ‌ିବା ବସ୍ତୁର ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଏକ ସରଳରେଖ୍ ଗ୍ରାଫ୍ ଅଟେ ।
  • ଗତିଜନିତ ଅନେକ ପ୍ରଶ୍ନ ସମାଧାନ କରିବାରେ ଗ୍ରାଫ୍ ସାହାଯ୍ୟ କରେ ।

→ ଆସ, ଜାଣିବା :

  • ଏକକ ସମୟରେ ଗୋଟିଏ ବସ୍ତୁ ଗତି କରିଥିବା ଦୂରତାକୁ ବସ୍ତୁର ବେଗ କୁହାଯାଏ ଏବଂ ତାହା ମି./ସେ.ରେ ପ୍ରକାଶ କରାଯାଏ ।
  • ଗୋଟିଏ ବସ୍ତୁ ଗତି କରିଥିବା ଦୂରତାକୁ ଗତି କରିଥିବା ସମୟରେ ଭାଗ କଲେ ତାହା ବସ୍ତୁର ହାରାହାରି ବେଗ ଦେଇଥାଏ । ବେଗର ମୌଳିକ ଏକକ ମି.|ସେ. ଅଟେ ।
  • ବିଭିନ୍ନ ଗତିଶୀଳ ବସ୍ତୁ ମଧ୍ଯରେ କେଉଁ ଗୋଟିକ କ୍ଷିପ୍ରତର ବା କ୍ଷିପ୍ରତମ ଜାଣିବାରେ ଆମକୁ ବେଗ ସାହାଯ୍ୟ କରିଥାଏ ।
  • ସମୟ ମାପିବା ପାଇଁ ଦୋଳନ ପ୍ରକ୍ରିୟା ବ୍ୟବହାର କରାଯାଏ । ସରଳ ଦୋଳକର ଦୋଳନ ଗତି ବ୍ୟବହାର କରି ଘଣ୍ଟା ସାଧାରଣତଃ ତିଆରି କରଯାଏ ।
  • ଗୋଟିଏ ବସ୍ତୁର ଗତିକୁ ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଦ୍ବାରା ଉପସ୍ଥାପନ କରାଯାଏ ।
  • ଏକ ସମାନ ବେଗରେ ଗତି କରୁଥିବା ବସ୍ତୁର ସମୟ-ଦୂରତା ଗ୍ରାଫ୍ ଏକ ସରଳରେଖ୍ ଗ୍ରାଫ୍ ଅଟେ ।

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a)

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a)

Question 1.
ଉକ୍ତିଟି ଠିକ୍ ଥିଲେ T ଏବଂ ଭୁଲ ଥିଲେ F ଲେଖ ।
(i) ଏକ ସମତଳରେ ଥିବା ଏକ ବକ୍ରରେଖାର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁ ଉକ୍ତ ସମତଳ ଉପରିସ୍ଥ ଏକ ଦତ୍ତ ବିନ୍ଦୁଠାରୁ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଦୂରତାରେ ଥିଲେ ବକ୍ରରେଖାଟିକୁ ବୃତ୍ତ କୁହାଯାଏ ।
(ii) ବୃତ୍ତର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁ କୌଣସି ଏକ ବ୍ୟାସାର୍କର ଏକ ପ୍ରାନ୍ତ ବିନ୍ଦୁ ଅଟେ ।
(iii) ଏକ ବୃତ୍ତର ଅସଂଖ୍ୟ ବ୍ୟାସ ରହିଛି ।
(iv) କେନ୍ଦ୍ର, ବୃତ୍ତର ଏକମାତ୍ର ବିନ୍ଦୁ ଯାହା ବୃତ୍ତର ପ୍ରତ୍ୟେକ ବ୍ୟାସ ଉପରେ ଅବସ୍ଥିତ ।
(v) ଏକ ଜ୍ୟା ବୃତ୍ତର ଅନ୍ତର୍ଦେଶକୁ ଯେଉଁ ଦୁଇ ଅଂଶରେ ବିଭକ୍ତ କରେ ସେମାନେ ପ୍ରତ୍ୟେକ ଉତ୍ତଳ ସେଟ୍ ଅଟନ୍ତି ।
(vi) ବୃତ୍ତର ଏକ ବ୍ୟାସ ଗୋଟିଏ ଜ୍ୟାକୁ ସମଦ୍ବିଖଣ୍ଡ କଲେ ସେମାନେ ପରସ୍ପର ପ୍ରତି ଲମ୍ବ ଅଟନ୍ତି ।
(vii) ପ୍ରତ୍ୟେକ ତ୍ରିଭୁଜର ପରିକେନ୍ଦ୍ର ଏହାର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ ।
(vii) ଏକ ବୃତ୍ତର କେନ୍ଦ୍ର, ଏହାର ଏକମାତ୍ର ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ ଯାହାଠାରୁ ବୃତ୍ତର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁର ଦୂରତା ସମାନ ।
(ix) ଏକ ରଶ୍ମି ବୃତ୍ତକୁ ଗୋଟିଏ ମାତ୍ର ବିନ୍ଦୁରେ ଛେଦ କରେ । ତେବେ ରଶ୍ମିର ଆଦ୍ୟ ବିନ୍ଦୁଟି ବୃତ୍ତର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ
(x) ଏକ ତ୍ରଭରେ \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{BC}}\) ହୁକଟି ସବସମ କ୍ୟା 6ହ6ଲ B ଦିନ୍ଦନାମ କ୍ୟାମଣ ∠ABC କୁ ସମଦ୍ୱଖଣ୍ଡ ହେବ ।
(xi) ଗୋଟିଏ ବିନ୍ଦୁ ଦୁଇ ବା ତତୋଽଧ୍ଵକ ବୃତ୍ତର କେନ୍ଦ୍ର ହୋଇପାରିବ ନାହିଁ ।
(xii) ଗୋଟିଏ ସରଳରେଖା ଗୋଟିଏ ବୃତ୍ତରକୁ ସର୍ବଦା ଦୁଇଟି ବିନ୍ଦୁରେ ଛେଦ କରେ ।
Solution:
(i) F (ବକ୍ରରେଖାଟି ଏକ ବୃତ୍ତର ଚାପ ହୋଇପାରେ ।)
(ii) T (ଏକ ବ୍ୟାସାର୍କର ଦୁଇଟି ପ୍ରାନ୍ତ ବିନ୍ଦୁ ମଧ୍ୟରୁ ଗୋଟିଏ ବିନ୍ଦୁ କେନ୍ଦ୍ର ଓ ଅନ୍ୟଟି ବୃତ୍ତ ଉପରିସ୍ଥ ।)
(iii) T (ବୃତ୍ତ ଉପରିସ୍ଥ )
(iv) F
(v) T
(vi) T (ବ୍ୟାସରେ କେନ୍ଦ୍ର ଅବସ୍ଥିତ । କେନ୍ଦ୍ର ଓ ଜ୍ୟାର ମଧ୍ୟବିନ୍ଦୁକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡ ଜ୍ୟା ପ୍ରତି ଲମ୍ବ ।)
(vii) F (ସ୍ଥୂଳକୋଣୀ ତ୍ରିଭୁଜ ଓ ସମକୋଣୀ ତ୍ରିଭୁଜର ପରିକେନ୍ଦ୍ର ତ୍ରିଭୁଜର ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ ନୁହେଁ ।)
(viii) T (ବୃତ୍ତର ସଂଜ୍ଞା ଅନୁସାରେ ।)
(ix) F (ରଶ୍ମିଟି ବୃତ୍ତପ୍ରତି ସ୍ପର୍ଶକ ହେବ ।)
(x) T (O ବୃତ୍ତର କେନ୍ଦ୍ର ହେଲେ A OAB = A OCB ହେବ ।
(xi) F (ଏକକୈନ୍ଦ୍ରିକ ବୃତ୍ତମାନଙ୍କର ଗୋଟିଏ କେନ୍ଦ୍ରବିନ୍ଦୁ ।)
(xii) F (ଆଦୌ ଛେଦ ନକରିପାରେ ବା ଗୋଟିଏ ବିନ୍ଦୁରେ ଛେଦ କରିପାରେ ।)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a)

Question 2.
ପ୍ରଦତ୍ତ ସମ୍ଭାବ୍ୟ ଉତ୍ତରରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

(i) ଦୁଇଟି ଅସମାନ୍ତର ଜ୍ୟାର ଛେଦବିନ୍ଦୁ …………………. ଅଟେ ।
(a) ବୃତ୍ତର ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ
(b) ବୃତ୍ତର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ
(c) ବୃତ୍ତ ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ
(d) ବୃତ୍ତ ଉପରିସ୍ଥ କିମ୍ବା ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ
Solution:
ହଉ ଭପରିମ୍କ କିମ୍ଵା ଆନ୍ତଃମ ଦିନ୍ଦୁ

(ii) P ବିନ୍ଦୁ ବୃତ୍ତର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ ହେଲେ ବୃତ୍ତ ଉପରେ P ଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ……………………… ଯୋଡ଼ା ବିନ୍ଦୁ ଅଛି ।
(a) 1
(b) 2
(c) 8
(d) ଅସଂଖ୍ୟ
Solution:
ଅସଂଖ୍ୟ

(iii) 6ଗାଟିଏ ରେଖାଖଣ୍ଡ ସଦାଧକ ………….. ଟି ବୃତ୍ତର ଜ୍ୟା ହୋଇପାରିବ ।
(a) 1
(b) 2
(c) 4
(d) ଅସଂଖ୍ୟ
Solution:
2

(iv) 6ଗାଟିଏ ରେଖାଖଣ୍ଡ ସଦାଧକ ………….. ବ୍ୟାସାର୍ଦ୍ଧ ହୋଇ ପାରିବ ।
(a) 1
(b) 2
(c) 4
(d) ଅସଂଖ୍ୟ
Solution:
ଅଫଖ୍ୟ

(v) ଗୋଟିଏ ବୃତ୍ତରେ ଏକ ଜ୍ୟାର ଗୋଟିଏ ପ୍ରାନ୍ତବିନ୍ଦୁ କେନ୍ଦ୍ରଠାରୁ 5 ସେ.ମି. ଦୂରରେ ଏବଂ ଜ୍ୟାଟିର ମଧ୍ୟବିନ୍ଦୁ କେନ୍ଦ୍ରଠାରୁ 3 ସେ.ମି. ଦୂରରେ ଅଛି । ଜ୍ୟାଟିର ଦୈର୍ଘ୍ୟ …………….. 6ପ.ମି.
(a) 8
(b) 12
(c) 16
(d) 20
Solution:
8

Question 3.
ଏକଭର 16 6ସ.ମି. ଦେଣ୍ୟ ବିଶିଷ୍ଟ 6ଟାଟିଏ ବ୍ୟା ଏକ ବ୍ୟାସାଶୁ \(\overline{OP}\) ଦାରା D ଜିନ୍ଦୁ 6ର ପମଦ୍ୱିଖଣ୍ଡତ ହୁଏ | ହୁଇର ଦ୍ୟାସାରୁ 10 6ସ.ମି. 6ଦ୍ର6କ DP ର 6ଦିଶ୍ୟ ନିଣ୍ଟଯ କର |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 1
Solution:
ଦକ ହଉ6ର AB ର୍ଯ୍ୟର 6ଦିଣ୍ୟ = 16 ପେ.ମି.
⇒ AD = \(\frac { 16 }{ 2 }\) ପେ.ମି. = 8 ପେ.ମି. (AD = \(\frac { 1 }{ 2 }\) AB)
ହଉର ଦ୍ୟାସାର୍ଦ (OA) = 10 ପେ.ମି. = OP
∴ △ODA 6ର OD = \(\sqrt{\mathrm{OA}^2-\mathrm{AD}^2}\) = \(\sqrt{\mathrm{10}^2-\mathrm{8}^2}\) = \(\sqrt{100-64}\) = \(\sqrt{36}\) = 6 6ସ.ମି. |
DP = OP – OD = 10 6ସ.ମି. – 6 6ସ.ମି. = 4 6ସ.ମି. |
∴ \(\overline{DP}\) ର 6ଦିଖ୍ୟ 4 6ସ.ମି. |

Question 4.
6ଟାଟିଏ ତ୍ରଭର 6କହ୍ O | ଏକ ଲ୍ୟା \(\overline{\mathbf{AB}}\) ର ମଧ୍ୟଦିନୁ D 6ହ6ଲ ପ୍ରମାଣୀ କର ଯେ \(\overline{\mathbf{OD}}\) , ∠AOB କୁ ସମଦ୍ୱିଖଣ୍ କ6ର |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 2
ଦଇ : ହଉର 6କହ O | ଖ୍ୟା AB ର ମଧ୍ୟଦିନ୍ଦୁ D |
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathbf{OD}}\), ∠AOB କୁ ସମଦିଖଣ୍ଡକ କ6ର |
ପ୍ରମାଣ : △AOD ଓ △BOD ମଧ୍ୟ6ର
AO = BO (ଏକା ଦୃତ୍ତର ବନ୍ଦ୍ୟାସାଦଁ)
AD = BD (ଦଇ)
\(\overline{\mathbf{OD}}\) ସାଧାରଣ ଦାନ୍ଦୁ
△AOD ≅ △BOD (ଦା.ଦା.ଦା. ସଦପମତା)
⇒ ∠AOD ≅ ∠BOD (ର୍ଥ ନୁର୍ପ କୋଣ)
⇒ \(\overline{\mathbf{OD}}\) , ∠AOB କୁ ସମଦ୍ୱିଖଣ୍ଡ କରେ |

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a)

Question 5.
6ଟାଟିଏ ତ୍ରଭର 6କହ୍ O | ଏକ ଲ୍ୟା \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{AC}}\) ର ମଧ୍ୟଦିନୁ D 6ହ6ଲ ପ୍ରମାଣୀ କର ଯେ \(\overline{\mathbf{OA}}\) , ∠BAC କୁ ସମଦ୍ୱିଖଣ୍ କ6ର |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 3
Solution:
ଦଇ : O, ABC ହଉର 6କହ ଏକ AB = AC | (ତ୍ୟାଦୟ ସଦଂସମ)
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathbf{OA}}\), ∠BAC କୁ ସମଦିଖଣ୍ଡ କ6ର |
ଆଥାତ୍ m∠OAB = m∠OAC
ର୍ଥଙନ : \(\overline{\mathbf{OB}}\) ଏବଂ \(\overline{\mathbf{OC}}\) ଅଳନ କର |
ପ୍ରମାଣ : △AOB ଓ △AOC ମଧ୍ୟରେ
AB = AC (ଦଇ)
OB = OC (ଗୋଟିଏ ହଭର ବ୍ୟାପୀ ବଂ)
\(\overline{\mathbf{OA}}\) ସାଧାରଣ ଦାନ୍ଦୁ
△AOB ≅ △AOD (ଦା.ଦା.ଦା. ସଦପମତା)
⇒ m∠OAB = m∠OAC (ର୍ଥ ନୁର୍ପ କୋଣ)

Question 6.
ଗୋଟିଏ ବୃତ୍ତର କେନ୍ଦ୍ର O ଏବଂ \(\overline{\mathbf{AB}}\) ଓ CD ଏହାର ଦୁଇଟି ସମାନ୍ତର ଜ୍ୟା । P ଓ Q ଯଥାକ୍ରମେ AB ଓ CDର ମଧ୍ୟବିନ୍ଦୁ ହେଲେ ପ୍ରମାଣ କର ଯେ ( ବିନ୍ଦୁ, \(\stackrel{\leftrightarrow}{P}\) ଉପରିସ୍ଥ ହେବ ।
Solution:
ଦତ୍ତ : AB ଓ CD ବୃତ୍ତର ଯେକୌଣସି ଦୁଇଟି ସମାନ୍ତର ଜ୍ୟା । P ଓ Q ଯଥାକ୍ରମେ AB ଏବଂ CD ଜ୍ୟାର ମଧ୍ୟବିନ୍ଦୁ ।
ପ୍ରାମାଣ୍ୟ : ବୃତ୍ତର କେନ୍ଦ୍ର ‘O’, \(\overline{\mathbf{PQ}}\) ଉପରିସ୍ଥ ହେବ ।
ପ୍ରମାଣ : ମନେକର ବୃତ୍ତର କେନ୍ଦ୍ର ‘O’, PQ ଉପରିସ୍ଥ ନୁହେଁ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 4
OP ଓ \(\overline{\mathbf{OQ}}\) ଅଙ୍କନ କର । O ବିନ୍ଦୁରେ AB ସହ ସମାନ୍ତର କରି \(\overrightarrow{\mathrm{OR}}\) ଅଙ୍କନ କର।
m∠APO = 90° (∵ 0 କେନ୍ଦ୍ର ଏବଂ AB ଜ୍ୟାର ମଧ୍ୟବିନ୍ଦୁ P)
∴ m∠POR = 90° (∵ AB || OR ଏବଂ \(\overline{\mathbf{PO}}\) ହେଲେ)
ସେହିପରି m∠CQO = 90° ଏବଂ m∠ROQ = 90° (\(\overline{\mathbf{CD}}\) || \(\overline{\mathbf{OR}}\) ଏବଂ QO ଛେଦକ)
∴m∠POR + m∠ROQ = 180°
⇒ P, O ଓ Q ଏକରେଖୀୟ ।
⇒ O, \(\overline{\mathbf{PQ}}\) ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ ହେବ ।

Question 7.
ଗୋଟିଏ ସମବାହୁ ତ୍ରିଭୁଜର ପରିକେନ୍ଦ୍ରଠାରୁ ତ୍ରିଭୁଜର ବାହୁମାନେ ସମଦୂରବର୍ତ୍ତୀ – ପ୍ରମାଣ କର ।
Solution:
ଆମେ ଜାଣୁ ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ସମାନ ।
ପୁନଶ୍ଚ ଏକ ବୃତ୍ତରେ ସମାନ ଦୈର୍ଘ୍ୟବିଶିଷ୍ଟ ଜ୍ୟାମାନ କେନ୍ଦ୍ରଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।
ତେଣୁ ସମବାହୁ ତ୍ରିଭୁଜର ପରିକେନ୍ଦ୍ରଠାରୁ ତ୍ରିଭୁଜର ବାହୁମାନ ସମଦୂରବର୍ତ୍ତୀ । (ପ୍ରମାଣିତ)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 5
ବିକଳ୍ପ ପ୍ରମାଣ :
ଦତ୍ତ : A ABCର ବାହୁମାନ କେନ୍ଦ୍ରଠାରୁ ସମୟଦୂରବର୍ତ୍ତୀ ।
ପ୍ରମାଣ : \(\overline{\mathbf{AB}}\), BC ଓ CA ର ସମଦ୍ଵିଖଣ୍ଡକ ଲମ୍ବମାନଙ୍କର ଛେଦବିନ୍ଦୁ O ।
ତେଣୁ O ପରିବୃତ୍ତର କେନ୍ଦ୍ର ।
AB, BC ଓ CA ହେତୁ ଜ୍ୟାମାନ କେନ୍ଦ୍ରଠାରୁ ସମଦୂରବର୍ତ୍ତୀ
କାରଣ AB = BC = CA (ଉପପାଦ୍ୟ – 8)

Question 8.
ପ୍ରମାଣ କର ଯେ ବୃତ୍ତରେ ଏକ ବ୍ୟାସ ଏହାର ବୃହତ୍ତମ ଜ୍ୟା । (ସୂଚନା : ଏକ କ୍ୟାର କେନ୍ଦ୍ରଠାରୁ ଦୂରତା d ≥ 0 ) ଏବଂ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r ହେଲେ, ଜ୍ୟାର ଦୈର୍ଘ୍ୟ \(2 \sqrt{r^2-d^2} \leq 2 r\) = ବ୍ୟାସ) ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 6
ଦତ୍ତ : ABC ବୃତ୍ତର \(\overline{\mathbf{AC}}\) ବ୍ୟାସ । O ବୃତ୍ତର କେନ୍ଦ୍ର ।
ପ୍ରାମାଣ୍ୟ : AC ବୃହତ୍ତମ ଜ୍ୟା ।
ଅଙ୍କନ : AB ଜ୍ୟା ଅଙ୍କନ କର । OD ⊥ AB ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △AODରେ m∠ADO = 90°
⇒ AO > AD
⇒ 2AO > 2AD ⇒ AC > AB
ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରେ ଯେ, \(\overline{\mathbf{AC}}\) ର ଦୈର୍ଘ୍ୟ
ଅନ୍ୟ ଯେକୌଣସି ଜ୍ୟାର ଦୈର୍ଘ୍ୟଠାରୁ ବୃହତ୍ତର ।
∴ AC ବ୍ୟାସ ବୃତ୍ତର ବୃହତ୍ତମ ଜ୍ୟା । (ପ୍ରମାଣିତ)

Question 9.
ଗୋଟିଏ ବୃତ୍ତରେ ଦୁଇଟି ସମାନ୍ତର ଜ୍ୟାର ଏକ ପାର୍ଶ୍ବରେ ବୃତ୍ତର କେନ୍ଦ୍ର ଅବସ୍ଥିତ । ପ୍ରମାଣ କର ଯେ ଜ୍ୟା ଦ୍ଵୟ ସର୍ବସମ ନୁହଁନ୍ତି ।
Solution:
S ବୃତ୍ତର O କେନ୍ଦ୍ର । କେନ୍ଦ୍ରର ଏକ ପାର୍ଶ୍ବରେ AB ଓ CD ଦୁଇଟି ଜ୍ୟା | AB || CD |
ପ୍ରାମାଣ୍ୟ : AB ≠ CD ଅର୍ଥାତ୍ AB ଓ CD ଜ୍ୟା ଦ୍ଵୟ ସର୍ବସମ ନୁହଁନ୍ତି ।
ଅଙ୍କନ : OM ⊥ CD ଅଙ୍କନ କର ଏବଂ OA ଓ OC ଅଙ୍କନ କର ।
ପ୍ରମାଣ : OM ⊥ AB ⇒ OM ⊥ CD
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 7
△OAM ରେ OA2 = OM2 + AM2 ….(i)
△ONC ରେ OC2 = ON2 + CN2 ….(ii)
OA = OC 6ହତ୍ର (i) ଓ (ii) ବ୍ ପାଲାଟା OM2 + AM2 = ON2 + CN2
⇒ AM2 – CN2 = ON2 – OM2 > 0 (∵ ON > OM)
⇒ CN < AM = \(\frac { 1 }{ 2 }\) CD < \(\frac { 1 }{ 2 }\) Ab
⇒ CD < AM ⇒ AB ≠ CD (ପ୍ରମାଣିତ) (∵ ON > OM ⇒ CD < AB)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a)

Question 10.
AB ଓ CD ଗୋଟିଏ ବୃତ୍ତର ଦୁଇଟି ସମାନ୍ତର ଜ୍ୟା । AB = CD = 8 ସେ.ମି. । ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 5 ସେ.ମି. ହେଲେ ଜ୍ୟାଦ୍ଵୟର ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା ନିର୍ଣ୍ଣୟ କର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 8
AB = CD ଓ AB || CD
ହେତୁ ଜ୍ୟାଦ୍ଵୟ କେନ୍ଦ୍ରର ବିପରୀତ ପାର୍ଶ୍ଵରେ ରହିବେ ।
AB = 8 6ସ.ମି. ⇒ BP \(\frac { 1 }{ 2 }\) × 8 ସେ.ମି. = 4 ସେ.ମି. |
OB = 5 ସେ.ମି., OP ⊥ AB ଓ OQ ⊥ CD ହେଉ ।
POB ସମ6କାଣା ତ୍ରିକୁଲ6ର OP = \(\sqrt{\mathrm{OB}^2-\mathrm{BP}^2}\) = \(\sqrt{5^2-4^2}\) = \(\sqrt{9}\) = 3
∴ PQ = OP + OQ = 3 ସେ.ମି. + 3 ସେ.ମି. = 6 ସେ.ମି. |
∴ ଜ୍ୟାଦ୍ବୟର ମଧ୍ୟବର୍ତ୍ତୀ ଦୂରତା 6 ସେ.ମି. |

Question 11.
10 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଗୋଟିଏ ବୃତ୍ତର ଦୁଇଟି ସମାନ୍ତର ଜ୍ଯା \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{CD}}\) ମଧ୍ୟରେ ଦୂରତା 10 ସେ.ମି. | \(\overline{\mathbf{AB}}\) କ୍ୟା କେନ୍ଦ୍ରଠାରୁ 6 ସେ.ମି. ଦୂରରେ ଅବସ୍ଥିତ ହେଲେ \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{CD}}\) ର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 9
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = 10 ସେ.ମି.
\(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{CD}}\) ମଧ୍ୟରେ ଦୂରତା = 10 ସେ.ମି. |
∴ \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{CD}}\) କେନ୍ଦ୍ରର ବିପରୀତ ପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ ।
ମନେକର O ବୃତ୍ତର କେନ୍ଦ୍ର ।
OP ⊥ AB ଓ OQ ⊥ CD |
PQ = 10 ସେ.ମି. , OP = 6 ସେ.ମି.
∴ OQ = (10 – 6) 6 ସେ.ମି. . = 4 ସେ.ମି. |
A OBP ରେ BP = \(\sqrt{\mathrm{OB}^2-\mathrm{OP}^2}\) = \(\sqrt{10^2-6^2}\) ସେ.ମି. = \(\sqrt{64}\) ସେ.ମି. = 8 ସେ.ମି.
⇒ AB = 2BP = 16 ସେ.ମି. |
△OQD ରେ QD = \(\sqrt{\mathrm{OD}^2-\mathrm{OQ}^2}\) = \(\sqrt{10^2-4^2}\) ସେ.ମି. = \(\sqrt{84}\) ସେ.ମି.
= 2√21 ସେ.ମି. |
∴CD = 2QD = 4√21 ସେ.ମି. |

Question 12.
ଗୋଟିଏ ବୃତ୍ତରେ △ABC ଅନ୍ତର୍ଲିଖ୍ ହୋଇଛି । ଯଦି AB = ÀC ହୁଏ, ପ୍ରମାଣ କର ଯେ ∠BACର ସମଦ୍ବିଖଣ୍ଡକ ରଶ୍ମି ବୃତ୍ତର କେନ୍ଦ୍ର ବିନ୍ଦୁଗାମୀ ଅଟେ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 10
: O, ABC ଦ୍ରଦ୍ଭର କେନ୍ଦ୍ର ଏବଂ AB = AC |
ପ୍ରାମାଣ୍ୟ : AO, ZBACର ସମଦ୍ଵିଖଣ୍ଡକ ।
ଅଙ୍କନ : OB ଓ OC ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △ABO ଓ △ACO ମଧ୍ୟରେ
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 11
AB = AC (ଦତ୍ତ)
AO (ସାଧାରଣ ବାହୁ)
OB = OC (ଏକା ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ)
⇒ △ABO ≅ △ACO (ବା. ବା. ଦା . ସବଂସମତା)
m∠BAO = m∠CAO (ଅନୁରୁପ କୋଣ)
⇒ ∠BACର ସମଦ୍ବିଖଣ୍ଡକ ରଶ୍ମି \(\overrightarrow{\mathrm{AO}}\) ବୃତ୍ତର କେନ୍ଦ୍ରବିନ୍ଦୁଗାମୀ ।

Question 13.
ଗୋଟିଏ ବୃତ୍ତର ଦୁଇଟି ଜ୍ୟା ଏକ ବ୍ୟାସ ଦ୍ବାରା ସମଦ୍ବିଖଣ୍ଡିତ ହେଲେ ପ୍ରମାଣ କର ଯେ ଜ୍ୟା ଦୁଇଟି ସମାନ୍ତର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 12
ଦତ୍ତ : ବୃତ୍ତର କେନ୍ଦ୍ର O । PQ ଓ RS ଜ୍ୟା ଦ୍ବୟ XY ବ୍ୟାସଦ୍ବାରା ଯଥାକ୍ରମେ M ଓ N ବିନ୍ଦୁରେ ସମଦ୍ବିଖଣ୍ଡିତ
ପ୍ରାମାଣ୍ୟ : PQ||\(\overline{\mathbf{RS}}\)
ପ୍ରମାଣ : PQ ର ମଧ୍ୟବିନ୍ଦୁ M ।
⇒ OM ⊥ PQ ⇒ m∠QMO = 90°
ସେହିପରି RS ର ମଧ୍ୟବିନ୍ଦୁ N |
⇒ ON ⊥ RS ⇒ m∠ONR = 90° |
∴ m∠QMO = m∠ONR = 90° |
ମାତ୍ର ଏହି କୋଣଦ୍ଵୟ ଏକାନ୍ତର, ତେଣୁ PQ || RS |

Question 14.
ପ୍ରମାଣ କର ଯେ ଗୋଟିଏ ବୃତ୍ତର ଦୁଇଟି ଜ୍ୟା କେନ୍ଦ୍ର ହେବ । (ସୂଚନା : ଅସମ୍ଭବାୟବ ପ୍ରଣାଳୀ ପରସ୍ପରକୁ ସମଦ୍ବିଖଣ୍ଡ କଲେ ସେମାନଙ୍କ ଛେଦବିନ୍ଦୁ ବୃତ୍ତର (Method of contradiction) ବ୍ୟବହାର କର)
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 13
ଦତ୍ତ : PQ ଓ RS ଜ୍ୟା ଦ୍ବୟ ପରସ୍ପରକୁ O ବିନ୍ଦୁରେ ସମଦ୍ଵିଖଣ୍ଡ କରନ୍ତି ।
ପ୍ରାମାଣ୍ୟ : ‘O’ ବୃତ୍ତର କେନ୍ଦ୍ର ।
ପ୍ରମାଣ : ମନେକର ‘O’ ବୃତ୍ତର କେନ୍ଦ୍ର ନୁହେଁ । O’ ବୃତ୍ତର କେନ୍ଦ୍ର ହେଉ । O’ O ଅଙ୍କନ କର ।
O, RS ଜ୍ୟାର ମଧ୍ୟବିନ୍ଦୁ ⇒ m∠O′OS = 90°
ସେହିପରି O, PQ ଜ୍ୟାର ମଧ୍ୟବିନ୍ଦୁ ⇒ m∠O’OQ = 90°
∴ m∠O’OS = m∠O’OQ = 90°
କିନ୍ତୁ ଏହା ଅସମ୍ଭବ ।
କାରଣ Q ଓ S ବିନ୍ଦୁ O’O ର ଏକ ପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ ।
∴ O’ ଓ O ବିନ୍ଦୁଦ୍ଵୟ ଏକ ଓ ଅଭିନ୍ନ ।
⇒ PQ ଓ RS ଜ୍ୟା ଦ୍ଵୟର ଛେଦବିନ୍ଦୁ, ‘O’ ବୃତ୍ତର କେନ୍ଦ୍ର ହେବ ।

15. ଗୋଟିଏ ବୃତ୍ତର ଦୁଇଟି ଜ୍ଯା AB ଓ BC, B ଠାରେ ୨୦ କୋଣ ଉତ୍ପନ୍ନ କରନ୍ତି । ବୃତ୍ତର କେନ୍ଦ୍ର O ହେଲେ ପ୍ରମାଣ କର ଯେ A, O ଏବଂ C ଏକ ଏକରେଖୀୟ ।
Solution:
ଦତ୍ତ : ( କେନ୍ଦ୍ର ବିଶିଷ୍ଟ ଏକ ବୃତ୍ତର \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{BC}}\) ଦୁଇଟି ଜ୍ୟା ।
m∠ABC = 90°
ପ୍ରାମାଣ୍ୟ : A – 0 – C ଅର୍ଥାତ୍ \(\overline{\mathrm{AC}}\) ବ୍ୟାସ ।
ଅଙ୍କନ : \(\overline{\mathrm{BO}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △ABO ରେ AO = BO ⇒ m∠OAB = m∠ABO = θ (ମ6ନକର)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 14
△BOC ରେ BO = CO ⇒ m∠OBC = m∠OCB = α (ମ6ନକର)
∴ θ + α = 90° (·.· m∠ABC = 90°)
△ABO ରେ m∠AOB = 180° – 2θ ଏବଂ △BOC ରେ m∠BOC = 180° – 2α
∴ m∠AOB +m∠BOC = 360 – 2(θ + α) = 360° – 2(90°) (∵ θ + α = 90°) = 180°
⇒ A – O – C = \(\overline{\mathrm{AC}}\) ଏକ ବ୍ୟାସ ।

Question 16.
ପ୍ରମାଣ କର ଯେ ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜରେ କର୍ପୂର ମଧ୍ୟବିନ୍ଦୁ, ଏହାର ପରିବୃତ୍ତର କେନ୍ଦ୍ର ଅଟେ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 15
ଦତ୍ତ : △ABC ର m∠ABC = 90° । ‘O’ \(\overline{\mathrm{AC}}\) କର୍ପୂର ମଧ୍ୟବିନ୍ଦୁ ।
ପ୍ରାମାଣ୍ୟ : A ABC ର ପରିବୃତ୍ତର କେନ୍ଦ୍ର O ।
ଅଙ୍କନ : \(\overrightarrow{\mathrm{BO}}\) ଅଙ୍କନ କର । \(\overrightarrow{\mathrm{BO}}\) ଉପରିସ୍ଥ ‘D’ ଏପରି ଏକ ବିନ୍ଦୁ ଯେପରିକି BO = OD |
DC ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △OBC ଓ △ODC ଦ୍ଵୟରେ AO = OC (∵ O, \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟବିନ୍ଦୁ)
BO = OD (ଅଙ୍କନ), m∠AOB = m∠COD (ପ୍ରତୀପ)
△ABO = △CDO ⇒ m∠BAO = m∠OCD 19° AB = CD
⇒ ମାତ୍ର ଏହି କୋଣଦ୍ଵୟ ଏକାନ୍ତର ⇒ \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{CD}}\)
ପୁନଶ୍ଚ, m∠ABC = 90° ହେତୁ ABCD ଏକ ଆୟତଚିତ୍ର ।
∴ AC = BD = \(\frac { 1 }{ 2 }\) AC = \(\frac { 1 }{ 2 }\) BD ⇒ AO = BO
∴ AO = BO = CO
ଏଠାରେ ଠ ବିନ୍ଦୁଠାରୁ A, B ଓ C ବିନ୍ଦୁତ୍ରୟ ସମଦୂରବର୍ତ୍ତୀ ।
⇒ O ବିନ୍ଦୁ △ABC ର ପରିବୃତ୍ତର କେନ୍ଦ୍ର ।

Question 17.
PQ ବୃତ୍ତର ଜ୍ୟା । P ଓ Q ଠାରେ ଉକ୍ତ ଜ୍ୟା ରେ ଅଙ୍କିତ ଲମ୍ବଦ୍ଵୟ R ଓ S ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । ପ୍ରମାଣ କର ଯେ PQSR ଏକ ଆୟତ ଚିତ୍ର |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 16
ଦତ୍ତ : \(\overline{\mathrm{PQ}}\) ଜ୍ୟାର P ଓ Q ଠାରେ ଅଙ୍କିତ ଲମ୍ବଦ୍ଵୟ ବୃତ୍ତକୁ R ଓ S ବିନ୍ଦୁରେ ଛେଦ କ6ର |
ପ୍ରାମାଣ୍ୟ : PQSR ଏକ ଆୟତଚିତ୍ର ।
ଅଙ୍କନ : \(\overline{\mathrm{RQ}}\) ଓ \(\overline{\mathrm{PS}}\) ଅଙ୍କନ କର ।
ପ୍ତମାଣ : △PRQ ରେ m∠RPQ = 90° ⇒ RQ ଏବ ଦ୍ୟାସ ….(i)
△SPQ ରେ m∠PQS = 90° ⇒ PQ ଏବ ଦ୍ୟାସ ….(ii)
ଆମେ ଜାଣିଛେ, ବ୍ୟାସଦ୍ଵୟ ପରସ୍ପରକୁ କେନ୍ଦ୍ର ‘O’ରେ ସମଦ୍ଵିଖଣ୍ଡ କରନ୍ତି ।
∴ PORS ଏକ ଆୟତଚିତ୍ର ।

ବିକଳ୍ପ ପ୍ରମାଣ:
PQRS ଦ୍ରଭନ୍ନକଖତ ଚତୁରୁଢର m∠P + m∠S = 180°
⇒ m∠S = 90° (∴ \(\overline{\mathrm{RQ}}\) ବୃତ୍ତର ଏକ ବ୍ୟାସ)
⇒ m∠P = 90°, 6ସଦ୍ରପତି m∠R = 90° |
∴ PQSR ଏକ ଆୟତଚିତ୍ର ।

BSE Odisha 10th Class Maths Solutions Geometry Chapter 2 ବୃତ୍ତ Ex 2(a)

Question 18.
ଚିତ୍ରରେ A ଓ B ଦୁଇଟି ପରସ୍ପର ଛେଦୀ ବୃତ୍ତର କେନ୍ଦ୍ର ଏବଂ P ଓ Q ବୃତ୍ତ ଦ୍ଵୟର ଛେଦବିନ୍ଦୁ ଅଟନ୍ତି । ପ୍ରମାଣ କର ଯେ,
(i) \(\stackrel{\leftrightarrow}{\mathbf{AB}}\), \(\overline{\mathrm{PQ}}\) ସାଧାରଣ ଜ୍ୟାକୁ ସମର୍ଦ୍ଦିଖଣ୍ଡ କରେ ।
ଏବଂ (ii) \(\stackrel{\leftrightarrow}{\mathbf{A B}}\) ⊥ \(\overline{\mathrm{PQ}}\)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 17
Solution:
ଦତ୍ତ : A ଓ B ଦୁଇଟି ପରସ୍ପରଛେଦୀ ବୃତ୍ତର କେନ୍ଦ୍ର ।
P ଓ Q ବୃତ୍ତଦ୍ଵୟର ଛେଦବିନ୍ଦୁ ।
ପ୍ରାମାଣ୍ୟ : (i) \(\stackrel{\leftrightarrow}{\mathbf{AB}}\), \(\overline{\mathrm{PQ}}\) କୁ ସମର୍ଦ୍ଦିଖଣ୍ଡ କରେ । (ii) \(\stackrel{\leftrightarrow}{\mathbf{A B}}\) ⊥ \(\overline{\mathrm{PQ}}\)
ଅଙ୍କନ : \(\overline{\mathrm{PA}}\), \(\overline{\mathrm{AQ}}\), \(\overline{\mathrm{BQ}}\) ଏବଂ \(\overline{\mathrm{BP}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : △APB ଏବଂ △AQB ଦ୍ବୟରେ
PA = AQ (ଏକା ଦ୍ୱଭର ବ୍ୟାସାକିଂ), BP = BQ
ଏବଂ \(\overline{\mathrm{AB}}\) ସାଧାରଣ ବାହୁ ।
∴△APB ≅ △AQB (ଦା .ଦା. ଦା)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 18
⇒ m∠PAM = m∠QAM (ଅନୁରୂପ କୋଣ)
ବର୍ତ୍ତମାନ △APM ଏବଂ △AQM ଦ୍ବୟରେ
PA = AQ, m∠PAM = m∠QAM ଏବଂ \(\overline{\mathrm{AM}}\) ସମତ୍ତିଖଣ୍ଡ କରିବ ।
∴ △APM ≅ △AQM (ଦା .ଦା. ଦା)
PM = MQ ⇒ \(\overleftrightarrow{\mathrm{AB}}\), \(\overline{\mathrm{PQ}}\) ….. (ii) (ପ୍ରମାଣିତ)
ଏବଂ m∠AMP = m∠AMQ
କିନ୍ତୁ ଏମାନେ ସନ୍ନିହିତ ପରିପୂରକ ହେତୁ m∠AMP = m∠AMQ = 90°
⇒ \(\overleftrightarrow{\mathrm{AB}}\) ⊥ \(\overline{\mathrm{PQ}}\) ….. (ii) (ପ୍ରମାଣିତ)

Question 19.
ଚିତ୍ରରେ ଦୁଇଟି ବୃତ୍ତ ପରସ୍ପରକୁ P ଓ Q ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । P ଠାରେ PQ ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ ବୃତ୍ତ ଦ୍ଵୟକୁ A ଓ B ଠାରେ ଛେଦ କରେ ଓ ସେହିପରି () ଠାରେ PQ ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ ବୃତ୍ତଦ୍ଵୟକୁ C ଓ D ଠାରେ ଛେଦ କରେ । ପ୍ରମାଣ କର ଯେ, AB = CD ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 19
Solution:
ଦତ୍ତ : ଦୁଇଟି ବୃତ୍ତ ପରସ୍ପରକୁ P ଓ Q ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । P ଠାରେ PQ ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ ବୃତ୍ତଦ୍ଵୟକୁ A ଓ B ଠାରେ ଛେଦକରେ Q ଠାରେ PQ ପ୍ରତି ଅଙ୍କିତ ଲମ୍ବ ବୃତ୍ତଦ୍ଵୟକୁ C ଓ D ଠାରେ ଛେଦ କରେ ।
ପ୍ରାମାଣ୍ୟ : AB = CD
ଅଙ୍କନ : ବୃତ୍ତଦ୍ଵୟର କେନ୍ଦ୍ର O1 ଓ O2 ନିଆ | O1X1 ⊥ AP ଏବଂ
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 20
O2 Y1 ⊥ PB ଅଙ୍କନ କର ।
X1 O1, ଓ Y1 O2, CDକୁ ଯଥାକ୍ରମେ X2 ଓ Y2 ରେ ଛେଦକରୁ ।
ପ୍ରମାଣ : AB || CD = X1 Y1 || X2 Y2
ପୁନଶ୍ଚ X1 X2 || Y1 Y2
∵ m∠X2X1Y1 +m∠X1Y1Y2 = 180°
⇒ X1Y1Y1X2 ଏକ ଆପ୍ତତତିତ୍ର |
X1Y1 = X2Y2
AB = AP + PB = 2PX1 + 2PY1
= 2(PX1 + PY1) = 2 X1Y1 = 2X2Y2
= 2(X2Q + QY2) = 2QX2 + 2QY2 = CQ + QD = CD (ପ୍ରମାଣିତ)

Question 20.
A ଓ B କେନ୍ଦ୍ର ବିଶିଷ୍ଟ ଦୁଇଟି ବୃତ୍ତ ପରସ୍ପରକୁ P ଓ ଠୁ ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି । P ମଧ୍ୟ ଦେଇ AB ସହିତ ସମାନ୍ତର ସରଳରେଖା ବୃତ୍ତ ଦ୍ଵୟକୁ M ଓ N ବିନ୍ଦୁରେ ଛେଦ କଲେ ପ୍ରମାଣ କର ଯେ, MN = 2AB | (ସୂଚନା : \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\), \(\overline{\mathbf{MN}}\) ପ୍ରତି ଲମ୍ବ ଅଙ୍କନ କରି ଦର୍ଶାଅ ଯେ, AB = CD)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 21
ସମାଧାନ :
ଦତ୍ତ : A ଓ B କେନ୍ଦ୍ର ବିଶିଷ୍ଟ ଦୁଇଟି ବୃତ୍ତ ପରସ୍ପରକୁ P ଓ Q ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି ।
P ବିନ୍ଦୁ ମଧ୍ୟଦେଇ ଅଙ୍କିତ \(\overline{\mathbf{MN}}\), \(\overline{\mathbf{AB}}\) ସହ ସମାନ୍ତର ।
ପ୍ରାମାଣ୍ୟ : MN = 2AB
ଅଙ୍କନ : AR ⊥ MP ଏବଂ BS ⊥ PN
ପ୍ରମାଣ : AR ⊥ MP ⇒ RP = \(\frac { 1 }{ 2 }\) MP
ସେହିପରି BS ⊥ NP ⇒ PS = \(\frac { 1 }{ 2 }\) PN
∴ RP + PS = \(\frac { 1 }{ 2 }\) (MP + PN) ⇒ RS = \(\frac { 1 }{ 2 }\) MN ……(i)
କିନ୍ତୁ RABS ଏକ ଆୟତଚିତ୍ର ⇒ RS = AB ….(ii)
(i) ଓ (ii) ରୁ AB = \(\frac { 1 }{ 2 }\), MN (ପ୍ରମାଣିତ)

Question 21.
ଚିତ୍ରରେ ଗୋଟିଏ ସରଳରେଖା ଦୁଇଟି ଏକ କେନ୍ଦ୍ରିକ ବୃତ୍ତ S1 ଓ S2 କୁ ଯଥାକ୍ରମେ A, C, D ଓ B ବିନ୍ଦୁରେ ଛେଦ କରୁଛି । ପ୍ରମାଣ କର ଯେ, AC = DB |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 22
Solution:
ଦତ୍ତ : S1 ଓ S2 ଦୁଇଟି ଏକ କେନ୍ଦ୍ରିକ ବୃତ୍ତ । ଏକ ସରଳରେଖା ବୃତ୍ତଦ୍ଵୟକୁ ଯଥାକ୍ରମେ A, C, D ଓ B ବିନ୍ଦୁରେ ଛେଦ କରୁଛି |
ପ୍ରାମାଣ୍ୟ : AC = DB
ଅଙ୍କନ : OM ⊥ AB ଅଙ୍କନ କର । \(\overline{\mathbf{OC}}\) ଏବଂ \(\overline{\mathbf{OA}}\) କୁ ଯୋଗକର ।
ପ୍ରମାଣ : \(\overline{\mathbf{OM}}\) ⊥ \(\overline{\mathbf{AB}}\) ଏବଂ \(\overline{\mathbf{OM}}\) ⊥ \(\overline{\mathbf{CD}}\)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 23
∴ AM = MB ଏବଂ CM = MD
⇒ AM – CM = MB – MD ⇒ AC = BD

Question 22.
ଗୋଟିଏ ବୃତ୍ତର ଏକ ବହିଃସ୍ଥ ବିନ୍ଦୁ P ମଧ୍ୟ ଦେଇ ଅଙ୍କିତ ଦୁଇଟି ଛେଦକ ବୃତ୍ତକୁ A, B ଏବଂ C, D ବିନ୍ଦୁରେ ଛେଦ କରନ୍ତି ଯେପରି P – A – B ଏବଂ P – C – D। ଯଦି AB = CD ହୁଏ, ପ୍ରମାଣ କର ଯେ, PA = PC ଏବଂ AC || BD |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 24
Solution:
ଦତ୍ତ : ବୃତ୍ତର ବହିଃସ୍ଥ ବିନ୍ଦୁ P ମଧ୍ୟଦେଇ ଏକ ଛେଦକ ବୃତ୍ତକୁ A, B ଏବଂ C, D ବିନ୍ଦୁରେ ଛେଦ କରୁଛନ୍ତି । AB = CD
ପ୍ରାମାଣ୍ୟ : (i) PA = PC (ii) \(\overline{\mathbf{AC}}\)||\(\overline{\mathbf{BD}}\)
ଅଙ୍କନ : କେନ୍ଦ୍ର O ଠାରୁ \(\overline{\mathbf{AB}}\) ଓ DC ପ୍ରତି ଯଥାକ୍ରମେ OM ଏବଂ ON ଲମ୍ବ ଅଙ୍କନ କର । OP ଅଙ୍କନ କର ।
ପ୍ରମାଣ : AB = CD = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) CD ⇒ AM = NC ଏବଂ MB = ND… (i)
△OMP ଏବଂ △ONP ଦ୍ୱୟରେ m∠OMP = m∠ONP (ପ୍ରତ୍ୟେକ ସମକୋଣ)
OM = ON ( ସମାନ ଦୈର୍ଘ୍ୟ ବିଶିଷ୍ଟ ଜ୍ୟା ମାନ କେନ୍ଦ୍ରଠାରୁ ସମଦୂରବର୍ତ୍ତୀ), \(\overline{\mathbf{OP}}\) ସାଧାରଣ ବାହୁ ।
△OMP ≅ △ONP
⇒ MP= NP ⇒ MP – AM = NP – NC [(i)ରୁ] → PA = PC (ପ୍ରମାଣିତ) …(ii)
ପୁନଶ୍ଚ MP + MB = NP + ND [(i)ରୁ]
⇒ PB = PD ⇒ △PBD ସମଦିବାହୁ (ii) ରୁ PA = PC → A PAC ଏକ ସମଦ୍ବିବାହୁ ।
△PBD ରେ PB = PD ⇒ m∠PBD = m∠PDB = θ (ମନେକର)
ସେହିପରି △PAC m∠PAC = m∠ACP = α (ମନେକର)
∴ △PBD ରେ θ + θ + m∠BPD = 180°
⇒ 2θ + m∠BPD = 180° ….(iii)
ପେଦ୍ୱିପରି △PAC ରେ, 2α + m∠APC = 180° …(iv)
(iii) 2θ = 2α

Question 23.
ABC ବୃତ୍ତର କେନ୍ଦ୍ର O । ଏହାର ଦୁଇଟି ସର୍ବସମ ଜ୍ୟା ପରସ୍ପରକୁ ଏକ ଅନ୍ତଃସ୍ଥ ବିନ୍ଦୁ P ଠାରେ ଛେଦ କରନ୍ତି । B ଓ C, \(\overline{\mathbf{OP}}\) ର ଏକ ପାର୍ଶ୍ୱସ୍ଥ ହେଲେ ପ୍ରମାଣ କର ଯେ, (i) PA = PC ଏବଂ (ii) \(\overline{\mathbf{AC}}\) || \(\overline{\mathbf{BD}}\) | (ସୂଚନା : \(\overline{\mathbf{OE}}\) L \(\overline{\mathbf{AB}}\) ଏବଂ \(\overline{\mathbf{OF}}\) ⊥ \(\overline{\mathbf{CD}}\) ଅଙ୍କନ କରି O, P ଯୋଗ କର)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 25
ସମାଧାନ :
ଦତ୍ତ : ବୃତ୍ତର କେନ୍ଦ୍ର ( 1 AB = CD, AB ଓ CD ଜ୍ୟା ଦ୍ଵୟର ଛେଦବିନ୍ଦୁ P ।
B ଓ C, \(\overline{\mathbf{OP}}\) ର ଏକ ପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ ।
ପ୍ତାମଣ୍ୟ : (i) PA = PC (ii) \(\overline{\mathbf{AC}}\) || \(\overline{\mathbf{BD}}\)
ଅଙ୍କନ : \(\overline{\mathbf{OE}}\) L \(\overline{\mathbf{AP}}\) ଏବଂ \(\overline{\mathbf{OD}}\) L \(\overline{\mathbf{CD}}\) ଅଙ୍କନ କର ।
ପ୍ରମାଣ : AB = CD ⇒ \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) CD ⇒ AE = CF ଏବଂ BE = FD
△OPE ଏବଂ OPF ଦଯ6ର m≤OEP = m≤OFP = 90°, OE = OF (∵ ଲ୍ୟା ଦ୍ବାଦାସମ)
\(\overline{\mathbf{OP}}\) ସାଧାରଣ ∴ △OPE ≅ △OPE ⇒ PE = PF
⇒ AE – PE = CF – PF ⇒ AP = CP ….(i)
⇒ △APC ସମଦିବାହୁ ।
ପୁନଶ୍ଚ BE + PE = DF + FP ⇒ PB = PD ⇒ △PBD ସମଦ୍ବିବାହୁ ।
△APC ସମଦିବାହୁ ।
m∠PAC = m∠PCA = θ (ମ6ନକର) ⇒ 2θ + m∠CPA = 180° ……(1)
△PBD ସମଙ୍ଗିବାହୁ = m∠PBD = m∠PDB = α (ମନେକର)
⇒ 2α + m∠BPD = 180° …..(2)
(1) ଓ (2)ରୁ 2θ + m∠CPA = 2α + m∠BPD (∵ m∠CPA = m∠BPD ପ୍ରତୀପ)
⇒ 2θ = 2α ⇒ θ = α ⇒ m∠PAC = m∠PBD
ମାତ୍ର ଏହି କୋଣଦ୍ଵୟ ଏକାନ୍ତର, ତେଣୁ \(\overline{\mathbf{AC}}\) || \(\overline{\mathbf{BD}}\) ……. (ii)

CHSE Odisha Class 11 English Grammar Additional Questions

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Additional Questions Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar Additional Questions

A. Rewrite the passages after correcting all grammatical errors in it.

(1) There is great excitement in the planet of Venus this week. For the first time, Venusia scientists manage to land an unmanned spacecraft in the planet Venus, and it is sending back signals, as well as photographs, ever since. The craft directed into an area known as Gonebay.
Answer:
There was great excitement on the planet of Venus this week. For the first time, Venusian scientists managed to land an unmanned spacecraft on the planet Venus, and it has been sending back signals, as well as photographs ever since. The craft was directed into an area known as Gonebay.
(2) How did birds know when to flew south for winter? How long do a bear sleep in winter? Do a porcupine really shoot its quills at an enemy? How do squirrels know where he buries nuts?
Answer:
How do birds know when to fly south for the winter? How long does a bear sleep in winter? Does a porcupine really shoot his quills at an enemy? How does a squirrel know where he buried a nut?

CHSE Odisha Class 11 English Grammar Additional Questions

B. Correct the errors.

(a) He is an European.
(b) I met the concerned clerk.
(c) It is high time you get up early.
(d) It has been five years since I last met you.
(e) I congratulate you for winning the prize.
Answer:
(a) He is a European.
(b) I met the clerk concerned.
(c) It is high time you got up early.
(d) It is five years since I last met you.
(e) I congratulate you on winning the prize.

C. Supply the correct tense of the verbs given in brackets.

1. Water always (freeze) at 0 degrees centigrade.
2. Students frequently (make) mistakes of tense usage when they do this exercise.
3. I (have) my hair cut whenever it gets too long.
4. I (take) my dog for a walk every evening before it died.
5. He (come) to my office whenever he needed money.
6. Last year she (wear) the same dress at every party.
7. Whenever I climb a hill, my ear (boil).
8. She (sing) very beautifully before she was married, but nowadays she (not sing) anymore.
9. I seldom (see) him at concerts these days. He (go) to them regularly before the war.
10. She cooks very well but her sister (cook) much better when I knew her.
11. Every time he opens his mouth, he (say) something foolish.
12. He occasionally makes a big effort, but usually he (not bother).
13. Whenever I (go) to see him, he was out.
14. In the past men frequently (fight) duels. Nowadays they seldom (do).
15. How often you (go) to the theatre when you were in London?
16. You (play) with dolls when you were a little girl?
17. The ancient Egyptians (build) pyramids as tombs for their kings.
18. When I was young, my father always (give) me some money on Saturdays.
19. If he is wise, a pianist (practise) four hours a day.
20. His parents don’t know what to do with their child. He (lie) habitually.
21. My aunt Jane (hate) girls who made up.
22. We all (study) Latin when we were at school.
23. Wood always (float).

Answer:
1. Water always freezes at 0 degrees centigrade.
2. Students frequently make mistakes of tense usage when they do this exercise.
3. I have my hair cut whenever it gets too long.
4. I took my dog for a walk every evening before it died.
5. He came to my office whenever he needed money.
6. Last year she wore the same dress at every party.
7. Whenever I climb a hill, my ear boils.
8. She sang very beautifully before she was married, but nowadays she does not sing anymore.
9. I seldom see him at concerts these days. He went to them regularly before the war.
10. She cooks very well but her sister cooked much better when I knew her.
11. Every time he opens his mouth, he says something foolish.
12. He occasionally makes a big effort, but usually, he does not bother.
13. Whenever I went to see him, he was out.
14. In the past men frequently fought duels. Nowadays they seldom do.
15. How often you go to. the theatre when you were in London?
16. Did you play with dolls when you were a little girl?
17. The ancient Egyptians built pyramids as tombs for their kings.
18. When I was young, my father always gave me some money on Saturdays.
19. If he is wise, a pianist practices four hours a day.
20. His parents don’t know what to do with their child. He lies habitually.
21. My aunt Jane hated girls who made up.
22. We all studied Latin when we were at school.
23. Wood always floats.

CHSE Odisha Class 11 English Grammar Additional Questions

D. Put the verbs in brackets into the correct Present Tense, Continuous, or Simple.

1. Buses usually (run) along this street. but today they (not run) because it is under repair.
2. John (pass) the post office on his way to work every day.
3. She usually (sit) at the back of the class, but today she (sit) in the front row.
4. I rarely (carry) an umbrella, but I (carry) one now because it is raining.
5. What you generally (do) for a living?
6. You (enjoy) your English class today?
7. You (enjoy) washing dishes as a rule?
8. We nearly always (spend) our holidays at the seaside, but this year we are going to France.
9. Mr. Jones usually (sell) only newspapers, but this week he (sell) magazines as well.
10. You (wash) your hands before every meals.
11. Mary generally (begin) cooking at 11, but today she came home early and (cook) now, although il is only 10.30.
12. I’m sorry you can’t see her. She (sleep) still. She usually (wake) much earlier.
13. Why you (wear) a coat this morning? I never (wear) one till October.
14. Joan still (do) her homework. Her sister, who always (work) quicker, (play) already in the garden.
15. These builders generally (build) very rapidly. They (work) at present on two separate contracts.
16. What (do) you at this moment? If you (not do) anything, please help me.
17. John, who (study) medicine at present, hopes to go abroad after graduation.
18. He generally (come) to my office every clay, but today he (visit) his parents in the country.
19. You (watch) television often? The electrician (install) ours at this moment.
20. Mary usually (wear) a hat to go shopping. but today, as the sun (shine) she (not wear) one.

Answer:
1. Buses usually run along this Street, but today they are not running because it is under repair.
2. John passes the post office on his way to work every day.
3. She usually sits at the back of the class, but today she is sitting in the front row.
4. I rarely carry an umbrella, but I am carrying one now because it is raining.
5. What do you generally do for a living?
6. Are you enjoying your English class today?
7. Do you enjoy washing dishes as a rule?
8. We nearly always spend our holidays at the seaside, but this year we are going to France.
9. Mr. Jones usually sells only newspapers, but this week he is selling magazines as well.
10. Do you wash your hands before every meal?
11. Mary generally begins cooking at 11, but today she came home early and is cooking now, although it is only 10.30.
12. I’m sorry you can’t see her. She is still sleeping. She usually wakes much earlier.
13. Why are you wearing a coat this morning? I never wear one till October.
14. Joan is still doing her homework. Her sister, who always works quicker, is already playing already in the garden.
15. These builders generally build very rapidly. They are working at present on two separate contracts.
16. What are you doing at this moment? If you are not doing anything, please help me.
17. John, who is studying medicine at present, hopes to go abroad after graduation.
18. He generally comes to my office every day, but today he (is visiting) his parents in the country.
19. Do you often watch television? The electrician is installing ours at this moment.
20. Mary usually wears a hat to go shopping, but today, as the sun is shining she is not wearing one.

CHSE Odisha Class 11 English Grammar Additional Questions

E. Put the verbs in brackets into the correct tense, Continuous or Simple.

1. You (see) the house on the corner? That is where I was born.
2. You (listen) to what I am saying You (understand) me?
3. I (notice) Mary (wear) a new hat today.
4. She (not understand) what you (mean).
5. I (need) a new suit. They (offer) special prices at the tailor this week.
6. You (smell) gas? I (think) the new stove is leaking.
7. Look at Mary! She (drink) up her medicine, but I can say that she (hate) it.
8. John (seem) rather tired today.
9. it still (rain), but it (look) as if it will soon stop.
10. You (mind) helping me a moment? I (try) to mend this table.
11. Ask him what he (want).
12. You (remember) the name of that girl who (walk) on the other side of the street?
13. ‘Will you have some tea?’ ‘I (prefer) coffee, please.’
14. I (suppose) I must go now. My wife (wait) for me at home.
15. You (see) this box? It (contain) matches.
16. These twins, who (resemble) one another so strongly, (study) art at present.
17. After what has happened, you really (mean) to say that you still (believe) him?
18. You (suppose) the children still (sleep)?
19. ‘The train still (stand) in the station. You (think) we can just catch it?
20. I (notice) you (possess) a copy of Waugh’s latest book. Will you lend it me?

Answer:
1. Do you see the house on the corner? That is where I was born.
2. Are you listening to what I am saying? Do you understand me?
3. I notice Mary is wearing a new hat today.
4. She does no: understand what you mean.
5. I need a new suit. They are offering special prices at the tailor this week.
6. Do you smell gás? I think the new stove is leaking.
7. Look at Mary! She is drinking up her medicine, but I can say that she hates it.
8. John seems rather tired today.
9. It is still raining, but it looks as if it will soon stop.
10. Do you mind helping me a moment? I urn Irving to mend this table.
11. Ask him what he wants.
12. Do you remember the name of that girl who is walking on the other side of the street?
13. ‘Will you have some tea?’ ‘I prefer coffee, please.’
14. 1 suppose I must go now. My wife is waiting for me at home.
15. Do you see this box? It contains matches.
16. These twins, who resemble one another so strongly, are studying art at present.
17. After what has happened, do you really mean to say that you still believe him?
18. Do you suppose the children still sleeping?
19. ‘The train is still standing in the station. Do you think we can just catch it
20. I notice you possess a copy of Waugh’s latest book. Will you lend it me?

CHSE Odisha Class 11 English Grammar Additional Questions

F. Supply the correct form of the Present Perfect tense, Continuous or Simple in the place of the verbs in brackets.

1. They just (arrive) from New York.
2. They still (not succeed) in reaching the summit.
3. I this very minutes (receive) a telegram from my brother in India.
4. We already (have) breakfast.
5. I now (study) your proposals and regret I cannot accept them.
6. They (live) here since January.
7. We (wait) on the platform since three o’ clock.
8. She already (ring) the bell twice.
9. I see you just (have) your hair cut.
10. She (write) letters all morning, but I (not start) to write any yet.
11. The children (sleep) all this afternoon.
12. How long you (stay) in that old hotel?
13. They (work) in the same factory for twenty years now.
14. Since when you (have) that new car?
15. I (knock) on the door for ten minutes now without an answer.
16. They (build) that bridge for over a year and still it isn’t finished.
17. I (try) three times and (be) successful only once.
18. How many times you (be) to the cinema this week?
19. He (go) to the dentist off and on for six months.
20. He (take) the exam. three times and (fail) every time.
21. William (marry) the eldest Jones girl at last.
22. I (try) to get in touch with you for several days now.
23. She just (spend) three weeks at her grandmother’s.
24. He (work) hard on his book for some time and (finish) it at last.
25. You ever (read) ‘War and Peace’?

Answer:
1. They have just arrived from New York.
2. They have still not succeeded in reaching the summit.
3. I have this very minutes received a telegram from my brother in India.
4. We have already had breakfast.
5. I have now studied your proposals and regret I cannot accept them.
6. They have been living here since January.
7. We have been waiting on the platform since three o’ clock.
8. She has already rung the bell twice.
9. I see you have just had your hair cut.
10. She has been writing letters all morning, but I have no: starred to write any yet.
11. The children have been sleeping all this afternoon.
12. How long have you been staying in that old hotel?
13. They have been working in the same factory for twenty years now.
14. Since when have you had that new car?
15. I have been knocking on the door for ten minutes now without an answer.
16. They have been building that bridge for over a year and still it isn’t finished.
17. I have tried three times and have been successful only once.
18. How many times have you been to the cinema this week?
19. He has been going to the dentist off and on for six months.
20. He has taken the exam. three times and has failed every time.
21. William has married the eldest Jones girl at last.
22. I have been trying to get in touch with you for several days now.
23. She has just spent three weeks at her grandmother’s.
24. He has been working hard on his book for some time and has just finished it at last.
25. Have you ever read ‘War and Peace’?

CHSE Odisha Class 11 English Grammar Additional Questions

G. Supply the missing prepositions using only to or at.

1. He is quite blind _____ her faults.
Answer:
to

2. He is extraordinarily clever _____ mimicking others.
Answer:
at

3. She, on the other hand, is very efficient _____ her work.
Answer:
at

4. He is an expert making himself understood in foreign languages.
Answer:
at

5. Contrary my expectations, I quite enjoyed myself at the party.
Answer:
to

6. She was standing too close _____ the fire and got burned.
Answer:
to

7. That fellow’s no good games at all.
Answer:
at

8. He carried out the project which had always been dear _____ his heart.
Answer:
to

9. We are all very indignant _____ the injustice done to him.
Answer:
at

10. I’m sorry! I’m very bad _____ explaining myself.
Answer:
at

CHSE Odisha Class 11 English Grammar Additional Questions

11. He’s the one who’s so very lucky _____ cards.
Answer:
at

12. That device is entirely new _____ me.
Answer:
to

13 He’s not equal _____ the job they’ve given him.
Answer:
to

14. He remained faithful _____ his principles in spite of great pressure.
Answer:
to

15. The delay proved fatal _____ our plans.
Answer:
to

16. She was overjoyed _____ the prospect of meeting him again.
Answer:
at

17. His activities are very harmful _____ my interests.
Answer:
to

18. The government showed itself hostile _____ any progress.
Answer:
to

19. She’s no terribly cruel _____ her dog.
Answer:
to

20. I engaged him because he was so prompt in understanding my instructions.
Answer:
at

CHSE Odisha Class 11 English Grammar Additional Questions

21. He was heartbroken _____ her indifference to him.
Answer:
at

22. He doesn’t like me although I’ve always been kind _____ him.
Answer:
to

23. Yes, he’s the kind of person who is always quick figures.
Answer:
at

24. This is much inferior _____ the one I bought last week.
Answer:
to

25. You will be liable _____ a heavy fine if you do that.
Answer:
to

26. This flower is not native _____ England.
Answer:
to

27. Naturally she was sad _____ the death of her parrot.
Answer:
at

28. He does his work carefully but he’s terribly slow ______ it.
Answer:
at

29. Who wouldn’t be triumphant _____ their success in the examination?
Answer:
at

30. A dutiful daughter and obedient _____ her parents.
Answer:
to

CHSE Odisha Class 11 English Grammar Additional Questions

31. You shouldn’t be surprised _____ a thing like that.
Answer:
at

32. It should be obvious _____ the meanest intelligence.
Answer:
to

33. The seeds are peculiar ______ this genus of plant.
Answer:
to

34. Can’t you manage to be a little more polite your aunt?
Answer:
to

35. He works in a factory. Previous _____ that he was in a laundry.
Answer:
to

36. I have been truly astonished _____ the number of people who believe it.
Answer:
at

37. This is quite irrelevant ______ the matter we are discussing.
Answer:
to

38. They were shocked _____ his apparent lack of appreciation.
Answer:
at

39. He was so rude _____ her that she never spoke to him again.
Answer:
to

40. Sacred _____ the memory of Mary Jones.
Answer:
to

CHSE Odisha Class 11 English Grammar Additional Questions

41. They are very sensitive _____ people’s opinion of them.
Answer:
to

42. I’ve got one similar _____ yours.
Answer:
to

43. Subject _____ the exigencies of the service.
Answer:
to

44. The people of this country are very skilful ______ making dolls.
Answer:
at

45. It’s useful _____ me to have him about the house.
Answer:
to

46. It’s vital _____ a proper understanding of the problem.
Answer:
to

H. Supply the missing preposition using with, for or of.

1. Don’t be afraid _____ the dog! It won’t bite you.
Answer:
of

2. I can’t be angry _____ him now that he’s apologized _____ what he has done.
Answer:
with, for

3. He’s far ahead _____ the others in arithmetic.
Answer:
of

CHSE Odisha Class 11 English Grammar Additional Questions

4. You ought to be ashamed _____ yourself.
Answer:
of

5. Are you aware _____ the fact that it is half past ten.
Answer:
of

6. I’m sorry. They are simply not capable doing it.
Answer:
of

7. Don’t disturb him! He’s busy _____ his accounts.
Answer:
with

8. He’s ambitious and eager _____ honours.
Answer:
for

9. For goodness sake! Do be careful _____ that vase. You could easily drop it.
Answer:
with

10. Children must be taught to be careful _____ traffic.
Answer:
of

11. I am not at all certain _____ the date of his arrival.
Answer:
of

12. His explanation was not consistent _____ the facts.
Answer:
with

13. Monsieur X was famous _____ his collection of pictures.
Answer:
for

14. I was conscious _____ a feeling of uneasiness.
Answer:
of

15. Kingston lies due west _____ London.
Answer:
of

CHSE Odisha Class 11 English Grammar Additional Questions

16. I’m sorry. I’m not content _____ your explanations.
Answer:
with

17. The soldier was pronounced fit _____ service.
Answer:
for

18. Mary was terribly envious _____ Joan’s new hat.
Answer:
of

19. We are all very fond _____ going to the theatre.
Answer:
of

20. John is very discontented _____ his salary.
Answer:
with

21. I shall be grateful any advice you can give me.
Answer:
for

22. This exercise is full ______ the most terrible mistakes.
Answer:
of

23. Are you familiar _____ the works of Milton?
Answer:
with

24. The manager is well qualified _____ his position.
Answer:
for

25. That is something I am profoundly glad.
Answer:
of

CHSE Odisha Class 11 English Grammar Additional Questions

26. This chair is not identical _____ the one I bought last year.
Answer:
with

27. He’s a sporting fellow, and always ready for anything.
Answer:
for

28. That student is ignorant _____ the first rules of grammar.
Answer:
of

29. Your explanation is incompatible _____ the story I heard.
Answer:
with

30. John, you will be responsible for providing the drinks.
Answer:
for

31. Judges must be independent of

political pressure.
Answer:
of

32. I know he’s a difficult child, but you must be patient _____ him.
Answer:
with

33. I am extremely sorry _____ the delay, but I was held up.
Answer:
for

34. He was jealous _____ his brother’s good fortune.
Answer:
of

CHSE Odisha Class 11 English Grammar Additional Questions

35. Comedians are always popular holiday crowds.
Answer:
with

36. I can’t bake a cake as we are short _____ eggs this week.
Answer:
of

37. His income is sufficient _____ his needs.
Answer:
for

38. He’s always very shy approaching his chief.
Answer:
of

39. Let us be thankful _____ small mercies.
Answer:
for

40. It is wise to be sure _____ your facts before you speak.
Answer:
of

41. One is generally tolerant _____ small faults.
Answer:
of

42. She is, unfortunately, devoid _____ a sense of humor.
Answer:
of

CHSE Odisha Class 11 English Grammar Additional Questions

43. They gave him a visa valid _____ all countries in Europe.
Answer:
for

44. I’m tired _____ arguing with you.
Answer:
of

45. They proved themselves unworthy _____ the trust which was placed in them.
Answer:
of

K. Supply the missing prepositions, from, about, on or in.

1. Keep away _____ the machine while it is running.
Answer:
from

2. He wsa singularly fortunate _____ his choice of wallpaper.
Answer:
in

3. He is intent _____ attending the football match on Saturday.
Answer:
on

4. I am very dubious _____ your chances of passing the examination.
Answer:
about

CHSE Odisha Class 11 English Grammar Additional Questions

5. They are proficient _____ the use of their fists.
Answer:
in

6. This is quite different _____ what I expected.
Answer:
from

7. The diet here is deficient _____ vitamins.
Answer:
in

8. That young man is very keen _____ cycling.
Answer:
on

9. We are all very enthusiastic ________ our next holiday.
Answer:
about

10. It was far ______ my intention to suggest that he was unintelligent.
Answer:
from

11. He was perfectly honest _____ his intentions to win the prize at all costs.
Answer:
about

12. The secretary was not well qualified _____ shorthand.
Answer:
in

13. Some people appear completely immune _____ this disease.
Answer:
from

14. I am very reluctant _____ asking him to do this.
Answer:
about

15. Our plans must remain dependent _____ the weather.
Answer:
on

16. Everybody was very uneasy _____ the outcome of the negotiations.
Answer:
about

CHSE Odisha Class 11 English Grammar Additional Questions

17. Of course, you are quite right _____ that.
Answer:
about

18. He was involved in an accident, resulting _____ the slippery condition of the road.
Answer:
from

19. The enemy is weak _____ artillery.
Answer:
in

20. Entomologists are still curious _____ the life-cycle of that month.
Answer:
about

21. I am not very interested _____ the story of your life.
Answer:
in

22. Put that cake back in the cupboard. where it will be safe _____ the cat.
Answer:
from

23. I’m afraid he is quite wrong _____ the date of the invasion.
Answer:
about

24. His father was very sad _____ his son’s failure in his final exams.
Answer:
about

25. I am extremely doubtful _____ the wisdom of pursuing that course of action.
Answer:
about

CHSE Odisha Class 11 English Grammar Additional Questions

L. Supply the prepositions at or to as appropriate.

1. What time did you arrive _____ your home?
Answer:
at

2. He finds it difficult to accustom himself _____ the climate.
Answer:
to

3. AIl the visitors exclaimed _____ the beauty of the place.
Answer:
at

4. Just glance _____ this for me, would you?
Answer:
at

5. His debts amount _____ a considerable sum.
Answer:
to

6 I can only guess _____ the extent of the damage.
Answer:
at

7. It is useless to appeal _____ his better nature.
Answer:
to

8. She hinted darkly _____ all sorts of wild actions in his youth.
Answer:
at

CHSE Odisha Class 11 English Grammar Additional Questions

9. If you want permission. you must apply _____ the caretaker.
Answer:
to

10. He was attached _____ the French Army during the war.
Answer:
to

11. He is too sick to attend _____ his duties.
Answer:
to

12. _____ what do you attribute your success in life?
Answer:
to

13. I’m sure this one doesn’t belong _____ me.
Answer:
to

14. He challenged him _____ a game of chess.
Answer:
to

15. It is very unkind to joke _____ the expense of the disabled.
Answer:
at

16. Shall I compare thee _____ a summer’s day?
Answer:
to

17. If you want them to hear, you’ll have to knock a good deal harder _____ the door.
Answer:
at

18. The prisoner was condemned _____ penal servitude for life.
Answer:
to

CHSE Odisha Class 11 English Grammar Additional Questions

19. The fire was confined _____ the kitchen regions.
Answer:
to

20. I will never consent _____ her marrying that man.
Answer:
to

21. Just have a look _____ this for me, would you?
Answer:
at

22. I have been entirely converted _____ the use of an electric razor.
Answer:
to

23. Employees who have twenty-five years’ service become entitled _____ a pension.
Answer:
to

24. You can safely entrust your little son _____ her care.
Answer:
to

25. People who heard her voice marvelled _____ it.
Answer:
at

26. Let’s invite them all _____ dinner.
Answer:
to

27. Listen _____ me!
Answer:
to

CHSE Odisha Class 11 English Grammar Additional Questions

28. Do you object my smoking?
Answer:
to

29. He peered _____ the exhibit on account of his short-sightedness.
Answer:
at

30. Such an idea would never occur _____ me!
Answer:
to

31. The patient is reacting very unsatisfactorily _____ the drug.
Answer:
to

32. The children peeped _____ the guests as they were arriving.
Answer:
at

33. I have been reduced ______ using oil for lack of fat.
Answer:
to

34. The children are playing _____ Red Indians again.
Answer:
at

35. She has been forced to resort _____ all sorts of devices to avoid him.
Answer:
to

36. The patient has not responded _____ treatment.
Answer:
to

37. It is very rude to point people in the street.
Answer:
at

CHSE Odisha Class 11 English Grammar Additional Questions

38. If you’ll bring the drinks, I’ll see _____ the food.
Answer:
to

39. They have been subjected _____ all sorts of indignities.
Answer:
to

40. I refuse to submit ______ that sort of treatment.
Answer:
to

41. Can you wonder _____ it, if they are reduced _____ begging.
Answer:
at, to

42. I’m sorry he finally succumbed _____ the temptation of stealing.
Answer:
to

43. If you want to pass your examination, you il have to work very hard _____ your Latin.
Answer:
at

44. We shall never surrender _____ that enemy.
Answer:
to

45. I don’t know him, but he has been starting _____ me for ten minutes.
Answer:
at

46. I have never subscribed _____ the general opinion of him.
Answer:
to

CHSE Odisha Class 11 English Grammar Additional Questions

47. Turn _____ page 22 and start reading!
Answer:
to

48. The government has again yielded _____ the pressure from outside.
Answer:
to

49. She always trusts _____ her neighbors to help her.
Answer:
to

50. Would you please reply my question?
Answer:
to

M. Turn the following into passive.

1. The government has called out troops.
Answer:
Troops have been called out.

2. Fog held up the trains. (agent required)
Answer:
Trains were held up by fog.

3. You are to leave this here. Someone will call for it later on.
Answer:
This is to be left here. k will be called for.

4. We called in the police.
Answer:
Police were called in.

CHSE Odisha Class 11 English Grammar Additional Questions

5. They didn’t look after the children properly.
Answer:
Children were not properly looked after.

6. They are flying in reinforcements.
Answer:
Reinforcements are being flown in.

7. Then they called up men of 28.
Answer:
Men of 28 were called up.

8. Everyone looked up to him. (agent required)
Answer:
He was looked up to by everyone.

9. All the ministers will see him off at the airport. (agent required)
Answer:
He will be seen off at the airport by all the ministers.

10. He hasn’t slept in his bed.
Answer:
Bed hasn’t been slept in.

11. We can build on more rooms.
Answer:
More rooms can be built on.

12. They threw him out.
Answer:
He was thrown out.

CHSE Odisha Class 11 English Grammar Additional Questions

13. They will have to adopt a different attitude.
Answer:
Different attitude will have to be adopted.

14. He’s a dangerous maniac. They ought to lock him up.
Answer:
He ought to be locked up.

15. Her story didn’t take them in. (agent required)
Answer:
They weren’t taken in by her story.

16. Burglars broke into the house.
Answer:
House was broken into.

17. The manufacturers are giving away small plastic toys with each packet of cereal.
Answer:
Small plastic toys are being given away.

18. They took down the notice.
Answer:
Notice was taken down.

19. They frown on smoking here.
Answer:
Smoking is frown on.

20. After the government had spent a million pounds on the scheme they decided that it was impracticable and gave it up. (Make only the first and last verbs passive)
Answer:
After a million pounds had been spent, the scheme was given up.

CHSE Odisha Class 11 English Grammar Additional Questions

21. When I returned I found that they had towed my car away. I asked why they had done this and they told me that it was because I had parked it under a No Parking sign. (four passives)
Answer:
My car had been towed away. I asked why this had been done and was told that it had been parked.

22. People must hand in their weapons.
Answer:
Weapons must be handed in.

23. The crowd shouted him down.
Answer:
He was shouted down.

24. People often take him for his brother.
Answer:
He is often taken for his brother.

25. No one has taken cut the cork.
Answer:
The cork hasn’t been taken out.

26. The film company were to have used the pool for aquatic displays, but now they have changed their minds about it and are filling it in. (Make the first and last verbs passive)
Answer:
Pool was to have been used it is being filled in.

27. This college is already full. We ne turning away students the whole time.
Answer:
Students are being turned away.

28. You will have to pull down this skyscraper as you have not complied with the town planning regulations.
Answer:
Skyscraper will have to be pulled down as the town planning regulations have not been complied with.

CHSE Odisha Class 11 English Grammar Additional Questions

N. Put the following sentences into passive, using infinitive construction where possible.

1. We added up the money and found that it was correct.
Answer:
Money was added up and found to be correct.

2. I’m employing a man to tile the bathroom.
Answer:
I am having the bathroom tiled.

3. Someone seems to have made a terrible mistake.
Answer:
A terrible mistake seem to have been made.

4. It is your duty to make tea at eleven o’ clock. (Use suppose.)
Answer:
You are supposed to make tea.

5. People know that he is armed.
Answer:
He is known to be armed.

6. Someone saw him pick up the gun.
Answer:
He was seen to pick up?

7. We know that you were in town on the night of the crime.
Answer:
You are known to have been.

CHSE Odisha Class 11 English Grammar Additional Questions

8. We believe that he has special knowledge which may be useful to the police. (one passive)
Answer:
He is believed to have special knowledge.

9. You needn’t have done this.
Answer:
This needn’t have been done.

10. It’s a little too loose: you had better ask your tailor to take it in. (one passive)
Answer:
You had better have it taken in.

11. He likes people to call him — ‘sir’.
Answer:
He likes to be called sir’.

12. Don’t touch this switch.
Answer:
This switch isn’t to be/mustn’t be touched.

13. You will have to get someone to see to it.
Answer:
You will have to have/get it seen to.
(Or) It will have to be seen to.

CHSE Odisha Class 11 English Grammar Additional Questions

14. It is impossible to do this. (Use can’t)
Answer:
This can’t be done.

15. Someone is following us.
Answer:
We are being followed.

CHSE Odisha Class 11 Math Notes Chapter 5 Principles of Mathematical Induction

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 5 Principles of Mathematical Induction will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 5 Principles of Mathematical Induction

Principles Of Mathematical Induction

(i) Principle – 1
Let P(n) is a statement , n ∈ Z

Step – 1: Verification step:
verify that P(1) is true.

Step – 2: Induction step – 1:
Assume that P(k) is true for any arbitrary k ∈ N.

Step – 3: Induction step – 2:
prove that P(k+1) is true using step – 1 and step – 2

Step – 4: Conclusion Step:
If P(k+1) is true then take a conclusion that by Principle of mathematical induction P(n) is true for all n ∈ N.

CHSE Odisha Class 11 Math Notes Chapter 5 Principles of Mathematical Induction

(ii) Principle – 2
Let P(n) be a statement, n ∈ N.

Step – 1: Verification step:
verify the P(1) is true

Step – 2: Induction step – 1:
Assume that P(2), P(3),….. P(k) is true.

Step – 3: Induction step – 2:
Using step – 1 and step – 2 prove that P(k + 1) is true.

Step – 4: Conclusion step:
If P(k + 1) is true, then take a conclusion that by Principle of mathematical induction P(n) is true for all n ∈ N.

CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 4 Trigonometric Functions

Angle:
If A, B, and C are three non-collinear points, then ∠ABC = \(\overrightarrow{\mathrm{BA}} \cup \overrightarrow{\mathrm{BC}}\)
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions
\(\overrightarrow{\mathrm{BA}}\) is the initial side, \(\overrightarrow{\mathrm{BC}}\) is called the terminal side and B is called the vertex of the angle.

Positive and negative angles:
If the direction of rotation is anti-clockwise then the angle is positive and if the direction of rotation is clockwise then the angle is negative.
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 1

Measure of an angle:
(a) Sexagesimal system or English System (Degree measure):
1 degree = 1° = \(\left(\frac{1}{360}\right) \text { th }\) of revolution from initial side to terminal side.

  • One revolution = 360°
  • 1° = 60′ (sixty minute)
  • 1′ = 60” (sixty seconds)

(b) Circular system(Radian measure):
One radian = 1c = The angle at the centre of the circle by an arc where the arc length equals to

Note:
(i) θ = \(\frac{l}{r}=\frac{\text { arc }}{\text { radius }}\) where θ is an radian.
(ii) θ in radian is created as a real number.

(c) Relation between Degree and radian measure:

  • 2π radians = 360°
    ⇒ π radian = 180°
  • We can convert radian to degree or degree to radian by using the identity.
    \(\frac{\mathrm{D}}{180}=\frac{\mathrm{R}}{\pi}\) where D is the degree measure and R is the radian measure of an angle.

CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions

Trigonometry Functions:
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 2

(i) Sign of trigonometry functions:

CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 3

(ii) \(\begin{array}{cccc}
\text { Add } & \text { Sugar } & \text { To } & \text { Coffee } \\
\downarrow & \downarrow & \downarrow & \downarrow \\
\text { all }+ & \sin + & \tan + & \cos +
\end{array}\)

(iii) Periodicity of trigonometry functions:

Trigonometric function Period
sin x
cos x
tan x π
cot x π
sec x
cosec x
sin2 x or cos2 x π
|sin x| or |cos x| π

Trigonometric functions of some standard angles
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 4

Fundamental trigonometric identities:
(a) sin θ = \(\frac{1}{{cosec} \theta}\)
(b) cos θ = \(\frac{1}{{sec} \theta}\)
(c) tan θ = \(\frac{1}{{cot} \theta}\)
(d) sin2 θ + cos2 θ = 1
(e) sec2 θ – tan2 θ = 1
(f) cosec2 θ – cot2 θ = 1
(g) sin (-θ) = -sin (θ)
(h) cos (-θ) = -cos (θ)

Trigonometric functions of allied angles:
(a) sin \(\left((2 n+1) \frac{\pi}{2} \pm \theta\right)\) = (±) cos θ choose + or – in (±) by using ASTC rule

(b) cos \(\left((2 n+1) \frac{\pi}{2} \pm \theta\right)\) = (±) sin θ choose + or – in (±) by using ASIC rule
Similar technique can be used for other trigonometric functions.

(c) sin (nπ ± θ) = (±) sin θ
cos (nπ ± θ) = (±) cos θ
tan (nπ ± θ) = (±) tan θ
choose + or – in (±) by using ASTC rule.

Sum And Difference Formulae:
(a) sin(A + B) = sin A . cos b + cos A . sin B

(b) sin(A – B) = sin A . cos B – cos A . sin B

(c) cos(A + B) = sin A . cos B – cos A . sin B

(d) cos(A – B) = sin A . cos B + cos A . sin B

(e) tan(A + B) = \(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\)

(f) tan(A – B) = \(\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}\)

(g) cot(A + B) = \(\frac{\cot A \cdot \cot B-1}{\cot A+\cot B}\)

(h) cot(A – B) = \(\frac{\cot A \cdot \cot B+1}{\cot A-\cot B}\)

(i) sin(A + B) + sin (A – B) = 2 sin A . cos B

(j) sin (A + B) – sin(A – B) = 2 cos A . sin B

(k) cos(A + B) + cos(A – B) = 2 cos A . cos B

(l) cos(A + B) – cos(A – B) = -2sin A . sin B

(m) sin(A + B) sin(A – B) = sin2 A – sin2 B = cos2 B – cos2 A

(n) cos(A + B) cos(A – B) = cos2 A – sin2 B = cos2 B – sin2 A

(o) sin 2A = 2 sin A cos A = \(\frac{2 \tan \mathrm{A}}{1+\tan ^2 \mathrm{~A}}\)

(p) cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
= 1 – 2 sin2 A
= \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\)

(q) tan 2A = \(\frac{2 \tan A}{1-\tan ^2 A}\)

(r) tan(A + B + C) = \(\frac{\tan A+\tan B+\tan C-\tan n A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}\)

(s) sin 3A = 3 sin A – 4 sin3 A
= 4 sin A sin(\(\frac{\pi}{3}\) – A) sin(\(\frac{\pi}{3}\) + A)

(t) cos 3A = 4 cos3 A – 3 cos A
= 4 cos A cos(\(\frac{\pi}{3}\) – A) cos(\(\frac{\pi}{3}\) + A)

(u) tan 3A = \(\frac{3 \tan A-\tan ^3 A}{1-3 \tan ^2 A}\)
= tan A. tan(\(\frac{\pi}{3}\) – A) tan(\(\frac{\pi}{3}\) + A)

CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions

Sum or Difference → Product:
(a) sin A + sin B = 2 sin(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\))
(b) sin A – sin B = 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\))
(c) cos A + cos B = 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\))
(d) cos A – cos B = -2 sin(\(\frac{A+B}{2}\)) sin(\(\frac{A-B}{2}\))

Submultiple Arguments:
(a)
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 5

(b) 2 sin2 \(\frac{\theta}{2}\) = 1 – cos θ
2 cos2 \(\frac{\theta}{2}\) = 1 + cos θ

(c) tan \(\frac{\theta}{2}\) = \(\frac{\sin \theta}{1+\cos \theta}=\frac{1-\cos \theta}{\sin \theta}\)

(d) sin θ = 3 sin \(\frac{\theta}{2}\) – 4 sin3 \(\frac{\theta}{2}\)
cos θ = 4 cos3 \(\frac{\theta}{2}\) – 3 cos \(\frac{\theta}{2}\)

(e) tan θ = \(\frac{3 \tan \frac{\theta}{2}-\tan ^3 \frac{\theta}{2}}{1-3 \tan ^2 \frac{\theta}{2}}\)

Trigonometric Equations:
(a) Equation involving trigonometric equations of unknown angles are called trigonometric function.
(b) Principle solution: The solution ‘x’ of a trigonometric equation is said to be a principle solution if x ∈ (0, 2π)
(c) The solution considered over the entire set R are called the general solution.
(d) General solution of some standard trigonometric equations.

  • sin x = 0 ⇒ x = nπ, n ∈ Z
  • cos x = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
  • tan x = 0 ⇒ x = nπ, n ∈ Z
  • sin x = sin α ⇒ x = nπ + (-1)n, n ∈ Z
  • cos x = cos α ⇒ x = 2nπ ± α, n ∈ Z
  • tan x = tan α ⇒ x = nπ + α, n ∈ Z
  • \(\left.\begin{array}{l}
    \sin ^2 x=\sin ^2 \alpha \\
    \cos ^2 x=\cos ^2 \alpha \\
    \tan ^2 x=\tan ^2 \alpha
    \end{array}\right]\) ⇒ x = nπ ± α
  • \(\left.\begin{array}{l}
    \cos x=\cos \alpha \\
    \text { and } \sin x=\sin \alpha
    \end{array}\right]\) ⇒ x = nπ ± α, n ∈ Z

Sine Formula:
In any Δ ABC, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) or, \(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}\) = 2R
∴ a = 2R sin A, b = 2R sin B and c = 2R sin C
Also, sin A = \(\frac{a}{2R}\), sin B = \(\frac{b}{2R}\) and sin c = \(\frac{c}{2R}\)

Cosine fromulae:
In any Δ ABC,
(i) a2 = b2 + c2 – 2bc cos A
(ii) b2 = c2 + a2 – 2ca cos B
(iii) c2 = a2 + b2 – 2ab cos C
or, → cos A = \(\frac{b^2+c^2-a^2}{2 b c}\)
→ cos B = \(\frac{c^2+a^2-a^2}{2 c a}\)
→ cos C = \(\frac{a^2+b^2-c^2}{2 a b}\)

Projection formulae:
In any Δ ABC,
(i) a = b sin C + c sin B
(ii) b = c cos A + a cos C
(iii) c = a cos B + b cos A

Tangent formulae (Napier’s Analogy);
In any Δ ABC
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 6

Area of Triangle (Heron’s formulae):
(i) Area of triangle ABC
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 7

(ii) Heron’s formulae:
In any Δ ABC Let 2S = a + b + c
Area of Δ ABC = Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Δ = \(\frac{1}{2}\) bc sin A = \(\frac{1}{2}\) ca sin B
= \(\frac{1}{2}\) ab sin C, Δ = \(\frac{abc}{4R}\)

CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions

Semi-Angle Formulae:
CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions 8

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a)

Question 1.
(i) ଗୋଟିଏ ଲୁଡୁ ଗୋଟିକୁ ଥରେ ଗଡ଼ାଗଲା । ‘ଫଳ 8’ ଆସିବାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(ii) ଗୋଟିଏ ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡ଼ାଗଲା । ‘ଫଳ 7ରୁ କମ୍’ ଆସିବାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(iii) ଗୋଟିଏ ଲୁଡୁଗୋଟି ଥରେ ଗଡ଼ାଗଲା । ‘ଫଳ ≤ 3’ ଆସିବାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(iv) ମିଲି ଓ ଲିମା ଟେନିସ୍ ଖେଳୁଥ‌ିଲେ । ଯଦି ଖେଳରେ ମିଲି ଜିଣିବାର ସମ୍ଭାବ୍ୟତା 0.62 ହୁଏ, ତେବେ ଲିମ୍ବା ହାରିବାର ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(v) ଦୁଇଟି ମୁଦ୍ରାକୁ ଥରେ ଟସ୍ କରାଗଲା । ‘ଫଳ ଅତିକମ୍‌ରେ ଗୋଟିଏ T’ ଆସିବାର ସମ୍ଭାବ୍ୟତା ସ୍ଥିର କର ।
(vi) ଗୋଟିଏ ପରୀକ୍ଷଣରେ ସମସ୍ତ ମୌଳିକ ବା ସରଳ ଘଟଣାଗୁଡ଼ିକର ସମ୍ଭାବ୍ୟତାର ସମଷ୍ଟି ସ୍ଥିର କର ।
(vii) P(E) = 0.05 ହେଲେ P(E) କେତେ ସ୍ଥିର କର ।
ସମାଧାନ:
(i) ଗୋଟିଏ ଲୁଡୁଗୋଟି ଥରେ ଗଡ଼ାଇଲେ ମୋଟ ଫଳାଫଳ ସଂଖ୍ୟା 6 ହେବ ।
8 ଲୁଡୁଗୋଟିରେ ଫଳର ବାରମ୍ବାରତା 0 ହେବ ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 1

(ii) ଗୋଟିଏ ଲୁଡୁଗୋଟି ଥରେ ଗଡ଼ାଇଲେ ମୋଟ ଫଳାଫଳ ସଂଖ୍ୟା 6 ହେବ ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 2
ଲୁଡୁଗୋଟିକୁ ଥରେ ଗଡାଗଲା । ଫଳ 7 ରୁ କମ୍ ନିଶ୍ଚିତ ଘଟଣା ହେତୁ ଉକ୍ତ ଘଟଣାର ସମ୍ଭାବ୍ୟତା = 1
P(1) = \(\frac{1}{6}\), P(2) = \(\frac{1}{6}\), P(3) = \(\frac{1}{6}\), P(4) = \(\frac{1}{6}\), P(5) = \(\frac{1}{6}\), P(6) = \(\frac{1}{6}\)
∴ P(<7) = 6 × \(\frac{1}{6}\) = 1

(iii) ଏଠାରେ ମୋଟ ଫଳାଫଳ ସଂଖ୍ୟା = 6
‘ଫଳ ≤ 3’ ଆସ।ର ସଂଖ୍ୟା = 3; (∵ ଫଳ ≤ 3 = {1, 2, 3})
∴ ଘଟଣାଉ ଉ।ଉପୃ।ରତା 3 ।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 3

(iv) ଖେଳଟିରେ ମିଲି ଜିଣିବାର ସମ୍ଭାବ୍ୟତା 0.62 ହେଲେ, ଲିମା ହାରିବାର ସମ୍ଭାବ୍ୟତା = 1 – 0.62 = 0.38
କାରଣ ହାରିବା ବା ଜିଣିବା ଉଭୟର ସମ୍ଭାବ୍ୟତାର ସମଷ୍ଟି ।
ବି. ତ୍ର.: P(E) = 0.62 ହେଲେ, P(E) = 1 – 0.62 = 0.38

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a)

(v) ଦୁଇଟି ମୁଦ୍ରାକୁ ଥରେ ଟସ୍କକଲେ ସମ୍ଭାବ୍ୟ ସମସ୍ତ ଫଳାଫଳଗୁଡ଼ିକ HH, HT, TH, TT ।
ଏଗୁଡ଼ିକର ସଂଖ୍ୟା 4 ।
ଘଟଣା E ଅତି କମ୍‌ରେ ଗୋଟିଏ I ଆସିବା ଏକ ଘଟଣାଦ୍ଵାରା ଅନୁଗୃହୀତ ଫଳାଫଳଗୁଡ଼ିକ TI, TH, HT । ଏଗୁଡ଼ିକର ସଂଖ୍ୟା = 3
∴ ଫଳ ଅତିକମରେ ଗୋଟିଏ ‘T’ ଆସିବାର ସମ୍ଭାବ୍ୟତା = P (E) = \(\frac{3}{4}\)
∴ ଦୁଇଟି ମୁଦ୍ରାର ଟସ୍‌ରେ ଅତିକମ୍‌ରେ ଗୋଟିଏ T ଆସିବାର ସମ୍ଭାବ୍ୟତା = \(\frac{3}{4}\)।

(vi) ଘଟଣା E ଘଟଣା Ē ର ପରିପୂରକ ଘଟଣା । ଅର୍ଥାତ୍ P(E) + P(E) = 1
∴ ଗୋଟିଏ ପରୀକ୍ଷଣରେ ସମସ୍ତ ମୌଳିକ ବା ସରଳ ଘଟଣାଗୁଡ଼ିକର ସମ୍ଭାବ୍ୟତାର ସମଷ୍ଟି = 1

(vii) P(E) = 0.05 ହେଲେ P(E) = 1 – \( 0.0 \overline{5}\) = 0.95
କାରଣ P(E) + P(E’) = 1)

Question 2.
ଗୋଟିଏ ବାକ୍ସରେ ତିନୋଟି ନୀଳ, ଦୁଇଟି ଧଳା ଓ ଚାରୋଟି ଲାଲ ମାର୍ବଲ ରହିଛି । ସେଥୁରୁ ଗୋଟିଏ ମାର୍ବଲ ବାକ୍ସରୁ ଯଦୃଚ୍ଛା (randomly) ବଛାଗଲା । ନିମ୍ନଲିଖତ କ୍ଷେତ୍ରରେ ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର।
(i) ଗୋଟିଏ ଧଳା ମାର୍ବଲ ଆସିବାର,
(ii) ଗୋଟିଏ ନୀଳ ମାର୍ବଲ ଆସିବାର ଓ
(iii) ଗୋଟିଏ ଲାଲ୍ ମାର୍ବଲ ଆସିବାର
ସମାଧାନ:
ବାକ୍ସରେ ଥ‌ିବା ବିଭିନ୍ନ ରଙ୍ଗର ସମୁଦାୟ ମାର୍ବଲ ସଂଖ୍ୟା
(i) ଧଳା ମାର୍ବଲ ସଂଖ୍ୟା = 2
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 4
(ii) ନୀଳ ମାର୍ବଲ ସଂଖ୍ୟା = 3
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 5
(iii) ଲାଲ୍ ମାର୍ବଲ ସଂଖ୍ୟା = 4
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 6

Question 3.
ଗୋଟିଏ ବ୍ୟାଗରେ ପାଞ୍ଚଟି ଧଳା, ଚାରୋଟି ଲାଲ୍ ଏବଂ ତିନୋଟି କଳା ଏକ ଆକୃତିବିଶିଷ୍ଟ ବଲ୍‌ ରହିଛି । ନିମ୍ନଲିଖତ କ୍ଷେତ୍ରରେ ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(i) ଗୋଟିଏ ଧଳାବଲ୍ ନ ଆସିବାର
(ii) ଗୋଟିଏ ଲାଲ୍ ବଲ୍‌ ନଆସିବାର
(iii) ଗୋଟିଏ ଧଳାବଲ୍ ନ ଆସିବାର
ସମାଧାନ:
ବ୍ୟାଗରେ ଥ‌ିବା ବିଭିନ୍ନ ରଙ୍ଗର ଏକ ଆକୃତି ବିଶିଷ୍ଟ ମୋଟ ବଲ୍‌ ସଂଖ୍ୟା = 5 + 4 + 3 = 12
(i) ବ୍ୟାଗରେ ଥ‌ିବା କଳା ବଲ୍‌ର ସଂଖ୍ୟା = 3
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 7

(ii) ଲାଲ୍ ବଲ୍‌ର ସଂଖ୍ୟା = 4
ଗୋଟିଏ ଲାଲ୍ ବଲ୍ ଆସିବାର ସମ୍ଭାବ୍ୟତା P(E) = \(\frac{4}{12}\) = \(\frac{1}{3}\)
ଗୋଟିଏ ଲାଲ୍ ବଲ୍ ନ ଆସିବାର ସମ୍ଭାବ୍ୟତା = P(E) = 1 – P(E) = 1 – \(\frac{1}{3}\) – \(\frac{2}{3}\)

(iii) ଧଳା ବଲ୍ ସଂଖ୍ୟା = 5
ଗୋଟିଏ ଧଳା ବଲ୍ ଆସିବାର ସମ୍ଭାବ୍ୟତା E = \(\frac{5}{12}\)
ଗୋଟିଏ ଧଳା ବଲ୍ ନ ଆସିବାର ସମ୍ଭାବ୍ୟତା E = 1 – E = 1 – \(\frac{5}{12}\) = \(\frac{7}{12}\)।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a)

Question 4.
ଗୋଟିଏ ବାକ୍ସରେ 60 ବୈଦ୍ୟୁତିକ ବଲ୍‌ବ ଅଛି । ସେଥ‌ିରୁ 12ଟି ଖରାପ ଏବଂ ଅନ୍ୟ ସମସ୍ତ ଭଲ ବଲ୍‌ବ । ସେଥ୍ ମଧ୍ୟରୁ ଗୋଟିଏ ବଲ୍‌ବ ଯଦୃଚ୍ଛା ବାହାର କରାଗଲା । ନିମ୍ନଲିଖତ କ୍ଷେତ୍ରରେ ସମ୍ଭାବ୍ୟତା ନିରୂପଣ କର ।
(i) ଗୋଟିଏ ଭଲ ବଲ୍‌ବ ବାହାରିବା
(ii) ଗୋଟିଏ ଖରାପ ବଲ୍‌ବ ବାହାରିବା
ସମାଧାନ:
ଗୋଟିଏ ବାକ୍ସରେ 60ଟି ବୈଦ୍ୟୁତିକ ବଲ୍‌ବ ଅଛି।
ସେଥୁରୁ 12ଟି ଖରାପ ବଲ୍‌ବ।
ଭଲ ବଲ୍‌ବର ସଂଖ୍ୟା 60 – 12 = 48
BSE Odisha 10th Class Maths Solutions Algebra Chapter 4 ସମ୍ଭାବ୍ୟତା Ex 4(a) - 8

CHSE Odisha Class 11 Math Notes Chapter 3 Relations And Function

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 3 Relations And Function will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 3 Relations And Function

Order Pairs
An ordered pair consists of a pair of objects, or elements or numbers or functions in order.
We denote order pairs as (a, b)

  • An order pair is not a set of two objects.
  • (a, b) = (c, d) ⇒ a = c and b = d
  • (a, b) × (b, a)

Cartesian Product Of Sets:
If A and B are non-empty sets, then their Cartesian product, denoted by A × B and defined by A × B = {(a, b): a ∈ A, b ∈ B} = Set of all ordered pairs (a,b) where a ∈ A and b ∈ B
Note:
1. For finite sets A and B |A × B| = |A| . |B|
2. A × B = Φ ⇔ A = Φ or B = Φ
3. A2 = A × A

Properties of Cartesian product:
1. A × B ≠ B × A (Cartesian product is non-commutative)
2. A × (B ∪ C) = (A × B) ∪ (A × C)
3. A × (B ∩ C) = (A × B) ∩ (A × C)
4. A × B = B × A ⇔ A = B
5. A × (B – C) = ( A × B) – (A × C)
6. A ⊂ B ⇒ A × A ⊂ (A × B) ∩ (B × A)
7. A ⊂ B ⇒ A × C ⊂ B × C
8. A ⊂ Band C ⊂ D ⇒ A × C ⊂ B × D
9. (A × B) ∩ (C × D) = ( A ∩ C) × (B ∩ D)

Relation
Let A and B be two arbitrary sets. A binary relation from A to B is a subset of A × B.
OR f is a relation from A to B if f ⊆ A × B
Note:

  • If a of A is related to b of B by relation ‘f’ then we write (a,b) ∈ f or a f b
  • As Φ ⊂ A × B we have Φ is a relation from A to B. This relation is known as a null of void or empty relation.
  • As A × B ⊆ A × B, A × B is also a relation from A to B. This relation is known as universal relation.
  • If |A| = m and |B| = n then number of relations from A to B is 2mn

Domain, co-domain, and Range of a relation:
Let f is a relation from A to B. Domain of f = Dom (f) or Df
={x ∈ A : (x, y) ∈ f for some y ∈ B) Co-domain of f = B
Range of ‘f’ = Rng (f) or Rf = {y ∈ B : (x, y) ∈ f for some x ∈ A}

CHSE Odisha Class 11 Math Notes Chapter 3 Relations And Function

Types Of Relation:
(a) One-many relation: A relation f from A to B is one many if (a, b) and (a, b’) ∈ f ⇒ b ≠ b’
(b) Many-one relations: A relation f from A to B is many-one if (a, b) and (a’, b) ∈ f ⇒ a ≠ a’
(c) One-one relation: A relation f from A to B is one-one if (a, b), (a, b’) ∈ f ⇒ b = b’ and (a, b), (a’, b) ∈ f ⇒ a = a’

Inverse of a relation: Let f is a relation from A to B. The inverse of f is denoted by f-1 is a relation from B to A defined as f-1 = {(b, a): (a, b) ∈ f}

Function:
A relation ‘f’ from X to Y is called a function if:
(a) Df = Dom (f) = X and
(b) (x, y) and (x, z) ∈ f ⇒ y = z or A relation from A to B is a function
if ⇒ Domain of f = X i.e All elements of X is engaged in the relation and
⇒ f is not one many.

Note:
(1) If a relation f from X to Y becomes a function then we write f: X → Y.
(2) If f is a function from A to B i.e. f: X → Y and (x, y) ∈ f then we write y = f(x)
(3) Mapping, map, transformation, transform, operator, and correspondence are different synonym terms of function.
(4) If f: X → Y is defined as y = f(x), then

  • y is called the value of the function at x or the image of x under f or the dependent variable.
  • x is called the independent variable or pre-image of y under f.

Domain, Co-domain or Range of a function:
Let f: X → Y defined as y = f(x)
(a) Domain of ‘f’ = Dom f = Df = {x ∈ X: y = f(x)}
(b) Range of f = Rng f = Rf = f(A) = {f(x) ∈ Y: x ∈ A } Clearly f(A) ⊆ y
(c) If |A| = m, |B| = n then number of functions from A to B = nm

Real valued function :
A function f: A → B is a real-valued function if B ⊆ R.
→ f is a real function if A ⊆ R and B ⊆ R

Techniques to find Domain and Range of a Real function:
(a) Techniques to find Domain: Let the function is defined as y = f(x).
Step -1: Check the values of x for which f(x) is well defined.
Step -2: The set of all values obtained from step -1 is the domain of ‘f.

(b) Techniques to find range: Let the function is y = f(x)

  • Method-1 (By inspection):
    → Step -1: Get values of y for all values x ∈ Dom f.
    → Step -2: Set of all these values of y = Rng f
  • Method-2:
    → Step -1: Write x in terms of y
    → Step -2: Get values of y for which x is well defined in Dom f.
    → Step -3: Rng (f) = The set of all y obtained from step 2.

Some Real Functions:
(a) Constant function: A function f: A → R defined as f(x) = k, for some k ∈ R is called a constant function.

(b) Identity function: Let A ⊆ R. The function f: A → A defined as f(x) = x, x ∈ A is called the identity function on A. We denote it by IA

(c) Polynomial function: A function f: A → R defined by f(x) = f(x) = a0 + a1x + a2x2 + anxn where a0, a1, a2, ….., an are real numbers and an ≠ 0 is called a polynomial function (polynomial) of degree n.

(d) Rational function: A function of form f(x) = \(\frac{\mathrm{P}(x)}{\mathrm{Q}(x)}\) where P(x) and Q(x) are polynomial functions of x is known as a rational function.

(e) Absolute value function OR modulus function: The function f: R → R defined as  f(x) = |x| = \(\begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}\) is called as the modulus function.
→ Rng f = [ 0, ∞] = R+U {0}

CHSE Odisha Class 11 Math Notes Chapter 3 Relations And Function

Properties Of Modulus Function:
1. For any real number x, we have \(\sqrt{x^2}=|x|\)
2. If a and b are positive real numbers then

  • x2 ≤ a2 ⇔ |x| ≤ a
  • x2 ≥ a2 ⇔ |x| ≥ a
  • a2 ≤ x2 ≤ b2 ⇔ a ≤ |x| ≤ b ⇔ x ∈ [-b, – a] ∪ [a, b]

(f) Signum function: The function f: R → R defined as f(x) = \(\begin{cases}\frac{x}{|x|}, & x \neq 0 \\ 0, & x=0\end{cases}\) is called signum function.
→ We denote a signum function as f(x) = sgn(x)
→ Range of a signum function = {-1, 0, 1}

(g) Greatest integer function: The function f: R → R defined by f(x) = [x] is called the greatest integer function. [x] = The greatest among all integers ≤ x. OR [x] = n for n ≤ x < n + 1

Properties of the greatest integer function :
Let n is an integer and x is a real number between n and n + 1
(i) [-n]= -[n]
(ii) [x + k] = [x] + k (for an integer ‘k’)
(iii) [-x] = -[x] – 1
(iv) [x] + [-x] = \(\begin{cases}-1, & \mathrm{x} \notin \mathrm{Z} \\ 0, & \mathrm{x} \in \mathrm{Z}\end{cases}\)
(v) [x] – [-x] = \(\begin{cases}2[\mathrm{x}]+1, & \mathrm{x} \notin \mathrm{Z} \\ 2[\mathrm{x}], & \mathrm{x} \in \mathrm{Z}\end{cases}\)
(vi) [x] ≥ k ⇒ x ≥ k for k ∈ Z
(vii) [x] ≤ k ⇒ x < k +1 for k ∈ Z
(viii) [x] > k ⇒ x > k + 1 for k ∈ Z
(ix) [x] < k ⇒ x < k for k ∈ Z

(h) Exponential Function: A function f: R → R defined as f(x) = ax where a > 0 and a ≠ 1 is called the exponential function.

Properties Of Exponential Function:
1. ax+y =  ax . ay
2. (ax)y = axy
3. ax = 1 if x = 0
4. If a > 1, ax > ay ⇒ x > y
5. If a < 1, ax > ay ⇒ x < y

Logarithmic Function:
Let a ≠ 1 is a positive real number. The function f: (0, ∞) → R defined by f(x) = logax is called the logarithmic function, where y = loga ⇔ ay = x
→ Domain of a logarithmic function = (0, ∞) and Range = R

Properties of logarithmic function:
1. loga (xy) = logax + logay
2. loga (x/y) = logax – logay
3. logaa = 1
4. loga(x)y = y logax
5. loga x = 0 ⇔ x = 1
6. logax = \(\frac{1}{\log _a{ }_a}\) , x ≠ 1
7. logab = \(\frac{\log _a b}{\log _c a}\)
8. \(\log _{a^n}\left(x^m\right)\) = \(\frac{m}{n}\) loga|x|

Different Categories of function:
(a) Algebraic Function: A function that can be generated by a variable by a finite number of algebraic operations such as addition, subtraction, multiplication, division, square root, etc. is called an algebraic function.

(b) Transcendental function: A non-algebraic function is a transcendental function.
⇒ Trigonometric, trigonometric, Exponential, and logarithmic functions are transcendental functions.

Even And Odd Functions:
A function ‘f’ is an even function  if f(-x) = x and is an odd function
if f(-x) = x and is an odd function: if f(-x) = -f(x)
Note:
1. If ‘f’ is any function f(x) + f(-x) is always an even function and f(x) – f(-x) is an odd function.
2. Every function f(x) can be expressed as the sum of an even and an odd function as f(x) = g(x) + h(x), where
g(x) = \(\frac{f(x)+f(-x)}{2}\)
h(x) = \(\frac{f(x)-f(-x)}{2}\)

Periodic Function:
A function is called a periodic function with period k if f(x + k) = f(x) for some constant k ≠ 0. The least positive value of k for which f(x + k) = f(x) holds is called the fundamental period of f.

Properties of periodic function :
(1) If k is the period of f then any non-zero integral multiples of k is also a period of f.
(2) If k is the period of f(x) then f(ax + b) is also periodic with period \(\frac{k}{a}\)
(3) If f1(x) + f2(x) and f3(x) are periodic functions with periods k1, k2, k3, respectively then the function a1f1(x) + a2f2(x) + a3f3(x) is also a periodic function with period, LCM (k1, k2, k3)

CHSE Odisha Class 11 Math Notes Chapter 3 Relations And Function

Algebra Of Real functions:
(a) Equality of two functions: Two functions f and g are equal iff ‘
(i) Dom f = Dom g
(ii) Co-Dom f = Co-Dom g
(iii) f(x) = g(x) for all x belonging to their common domain.

(b) Addition of two functions: Let f: D1 → R and g: D2 → R be two real functions.
The sum function f + g is defined by f + g: D1 ∩ D2 → R and (f + g)(x) = f(x) + g(x) ∀ x ∈ D1 n D2

(c) Subtraction of two functions: Let f: D1 → R and g: D2 → R. The difference function (f – g) is f – g: D1 ∩ D2) → R defined by (f – g) (x) = f(x) – g(x) ∀ x ∈ D1 ∩ D2

(d) Scalar multiplication: Let f: D → R and c is any scalar. The scalar multiple of f by the scalar c is cf: D → R defined as (cf)(x) = c. f(x) ∀ x ∈ D1.

(e) Multiplication of two functions: Let f: D1 → R and g: D2 → R are two real functions. The product function (fg) is (fg): D1 ∩ D2 → R defined as (fg)(x) = f(x)g(x) ∀ ∈ D1 ∩ D2

(f) The quotient of two functions: Let f: D1 → R and g: D2 → R are two real functions. the quotient function (\(\frac{f}{g}\)) i,e,. \(\frac{f}{g}\): D1 ∩ D2 → R, defined by (\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\), ∀ x ∈ D1 ∩ D2

CHSE Odisha Class 11 Math Notes Chapter 2 Sets

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 2 Sets will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 2 Sets

Set:
Set is an undefined term in mathematics. But we understand set as “a collection of well-defined objects”.

  • Set is a collection.
  • The objects (called elements) in a set must be well-defined.

Set Notation:
We denote set as capital alphabets like A, B, C, D…..and the elements by the small alphabets like x, y, z ….

  • If x is an element of set A we say “x belongs to A” and write ‘x ∈ A’.
  • If x is not an element of set A we say “x does not belong to A” and we write ‘x ∉ A’.

Set Representation:
(a) Extension or tabular or Roster Method: In this method, we describe a set listing the elements, separated by commas within curly brackets.
Note: While listing out the elements the repetition of an object has no effect. Thus, we don’t do this.

(b) Intention or set builder or set selector method: In this method: a set is described by a characterizing property p(x) of element x. In this case, the set is described as {x : p(x) holds}

Types Of Set:
(a) Empty of full or void set: It is a set with no element.

  • We denote empty set by ‘Φ’
  • There is only one empty set.

(b) Singleton set: It is a set with only one element.

(c) Finite set: A set is finite if it has a finite number of elements.

(d) Infinite Set: A set that is not finite is called an infinite set.

(e) Equal sets: Two sets A and B are equal if they have the same elements. Two sets A and B are equal if all elements of A are also elements of B and all elements of B are also elements of A.

(f) Equivalent set: Two finite sets A and B are equivalent if they have the same number of elements.

CHSE Odisha Class 11 Math Notes Chapter 2 Sets

Subsets: Let A and B be two sets. If every element of A is an element of B then A is called a. subset of B (we write A ⊂ B) and B is called a superset of A (We write B ⊃ A)
Thus A ⊂ B is x ∈ A ⇒ x ∈ B
Note.
(i) A set is a subset of itself.
(ii) Empty set Φ is a subset of every set.
(iii) A is called a proper subset of B if B contains at least one element that is not in A
(iv) If A has n elements then total number of subsets of A = 2n.

Universal set:
A set ‘U’ that contains all sets in a given context is called the universal set.

Power set:
Let A is any set. The collection (or set) of all subsets of A is called the power set of A. We denote it as P(A)
P(A) = { S: S ⊂ A }

Set Operations:
(a) Union of sets :
The union of two sets A and B is the set of all elements of A or B or both.
∴ A ∪ B = {x ∈ A or x ∈ B}

(b) Intersection of sets:
Intersection of two sets A and B is the set of all those elements that belong to both A and B . (or all common elements of A and B)
∴ A ∩ B = {x: x ∈ A and x ∈ B }
Two sets A and B are disjoint if A ∩ B = Φ. Otherwise, A and B are intersecting or overlapping sets.

(c) Difference of sets: The difference of two sets ‘A and B’ is the set of all elements of A which do not belong to B.
∴ A- B = {x: x ∈ A and x ∈ B)

(d) Symmetric difference of two sets: Symmetric difference of two sets A and B is the set (A – B) ∪ (B – A)
∴ A Δ B = (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)

(e) Complement of a set: Let the complement of a set A (denoted as A’ or Ac) be defined as U – A

  • A’ = {x ∈ U) : x ∉ A)
  • x ∈ A’ ⇔ x ∉ A

Laws Of Set Algebra:
(a) Idempotent law: For any set A we have
(i) A ∪ A = A
(ii) A ∩ A = A

(b) Identity laws: For any set A we have
(i) A ∪ Φ = A and
(ii) A ∩ U = A

(c) Commutative laws: For any three sets A, B, and C
(i) A ∪ B = B ∪ A
(ii) A ∩ B = B ∩ A

(d) Associative laws: For any three sets A, B, and C
(i) A ∪ (B ∪ C) = (A ∪ B) ∪ C
(ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C

(e) Distributive laws: For any three sets A, B, and C
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(f) De-morgans laws: For any two sets A and B
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’

CHSE Odisha Class 11 Math Notes Chapter 2 Sets

Some more properties of sets: For any three sets A, B, and C
(a) A ⊂ (A ∪ B) and (A ∩ B) ⊂ A
(b) A ∪ B = B ⇔ A ⊂ B
(c) A ∩ B = A ⇔ A ⊂ B
(d) B ⊂ A and C ⊂ A ⇒ (B ∪ C ) ⊂ A and A ⊂ B, A ⊂ C ⇒ A ⊂ (B ∩ C)
(e) B ⊂ C ⇒ A ∪ B ⊂ A ∪ C and A ∩ B ⊂ A ∩ C
(f) A – B = A ∩ B’
(g) A – B = A ⇔ A ∩ B = Φ
(h) (A – B) ∪ B = A ∪ B and (A – B ) ∩ B = Φ
(i) A ⊆ B ⇔  B’ ⊆ A’
(j) A Δ B = B Δ A

Cardinality or order of a finite set: The cardinality or order of a finite set A (denoted as |A| or O(A) or n (A)) is the number of elements in ‘A’.

Some important results on the cardinality of finite sets and applications of set theory:
If A, B, and C are finite sets and ‘U’ is the finite universal set then a number of elements belonging to at least one of A or B.
(a) |A ∪ B| = |A| + |B| – |A ∩ B|
(b) |A ∪ B| = |A| + |B| for A ∩ B = Φ i.e. for two disjoint sets A and B
(c) Number of elements belonging to at least one of A, B, or C
= |A ∪ B ∪ C|
= |A| + |B| + |C| – |A ∩ B| – |B ∩ C| – |C ∩ A| + |A ∩ B ∩ C|
(d) Number of elements belonging to exactly two of the three sets A, B, and C = |A ∩ B| + |B ∩ C| + |C ∩ A| – 3 |A ∩ B ∩ C|
(e) Number of elements belonging to exactly one of the three sets A, B, and C = |A| + |B| + |C| – 2 |A ∩ B| -2 |B ∩ C| – 2 |C ∩ A| + 3 |A ∩ B ∩ C|
(f) Number of elements belonging to A but not B = |A – B| = |A| – |A ∩ B|
∴ |A| = |A – B| + |A ∩ B|
(g) Number of elements belonging to exactly one of A or B
= |A Δ B| = |A – B| + |B – A|
= |A| + |B| – 2 |A ∩ B|
(h) |A’ ∪ B’| = |U| – |A ∩ B|
(i) |A’ ∩ B’| = |U| – |A ∪ B| = Number of elements belonging to neither A nor B.

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(a) \(\frac{1}{15×16}=…..- \frac{1}{16}\)
(b) \(\frac{1}{12×11}=- \frac{1}{11}\) – …….
(c) \(\frac{1}{n(n+1)}=…..- \frac{1}{n+1}\)
(d) \(\frac{1}{(n+1)n}=- \frac{1}{n}\) – …….
(e) 5 ଓ 9 ମଧ୍ୟରେ ଥ‌ିବା ସମାନ୍ତର ମଧ୍ଯକଟି …..
(f) x ଓ 7 ମଧ୍ଯସ୍ଥ ସମାନ୍ତର ମଧ୍ୟକଟି 5 ହେଲେ x = …..
(g) (a + b) 8 (a – b) ମଧ୍ୟରେ ଥ‌ିବା ସମାନ୍ତର ମଧ୍ଯକଟି ………
(h) ଦୁଇଟି ରାଶିର A.M. 11, ଯଦି ଗୋଟିଏ ରାଶି 7 ହୁଏ, ତେବେ ଅନ୍ୟଟି ……….
ଉ-
(a) \(\frac{1}{15}\)
(b) \(\frac{1}{12}\)
(c) \(\frac{1}{n}\)
(d) \(\frac{1}{n+1}\)
(e) 7
(f) 3
(g) a
(h) 15

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(a) \(\frac{1}{15 \times 16}=\frac{1}{15}-\frac{1}{16}\left[\text { R.H.S. }=\frac{1}{15}-\frac{1}{16}=\frac{16-15}{15 \times 16}=\frac{1}{15 \times 16}=\text { L.H.S. }\right]\)
(b) \(\frac{1}{12×11}=\frac{1}{11}-\frac{1}{12}\)
(c) \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
(d) \(\frac{1}{(n+1)n}=\frac{1}{n}-\frac{1}{n+1}\)
(e) 5 ଓ 9 ମଧ୍ୟରେ ଥ‌ିବା ସମାନ୍ତର ମଧ୍ଯକଟି = \(\frac{5+9}{2}=\frac{14}{2}=7\)
(f) x ଓ 7 ମଧ୍ଯସ୍ଥ ସମାନ୍ତର ମଧ୍ୟକଟି 5 ହେଲେ x = \(\frac{x+7}{2}\) = 5 ⇒ x = 10 – 7 = 3
(g) (a + b) 8 (a – b) ମଧ୍ୟରେ ଥ‌ିବା ସମାନ୍ତର ମଧ୍ଯକଟି \(\frac{a+b+a-b}{2}=\frac{2a}{2}\) = a
(h) ଦୁଇଟି ରାଶିର A.M. 11, ଯଦି ଗୋଟିଏ ରାଶି 7 ହୁଏ, ତେବେ ଅନ୍ୟଟି x
∴ \(\frac{x+7}{2}\) = 11 ⇒ x = 22 – 7 = 15

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Question 2.
ନିମ୍ନଲିଖୂତ ଅନୁକ୍ରମଗୁଡ଼ିକର ସମଷ୍ଟି ନିଶ୍ଚୟ କର ।
(a) \(\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}\) ……..20ଟି ପଦ ପର୍ଯ୍ୟନ୍ତ;
(b) \(\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}\) ……..16ଟି ପଦ ପର୍ଯ୍ୟନ୍ତ
ସମାଧାନ :
(a) \(\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}\) ……..20ଟି ପଦ ପର୍ଯ୍ୟନ୍ତ;
BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) -1
(b) \(\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}\) …….16ଟି ପଦ ପର୍ଯ୍ୟନ୍ତ ସମଷ୍ଟି।
BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) -2

Question 3.
(a) 7 × 15 + 8 × 20 + 9 × 25 + …..ର tn ନିର୍ଣ୍ଣୟ କର ।
(b) 6Σn²+4Σn³ ର ସରଳୀକୃତ ମାନ ନିର୍ଣ୍ଣୟ କର ।
(c) 1 × 2 + 2 × 3 + 3 × 4 ….. + n (n + 1) ପାଇଁ Sn ଓ S20 ନିର୍ଣ୍ଣୟ କର ।
(d) 1 × 3 + 2 × 4 + 3 × 5 …… tn, Sn ଓ S10 ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(a) 7 × 15 + 8 × 20 + 9 × 25 …….ର tn
ରାଶିମାଳାର ପ୍ରଥମ ଗୁଣନୀୟକଗୁଡ଼ିକ 7,8, 9, A.P.ରେ ଅଛନ୍ତି ।
a = 7, d = 8 – 7 = 9 – 8 = 1
∴ tn = a + (n – 1) d = 7 + (n – 1 ) 1 = 7 + n – 1 = n + 6
ରାଶିମାଳାଟିର ଦ୍ଵିତୀୟ ଗୁଣନୀୟକଗୁଡ଼ିକ 15, 20, 25 ……. A.P. ଅଛନ୍ତି ।
a = 15, d = 5, tn = a + (n – 1)d = 15 + (n – 1)5 = 15 + 5n – 5 = 5n + 10
∴ 7 × 15 + 8 × 20 + 9 × 25 + …..ର tn = (n + 6)(5n + 10)
= 5(n + 6)(n + 2) = 5(n² + 8n + 12)

(b) 6Σn²+4Σn³ = \(\frac{6 \times n(n+1)(2 n+1)}{6}+4\left\{\frac{n(n+1)}{2}\right\}^2\)
= n(n + 1)(2n + 1) + n²(n + 1)²
= n(n + 1){(2n+ I + n(n + 1)} = n(n + 1)(n² + 3n + 1)

(c) 1 × 2 + 2 × 3 + 3 × 4 ….. + n (n + 1)
ଏଠାରେ tn = n(n + 1)= n² + n
Sn = Σn² + Σn = \(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\)
= n(n+1){\(\frac{2 n +1}{6}+\frac{1}{2}\)}
= (n² + n){\(\frac{2 n +1+3}{6}\)} = \(\frac{(n² + n)(2 n +4)}{6}\)
= \(\frac{n (n+1)(n+2)}{3}\)
S20 = \(\frac{20×21×22}{3}\) = 3080

(d) 1 × 3 + 2 × 4 + 3 × 5 ……
ପ୍ରତ୍ୟେକ ପଦର ପ୍ରଥମ ଗୁଣନୀୟକ 1, 2, 3, 4 ……. । ଏହାର t = n
ପ୍ରତ୍ୟେକ ପଦର ଦ୍ଵିତୀୟ ଗୁଣନୀୟକଗୁଡ଼ିକ ହେଲେ 3, 4, 5, …
0166 a = 3, d=4-35-4=1
t =
= a + (n – 1 ) d = 3 + (n – 1) 1 = n + 2
∴ ରାଶିଟିର t = n(n + 2) = n² + 2n
Sn = Σn² + 2Σn = \(\frac{n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2}\)
= n(n+1)(\(\frac{2 n +1}{6}+1\)) = \(\frac{n(n+1)(2 n +7)}{6}\)
Sn = \(\frac{n(n+1)(2 n +7)}{6}\)
S10 = \(\frac{n(n+1)(2 n +7)}{6}\) = \(\frac{10×(10+1)(2×10 +7)}{6}\) = \(\frac{10×11×27}{6}\) \(\frac{2970}{6}\) = 495

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Question 4.
ନିମ୍ନଲିଖତ ଶ୍ରେଣୀଗୁଡ଼ିକର n ସଂଖ୍ୟକ ପଦ ପର୍ଯ୍ୟନ୍ତ ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
(a) 1.1 + 2.3 + 3.5 +4.7 + …….
(b) 1.3 +3.5 + 5.7 + 7.9 + …….
(c) 3.8 +6.11 + 9.14 + …….
(d) 1+ (1 + 3) + (1 + 3 + 5) +
(e) 1² + 4² + 7² + 10² + …….
(f) 2² + 4² +6² + 8² + …….
(g) 1 + 5 + 12 +22 + 35+…….
(h) 1² + (1² + 2²) + (1² + 2² + 3²) + (1² + 2² + 3² + 4²) + ……….
ସମାଧାନ :
(a) 1.1 + 2.3 + 3.5 + 4.7 + …..
ଦତ୍ତ ଶ୍ରେଣୀର ପ୍ରଥମ ଗୁଣନୀୟକଗୁଡ଼ିକ 1, 2, 3, 4 …….
ଏହାର tn = n
ସେହିପରି ଦ୍ୱିତୀୟ ଶ୍ରେଣୀର ଗୁଣନୀୟକଗୁଡ଼ିକ 1, 3, 5, 7 ……. । ଏହାର a = 1, d = 3 − 1 = 2
ଏହାର tn = 1 + (n – 1) × 2 = 2n – 1
ଦତ୍ତ ଶ୍ରେଣୀର tn = n(2n – 1) = 2n² – n
Sn = 2Σn² – Σn = \(\frac{2n (n + 1)(2n + 1)}{6}-\frac{n(n+1)}{2}\)
= n(n+1)(\(\frac{4 n +2}{6}-\frac{1}{2}\)) = \(\frac{n(n+1)(4 n +2-3)}{6}\) = \(\frac{n(n+1)(4 n -1)}{6}\)
Sn = \(\frac{n(n+1)(4 n -1)}{6}\)

(b) 1.3 + 3.5 + 5.7 + 7.9 + …….
ଦତ୍ତ ଶ୍ରେଣୀର ପ୍ରତ୍ୟେକ ପଦର ପ୍ରଥମ ଗୁଣନୀୟକ 1, 3, 5, 7 ……. A.P. ଅଟନ୍ତି ।
tn = 1 + (n – 1) 2 = 2n – 1
ଦତ୍ତ ଶ୍ରେଣୀର ପ୍ରତ୍ୟେକ ପଦର ଦ୍ୱିତୀୟ ଗୁଣନୀୟକ 3, 5, 7, 9,
ଏଠାରେ a = 3, d = 5 – 3 = 2
tn = 3 + (n – 1) × 2 = 3 + 2n – 2 = 2n + 1
BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) -3

(c) 3.8 +6.11 + 9.14 + …….
ଶ୍ରେଣୀର ପ୍ରତ୍ୟେକ ପଦର ପ୍ରଥମ ଗୁଣନୀୟକଗୁଡ଼ିକ 3, 6, 9, ……. A.P. ଅଟନ୍ତି ।
a = 3, d = 6 – 3 = 9 – 6 = 3, t = 3 + (n – 1 ) × 3 = 3n
ଦତ୍ତ ଶ୍ରେଣୀର ପ୍ରତ୍ୟେକ ପଦର ଦ୍ବିତୀୟ ଗୁଣନୀୟକଗୁଡ଼ିକ 8, 11, 14, …….
a = 8, d = 11-8 = 14 – 11 = 3
tn = a + (n – 1) d = 8 + (n – 1) 3 = 8 + 3n – 3 = 3n + 5
BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) -4

(d) Sn = 1 + (1 + 3) + (1 + 3 + 5) + …….. + tn
= 1 + (1 + 3) + (1 + 3 + 5) + …… +(1 + 3 + 5 + 7) …….
ଦତ୍ତ ଶ୍ରେଣୀର tn = 1 + 3 + 5 + 7 +……. + n-ତମ ପଦ ପର୍ଯ୍ୟନ୍ତ
⇒ tn = \(\frac{n}{2}\) {2·1 + (n – 1) 2} = \(\frac{n}{2}\) (2 + 2n – 2) = n²
⇒ Sn = Σn² = \(\frac{2n (n + 1)(2n + 1)}{6}\) = \(\frac{1}{6}\)n (n² + 3n + 1)

(e) 1² + 4² + 7² + 10² + …….
1, 4, 7, 10 ……… A.P.
a = 1, d= 4 – 1 = 7 – 4 = 3, tn = 1 + (n – 1) × 3 = 1 + 3n – 3 = 3n – 2
ଦତ୍ତ ଶ୍ରେଣୀର tn = (3n – 2)² = 9n² – 12n + 4
⇒ Sn = 9Σn² – 12Σn + 4Σ1 = \(\frac{9n (n + 1)(2n + 1)}{6}-12 \frac{n(n+1)}{2}+4n\)
= \(\frac{(9n²+9n) (2n+1)-36n²-36n+24n}{6}\)
= \(\frac{1}{6}\) (18n³ + 9n² + 18n² + 9n – 36n² – 36n + 24n)
= \(\frac{1}{6}\) (18n³ – 9n² – 3n)
= \(\frac{3}{6}\) (6n³ – 3n² – n)
= \(\frac{n}{2}\) (6n³ – 3n² – n)

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

(f) Sn = 2² + 4² +6² + 8² + ……. = 2²(1² + 2² +3² + 4² + ….. + tn)
= \(\frac{4n(n + 1)(2n + 1)}{6}\) (∵ 1² + 2² +3² + 4² + ….. + n² = \(\frac{4n(n + 1)(2n + 1)}{6}\))
⇒ Sn = \(\frac{2}{3}\) n(n + 1)(2n + 1)

(g)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) -5
⇒ tn = 1 + 4 +7 + 10 + 13 + ….. n ତମ ପଦ ପର୍ଯ୍ୟନ୍ତ
= \(\frac{n}{2}\) {2 + (n – 1) × 3}
= \(\frac{n}{2}\) {2 + 3n – 3}
= \(\frac{n}{2}\) {3n – 1}
= \(\frac{3}{2}\) n² – \(\frac{1}{2}\)n
Sn = \(\frac{3}{2}\) Σn² – \(\frac{1}{2}\) Σn
= \(\frac{3}{2}\) \(\frac{4n(n + 1)(2n + 1)}{6}\) – \(\frac{1}{2}\) \(\frac{n(n + 1)}{3}\)
= \(\frac{n(n + 1)(2n + 1)}{4}\) – \(\frac{n(n + 1)}{4}\)
= \(\frac{n(n + 1)}{4}\) (2n + 1 – 1) = \(\frac{n(n + 1)×2n}{4}\) = \(\frac{1}{2}\)n² (n+1)

(h) Sn = 1² + (1² + 2²) + (1² + 2² + 3²) + (1² + 2² + 3² + 4²) + ……….
tn = (1²+ 2²+ 3² + ………. + n²)
BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b) -6

Question 5.
15 ଓ 27 ମଧ୍ଯରେ (i) ଗୋଟିଏ ଓ (ii) ଦୁଇଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ ସ୍ଥାପନ କର ।
ସମାଧାନ :
(i) 15 ଓ 27 ମଧ୍ଯରେ ଗୋଟିଏ ସମାନ୍ତର ମଧ୍ୟକ x = \(\frac{a+b}{2}=\frac{15+27}{2}=\frac{42}{2}=21\)

(ii) ମନେକର 15 ଓ 27 ମଧ୍ୟରେ ଅବସ୍ଥିତ ଥ‌ିବା ଦୁଇଟି ସମାନ୍ତର ମଧ୍ୟକ x1 ଓ x2
ଏଠାରେ a = 15, b = 27, d = \(\frac{b-a}{3}=\frac{27-15}{3}=\frac{12}{3}=4\)
x1 = a + d = 15 + 4 = 19, x2 = a + 2d = 15 + 2 × 4 = 23
∴ 15 ଓ 27 ମଧ୍ୟରେ ଅବସ୍ଥିତ ଦୁଇଟି A.M. 19 ଓ 23।

ବିକଳ୍ପ ସମାଧାନ :
(i) 15 ଓ 27 ମଧ୍ଯସ୍ଥ ଗୋଟିଏ ସମାନ୍ତର ମଧ୍ୟକ = \(\frac{15+27}{2}=\frac{42}{2}=21\)
(ii) ମନେକର 15 ଓ 27 ମଧ୍ୟସ୍ଥ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ x, y ।
∴ 15, x, y, 27 A.P. ରେ ଆବସ୍ଥିତ ।
a = 15, a + d = x, a + 2d = y, a + 3d = 27
⇒ 15 + 3d = 27 ⇒ 3d = 27 – 15 = 12 ⇒ d = 4
x = a + d = 15 + 4 = 19, y = a + 2d = 15 + 2 × 4 = 23
∴ 15 ଓ 27 ମଧ୍ୟସ୍ଥ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ 19 ଏବଂ 23 ।

ବିକଳ୍ପ ପ୍ରଣାଳୀ :
ଏଠାରେ d = \(\frac{27-15}{3}=\frac{12}{3}=4\)
x1 = a + d = 15 + 4 = 19 ଏବଂ x2 = a + 2d = 15 + 2 × 4 = 23

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Question 6.
12 ଓ 36 ମଧ୍ଯରେ (i) ଦୁଇଗୋଟି ଓ (ii) ତିନିଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ ସ୍ଥାପନ କର ।
ସମାଧାନ :
(i) ମନେକର 12 ଓ 36 ମଧ୍ଯରେ ଅବସ୍ଥିତ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ p1 ଓ p2
ଏଠାରେ a = 12, b = 36, d = \(\frac{b-a}{3}=\frac{36-12}{3}=\frac{24}{3}=8\)
P1 = a + d = 12 + 8 = 20, p2 = 12 + 2d = 12 + 2 × 8 = 28
∴ 12 ଓ 36 ମଧ୍ୟରେ ଅବସ୍ଥିତ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ 20 ଓ 28 ।

(ii) ମନେକର 12 ଓ 36 ମଧ୍ଯରେ ଅବସ୍ଥିତ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ x1,x2 ଓ x3
ଏଠାରେ a = 12, b = 36, d = \(\frac{b-a}{4}=\frac{36-12}{4}=\frac{24}{4}=6\)
x1 = a + d = 12 + 6 = 18, x2 = a + 2d = 12 + 2 × 6 = 24.
x3 = a + 3d = 12 + 3 × 6 = 12 + 18 = 30
∴ 12 ଓ 36 ମଧ୍ୟରେ ଅବସ୍ଥିତ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ 18, 24 ଓ 30 ।

ବିକଳ୍ପ ସମାଧାନ :
(i) ମନେକର 12 ଓ 36 ମଧ୍ଯସ୍ଥ ଗୋଟି ସମାନ୍ତର ମଧ୍ୟକ x ଏବଂ y ।
∴ 12, x, y, 36 A.P. ରେ ଆସ୍ଥିତ ।
ଏଠାରେ a = 12 ଓ t4 = 36 ⇒ a + (4 – 1) d = 36
⇒ 12 + 3d = 36 ⇒ 3d = 36 – 12 = 24 ⇒ d = \(\frac{24}{3}=8\)
x = a + d = 12 + 8 = 20, y = a + 2d = 12 + 2 × 8 = 28

(ii) ମନେକର 12 ଓ 36 ମଧ୍ୟସ୍ଥ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ଯକ x, y, z । 12, x, y, z, 36 A.P. ରେ ଆସ୍ଥିତ ।
ଏଠାରେ a = 12 ଓ t5 = 36 ⇒ a + (5 – 1) d = 36
⇒ a + 4d = 36 ⇒ 4d = 36 – 12 = 24 ⇒ d = 6
x = a + d = 12 + 6 = 18, y = a + 2d = 12 + 2 × 6 = 12 + 12 = 24
z = a + 23d = 12 + 3 × 6 = 12 + 18 = 30

Question 7.
6 ଓ 46 ମଧ୍ଯରେ (i) ଦୁଇଗୋଟି ଓ (ii) ତିନିଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ ସ୍ଥାପନ କର ।
ସମାଧାନ :
(i) ମନେକର 6 ଓ 46 ମଧ୍ୟରେ ଅବସ୍ଥିତ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ୟକ x1 ଓ x2
ଏଠାରେ a = 6, b = 46, d = \(\frac{46-6}{3}=\frac{40}{3}\)
x1 = a + d = 6 + \(\frac{40}{3}\) = \(\frac{18+40}{3}=\frac{58}{3}\)
x2 = a + 2d = 6 + 2 × \(\frac{40}{3}\) = \(6+\frac{80}{3}=\frac{18+80}{3}=\frac{98}{3}\)
∴ 6 ଓ 46 ମଧ୍ୟରେ ଦୁଇଟି ସମାନ୍ତର ମଧ୍ୟକ \(\frac{58}{3}\) ଓ \(\frac{58}{3}\) ।

(ii) ମନେକର 6 ଓ 46 ମଧ୍ୟରେ 4ଟି ସମାନ୍ତର ମଧ୍ୟକ x1,x2,x3 ଓ x4
ଏଠାରେ a = 6, b = 46,
d = \(\frac{b-a}{5}=\frac{46-6}{5}=\frac{40}{5}=8\)
x1 = a + d = 6 + 8 = 14
x2 = a + 2d = 6 + 2 × 8 = 22
x3 = a + 3d = 6 + 3 × 8 = 30
x4 = a + 4d = 6 + 4 × 8 = 38
∴ 6 ଓ 46 ମଧ୍ୟସ୍ଥ ଚାରିଗୋଟି ସମାନ୍ତର ମଧ୍ୟକ 14, 22, 30 ଓ 38 ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Question 8.
5 ଓ 65 ମଧ୍ଯରେ (i) ତିନିଗୋଟି ଓ (ii) ପାଞ୍ଚଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ ସ୍ଥାପନ କର ।
ସମାଧାନ :
(i) ମନେକର 5 ଓ 65 ମଧ୍ୟରେ ଅବସ୍ଥିତ ତିନିଗୋଟି ସମାନ୍ତର ମଧ୍ୟକ x1, x2 ଓ x3
ଏଠାରେ a = 5, b = 65
d = \(\frac{b-a}{4}=\frac{65-5}{4}=\frac{60}{5}=15\)
x1 = a + d = 5 + 15 = 20
x2 = a + 2d = 5 + 2 × 15 = 35
x3 = a + 3d = 5 + 3 × 15 = 50
∴ 5 ଓ 65 ମଧ୍ଯସ୍ଥ ତିନୋଟି ସମାନ୍ତର ମଧ୍ଯକ 20, 35, 50 ।

(ii) ମନେକର 5 ଓ 65 ମଧ୍ୟସ୍ଥ ପାଞ୍ଚାଗୋଟି ସମାନ୍ତର ମଧ୍ୟକ x1, x2, x3, x4 ଓ x5
ଏଠାରେ a = 5, b = 65
d = \(\frac{b-a}{6}=\frac{65-5}{6}=\frac{60}{6}=10\)
x1 = a + d = 5 + 10 = 15
x2 = a + 2d = 5 + 2 × 10 = 25
x3 = a + 3d = 5 + 3 × 10 = 35
x4 = a + 2d = 5 + 4 × 10 = 45
x5 = a + 3d = 5 + 5 × 10 = 55
∴ 5 ଓ 65 ମଧ୍ଯସ୍ଥ ପାଞ୍ଚଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ 15, 25, 35, 45 ଓ 55 ।

Question 9.
11 ଓ 71 ମଧ୍ୟରେ ପାଞ୍ଚଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ ସ୍ଥାପନ କର ।
ସମାଧାନ :
ମନେକର 11 ଓ 71 ମଧ୍ୟସ୍ଥ ପାଞ୍ଚଗୋଟି ସମାନ୍ତର ମଧ୍ଯକ x1, x2, x3, x4 ଓ x5
ଏଠାରେ a = : 11, b = 71, d = \(\frac{b-a}{6}=\frac{71-11}{6}=\frac{60}{6}=10\)
x1 = a + d = 11 + 10 = 21
x2 = a + 2d = 11 + 2 × 10 = 31
x3 = a + 3d = 11 + 3 × 10 = 41
x4 = a + 2d = 11 + 4 × 10 = 51
x5 = a + 3d = 11 + 5 × 10 = 11 + 50 = 61
∴ 11 ଓ 71 ମଧ୍ୟସ୍ଥ ପାଞ୍ଚଗୋଟି ସମାନ୍ତର ମଧ୍ୟକ 21, 31, 41, 51 ଓ 61 ।

ବିକଳ୍ପ ସମାଧାନ :
ମନେକର 11 ଓ 71 ମଧ୍ୟସ୍ଥ ପାଞ୍ଚଟି ସମାନ୍ତର ମଧ୍ଯକ p, q, r, s ଓ t ।
∴ 11, p, g, r, s, t, 71 A.P.ରେ ଅବସ୍ଥିତ ।
ଏଠାରେ a = 11, t, = 71
t7 = 71 ⇒ a + (7 – 1)= 71 ⇒ 11 + 6d = 71
⇒ 6d = 71 – 11 = 60 ⇒ d = \(\frac{60}{6}=10\)
p = a + d = 11 + 10 = 21
q = a + 2d = 11 + 2 × 10 = 31
r = a + 3d = 11 + 3 × 10 = 41
s = a + 4d = 11 + 4 × 10 = 51
t = a + 5d = 11 + 5 × 10 = 61
∴ 11 ଓ 71 ମଧ୍ୟସ୍ଥ ପାଞ୍ଚଟି ସମାନ୍ତର ମଧ୍ୟକ 21, 31, 41, 51 ଓ 61 ।

BSE Odisha 10th Class Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(b)

Question 10.
20 ଓ 80 ମଧ୍ୟରେ n ସଂଖ୍ୟକ A.M. ଅଛି । ଯଦି ପ୍ରଥମ ମଧ୍ୟକ : ଶେଷ ମଧ୍ଯକ = 1 : 3 ହୁଏ ତେବେ, nର ମାନ ସ୍ଥିର କର ।
ସମାଧାନ :
20 ଓ 80 ମଧ୍ୟରେ n ସଂଖ୍ୟକ A.M. ଅଛି । ଏଠାରେ a = 20
tn+2 = 80 ⇒ a + (n + 2 – 1) d = 80
⇒ 20 + (n+1)d = 80 ⇒ (n + 1) d = 80 – 20 ⇒ (n + 1) d = 60 …….(i)
ପ୍ରଥମ ମଧ୍ୟକ = 20 + d ଓ ଶେଷ ମଧ୍ଯକ = 80 – d
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{20+d}{80-d}=\frac{1}{3}\) ⇒ 60 + 3d = 80 – d ⇒ 80 – 60 = 20 ⇒ d = 5
∴ (i)ରୁ (n + 1) d = 60 ⇒ (n + 1) 5 = 60
⇒ n + 1 = \(\frac{60}{5}\) = 12 ⇒ n = 12 -1 = 11
∴ nର ମାନ 11 ଅଟେ ।

Question 11.
A.P.ରେ ଥିବା ଚାରିଗୋଟି ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର ଯାହାର ଯୋଗଫଳ 2 ଏବଂ ଆଦ୍ୟ ଓ ପ୍ରାନ୍ତ ରାଶିଦ୍ଧୟର ଗୁଣଫଳ ମଧ୍ଯକ ଦ୍ଵୟର ଗୁଣଫଳର 10 ଗୁଣ ସହ ସମାନ ହେବ ।
ସମାଧାନ :
ମନେକର A.P. ରେ ଥ‌ିବା ଚାରୋଟି ସଂଖ୍ୟା ଯଥାକ୍ରମେ a – 3d, a – d, a + d, a + 3d ।
ପ୍ରଶ୍ନନୁସାରେ, a – 3d + a – d + a + d + a + 3d = 2
⇒ 4a = 2 ⇒ a = \(\frac{2}{4}=\frac{1}{2}\)
ପୁନଶ୍ଚ, (a – 3d) (a + 3d) = 10 (a – d) (a + d)
⇒ a² – 9d² = 10(a² – d²) ⇒ 10a² – 10d² = a² – 9d²
⇒ 10a² – a² = 10² – 9d² ⇒ 9a² = d²
⇒ 9 × (\(\frac{1}{2}\))² = d² = \(\frac{9}{4}\) = d = ±\(\sqrt{\frac{9}{4}}\)
d = ±\(\frac{3}{2}\)
a = \(\frac{1}{2}\) ଓ d = \(\frac{3}{2}\) ହେଲେ
a – 3d = \(\frac{1}{2}\) – 3 × \(\frac{3}{2}\) = \(\frac{1}{2}-\frac{9}{2}=\frac{-8}{2}=-4\)
a – d = \(\frac{1}{2}\) – \(\frac{3}{2}\) = \(\frac{-2}{2}\) = -1
a + d = \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2
a + 3d = \(\frac{1}{2}\) + 3 × \(\frac{3}{2}\) = \(\frac{10}{2}\) = 5
a = \(\frac{1}{2}\) ଓ d = \(\frac{-3}{2}\) ହେଲେ A.P. ଚାରୋଟି ପଦ 5, 2, – 1, – 4 ହେବ ।
∴ A.P.ରେ ଥ‌ିବା ଚାରୋଟି ସଂଖ୍ୟା ଯଥାକ୍ରମେ -4, -1, 2, 5 ବା 5, 2, -1, -4 ।