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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}}\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}}\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}}\overline{\mathbf{CD}} ମଧ୍ୟରେ ଦୂରତା 10 ସେ.ମି. | \overline{\mathbf{AB}} କ୍ୟା କେନ୍ଦ୍ରଠାରୁ 6 ସେ.ମି. ଦୂରରେ ଅବସ୍ଥିତ ହେଲେ \overline{\mathbf{AB}}\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}}\overline{\mathbf{CD}} ମଧ୍ୟରେ ଦୂରତା = 10 ସେ.ମି. |
\overline{\mathbf{AB}}\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}}\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}}\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}}\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}\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 ।

CHSE Odisha Class 11 English Grammar Prepositions

Odisha State Board CHSE Odisha Class 11 Invitation to English 4 Solutions Grammar Prepositions Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar Prepositions

Look at the following examples :
1. He went to Puri.
2. He went Puri.
In the sentences written above, Sentence 1 is grammatical and acceptable but Sentence 2 is not grammatical and acceptable. Sentence 2 i.e., He went Puri, is a non-standard and unusual sentence. It is not written or spoken by educated native speakers. The small word to is left out in Sentence 2. The absence of the small word ‘to’ gives no meaning to the construction. So the sentence is not acceptable and recommendable.

In Sentence – 1, the small word ‘to’ is more attracted/associated towards the Noun Phrase‘ Puri’ than the verb went. So we call the small word a preposition.  Prepositions are words or word groups that are usually used before a noun phrase. Their normal position in a sentence is in front of a noun phrase with its determiner if any. They (prepositions) sometimes are used in other positions than the one in front of a ‘noun phrase’. We can see different types of prepositions like time, place, movement, etc.

CHSE Odisha Class 11 English Grammar Prepositions

Activity – 1

Use the prepositions on, in, at with the time expressions given below.

(a) February                               (b) midnight               (c) eight o’clock

(d) the afternoon                       (e) night                      (f) Christmas day

(g) the eighteenth century         (h) lunchtime             (i) 1999

(j) Wednesday                            (k) the morning         (l) March 15

(m) Saturday night                      (n) Easter

In the box below, write ‘on’ ‘in’ or ‘at’, whichever is appropriate.

a b c d e f g h i j k l m n

Answer:

a b c d e f g h i j k l m n
in at at in at on in at in on in on at on

Activity – 2

Fill in the blank spaces with on, in, or at :
1. ___________ the daytime the streets are crowded but ___________night they are quite deserted.
2. He’s always in a bad temper ___________ breakfast time.
3. We couldn’t offer him a room in our flat, because ___________ that time our guest was staying with us.
4. Children get presents ___________ Christmas and ___________ their birthdays.
5. We want three seats for ‘Boothnaath’ ___________ Friday night.
6. I will reach Bhubaneswar ___________ 3 o’clock ___________ the morning but the offices start working only ___________ 10 a.m.
7. Millions of Indian soldiers were killed ___________ the Second World War.
8. It was ___________ the spring of 1985 that my brother and sister-in-law met for the first time. However, they got married only ___________ 1995.
9. ___________ 1950 the cost of living was only a fraction of what it was ___________ 1990.
10. They get up ___________ dawn and reached the summit noon.
11. We didn’t want anything to go wrong ___________ our sister’s wedding day.
12. I’ll meet you ___________ 2.30 ___________ Friday, August 2001.
13. I’ll see you ___________ six days’ time.
14. Birds don’t find much to eat ___________ winter.
15. What will you be doing ___________ the holidays?

CHSE Odisha Class 11 English Grammar Prepositions

Answer:
1. In the daytime the streets are crowded but at night they are quite deserted.
2. He’s always in a bad temper in breakfast time.
3. We couldn’t offer him a room in our flat, because at that time our guest was staying with us.
4. Children get presents at Christmas and on their birthdays.
5. We want three seats for ‘Bhootnaath’ on Friday night.
6. I will reach Bhubaneswar at 3 o’clock in the morning, but the offices start at working only at 10 a.m.
7. Millions of Indian soldiers were killed in the Second World War.
8. It was in the spring of 1985 that my brother and sister-in-law met for the first time. However, they got married only in 1995.
9. In 1950 the cost of living was only a fraction of what it was in 1990.
10. They got up at dawn and reached the summit at noon.
11. We didn’t want anything to go wrong on our sister’s wedding day.
12. I’ll meet you at 2.30 on Friday, August 2001.
13. I”ll see you in six day’s time.
14. Birds don’t find much to eat in winter.
15. What will you be doing on the holidays?

Activity – 3

Insert suitable prepositions in the blank spaces from the following list, (during, since, for, from ___________ to ___________.)
1. They’ve lived in this town _____________ five years.
2. The farmers have been working in the field _____________10 o’clock.
3. The employees stop working _____________ the night.
4. The players practised in the field _____________ 4 p.m. _____________6 p.m.
5. He looked as if he hadn’t slept _____________weeks.
6. The carpenters worked _____________9 a.m. _____________5 p.m. with an hour for lunch.
Can you explain, in your own words, what the rules are for the use of these prepositions?

Answer:
1. They’ve lived in this town for five years.
2. The farmers have been working in the field since 10 o’clock.
3. The employees stop working during the night.
4. The players practiced in the field from 4 p.m. to 6 p.m.
5. He looked as if he hadn’t slept for weeks.
6. The carpenters worked from 9 a.m. to 5 p.m. with an hour for lunch.

The following are the rules for the use of the prepositions such as ‘from, to, for, since, during’.
For is used to talk about a period of time continuing up to the present. It is used to say how long something has lasted. It can be used to talk about the past, present, or future.
‘Since’ is used to say when the action/event started (the starting point of the actions and situations). Both ‘since’ and ‘for’ are usually used in the present perfect tense.
‘The use of ‘from’ in a sentence says when the action/situation starts and it finishes.
The idea is normally expressed with from….to or from…. till/until…………
‘During’ is used to say when something happened but ‘for’ is used to say how long it took

CHSE Odisha Class 11 English Grammar Prepositions

Activity – 4

Fill in the blanks using appropriate prepositions.
1. All the students are busy _____________ the moment.
2. Sunita and Mahima finished their homework _____________ the same time.
3. They are getting married _____________six months time.
4. Hemanta is in class 8. He’ll be leaving school _____________ two years time.
5. It was a very interesting book. I read it _____________ a day.
6. Hurry up! We’ve got to go _____________ten minutes.
7. My sister is a doctor but she is out of work _____________ present.
8. He’s busy just now but he’ll be with you a _____________ moment.
9. A: I” ll meet you at 10.30.
B: OK, but please be _____________ time.
10. A child ran across in front of the car, but the driver managed to stop just _____________ time.
11. At first I didn’t like getting up early, but _____________ the end I got used to it.
12. Workers normally get paid _____________the end of the month.

Answer:
1. All the students are busy at the moment.
2. Sunita and Mahima finished their homework at the same time.
3. They are getting married in six months time.
4. Hemanta is in class 8. He’ll be leaving school in two years time.
5. It was a very interesting book. I read it in a day.
6. Hurry up! We’ve got to go in ten minutes.
7. My sister is a doctor but she is out of work at present.
8. He’s busy just now but he’ll be with you in a moment.
9. A: I” ll meet you at 10.30.
B: OK, but please be in time.
10. A child ran across in front of the car, but the driver managed to stop just on time.
11. At first I didn’t like getting up early, but in the end, I got used to it.
12. Workers normally get paid at the end of the month.

Activity – 5

Match the expressions in A with their meanings in B.
Activity-5
Answer:
on time — punctual, not late
in time — soon enough
at the end — at the time when something ends
in the end — finally

CHSE Odisha Class 11 English Grammar Prepositions

SECTION – 2

(Place)
Activity – 6

Complete the sentences using the prepositions at, in, or on.
1. There’s some sugar ___________ the shelf ___________the cupboard.
2. Is Seema ___________the kitchen ?
3. Sambalpur is ___________ the west of Orissa, ___________the River Mahanadi.
4. There’s a grocery shop ___________the comer ___________the end of the street.
5. There are three books ___________ the table.
6. He is ___________ the roof.
7. The old man is standing ___________ the gate.
8. He put the money ___________ his pocket.
9. They’ll meet the players ___________ the dining table.
10. I waited ___________ the bus stop for two hours.

Answer:
1. There’s some sugar on the shelf in the cupboard.
2. Is Seema in the kitchen?
3. Sambalpur is in the west of Orissa, on the River Mahanadi.
4. There’s a grocery shop in the comer at the end of the street.
5. There are three books on the table.
6. He is on the roof.
7. The old man is standing at the gate.
8. He put the money in his pocket.
9. They’ll meet the players at the dining table.
10. I waited at the bus stop for two hours.

Now complete the following activity by matching the prepositions with their respective meanings.
Now complete the following activity by matching the prepositions with their respective meanings.
Answer:
Now complete the following activity by matching the prepositions with their respective meanings.

Activity – 7

Complete the sentences using at, in or on.
1. Our teacher’s flat is ___________the second floor.
2. The boy was holding a ball ___________ his hand.
3. The children spent an afternoon ___________ the zoo.
4. He traveled from Calcutta to Delhi. He is ___________ Delhi now.
5. They have gone to a wedding. They are probably ___________ the wedding.
6. He was taken to hospital. He is ___________ hospital now.
7. The baby was playing ___________ the floor.
8. There’s ink ___________ your shirt.
9. Hari met his friend ___________ the crossroad.
10. Gauhati is ___________ Assam.

Answer:
1. Our teacher’s flat is on the second floor.
2. The boy was holding a ball in his hand.
3. The children spent an afternoon in the zoo.
4. He traveled from Calcutta to Delhi. He is in Delhi now.
5. They have gone to a wedding. They are probably in the wedding.
6. He was taken to the hospital. He is in hospital now.
7. The baby was playing on the floor.
8. There’s ink in your shirt.
9. Hari met his friend at the crossroad.
10. Gauhati is in Assam.

CHSE Odisha Class 11 English Grammar Prepositions

Activity – 8

Complete the sentences using above, across, along, below, down, over, past, through, under, and up.
1. Jitu ran ___________ the road to meet his friend.
2. They took a shorter route ___________ the forest to save time.
3. He walked ___________ the road slowly, examining the shop windows.
4. They saw the car going ___________the house and shouted to the driver to turn back.
5. The woman climbed ___________ the stairs to the first floor.
6. We saw the new bridge which has been built ___________the river.
7. Madhu lives on the third floor. He came ___________ the stairs to greet his friends.
8. The temperature is ___________ zero in the poles.
9. The plane went up quickly. Soon it was ___________ the clouds.
10. We live on the earth, ___________ the sky.

Answer:
1. Jitu ran across the road to meet his friend.
2. They took a shortcut through the forest to save time.
3. He walked alone the road slowly, examining the shop windows.
4. They saw the car going past the house and shouted to the driver to turn back.
5. The woman climbed up the stairs to the first floor.
6. We saw the new bridge which has been built over the river.
7. Madhu lives on the third floor. He came down the stairs to greet his friends.
8. The temperature is below zero in the poles.
9. The plane went up quickly. Soon it was above the clouds.
10. We live on the earth, under the sky.

Activity – 9

Complete the sentences using the prepositions at, in, or on. (More than one answer is possible)
1. There’s chemist’s ___________ the comer ___________ the end of the street.
2. They had breakfast ___________ Khan’s Cafe ___________ the main road their way home.
3. Calcutta is ___________ the eastern part of India ___________ the River Ganga.
4. They were waiting ___________ the station.
5. He’d spend the whole day sitting ___________ a desk.
6. If you walk further, you’ll see a small shop ___________ the comer.
7. Is Mother ___________ the kitchen? No, she’s ___________ the back of the house.

Answer:
1. There’s chemists in the comer at the end of the street.
2. They had breakfast at Khan’s Cafe on the main road on their way home.
3. Calcutta is in the eastern part of India on the River Ganga.
4. They were waiting at the station.
5. He’d spend the whole day sitting at a desk.
6. If you walk further, you’ll see a small shop at the comer.
7. Is Mother in the kitchen? No, she’s at the back of the house.

CHSE Odisha Class 11 English Grammar Prepositions

SECTION – 3

Study the use of in, on, and at in the following sentences.
1. (a) They are not in town. They are on holiday in Darjiling.
(b) They heard the news on the radio.
(c) He has put on weight. He’ll have to go on a diet.
Some useful expressions with on are the following.

on business, on a tour, on a cruise, on television, on the phone, on strike, on fire, on the whole, on purpose.
2. (a) The old woman did not like the sun. She preferred to sit in the shade, (b) My friend always writes in pencil.
Some more expressions within are given below.
in the rain, in the sun, in the dark, in bad weather, in ink, in words, in figures, in block letters, in cash

3. He left school at the age of 14.
Some more expressions with at are given below.
at a speed of, at a temperature of

Activity – 10

Complete the sentence using on, in, or at.
1. When you write a cheque, you should write the amount ____ words as well as figures.
2. Water boils ____ 100 degrees Celsius.
3. He avoided meeting him ____ purpose.
4. My brother is going ____ a tour tomorrow.
5. The old man likes to keep warm, so he does not go out ____ cold weather.
6. Look! The train is ____ fire!
7. The workers are ____ strike.
8. He left home ____ the age of 10.
9. We pay for things ____ cash.
10. I watch the morning news ____ television.

Answer:
1. When you write a cheque, you should write the amount in words as well as in figures.
2. Water boils at 100 degree Celsius.
3. He avoided meeting him on purpose.
4. My brother is going on a tour tomorrow.
5. The old man likes to keep warm, so he does not go out in cold weather.
6. Look! The train is on fire!
7. The workers are on strike.
8. He left home at the age of 10.
9. We pay for things in cash.
10. I watch the morning news on television.

CHSE Odisha Class 11 English Grammar Prepositions

Activity – 11

Fill in the blanks using the prepositions in, on, or at.
We live ___________ a house ___________ Janpath. We live ___________ 85, Bapuji Nagar. Our house is ___________ a convenient location, just ___________ the main road. It is ___________ the intersection of Cuttack Road and Lewis Road.
Answer:
We live in a house on Janpath. We live on 85 Bapuji Nagar. Our house is in a convenient location, just on the main road. It is in the intersection of Cuttack Road and Lewis Road.

SECTION – 4

Study the following sentences.
(a) He called the stranger by mistake.
(b) My father always makes payments by cheque.
(c) My sister goes to college by bus.
(d) My friend did not use his car. He came in a taxi.

In (a) and (b) by is used with mistake and cheque. The following are some of the expressions that usually go with by.
by chance, design, accident
by letter, post, hand, cable, telegram
by heart
by day, night
In (c), by is used with bus. By is generally used to refer to some means of transport. Here are some more examples.
by land, by air, by road, by sea, by coach, by train, etc.
But by is replaced by in, on, etc. when a noun such as a ship, car, bus, etc. is used along with a determiner such as my, a, the. For example: in my car, on my bicycle, etc.

Activity – 12

Complete the sentences using by, on, or in.
1. A bat sleeps __________ day and flies __________ night.
2. The tourists have decided to travel to Australia __________ sea rather than __________ air.
3. Usha usually goes to college __________ her scooter, but sometimes she goes __________ bus.
4. It was only __________ chance that he passed the examination. He hadn’t worked at all.
5. The officer did not come to work __________ his car on Monday. His car had broken down and he had to come __________ taxi.
6. We decided not to go __________ bus. We went __________ my bike instead.
7. The journey takes 20 minutes __________bus and about 50 minutes __________foot.
8. They went for a ride __________ a motorbike.
9. I traveled to Delhi __________ train.
10. How long does it take to get to Sri Lanka __________ boat?

Answer:
1. A bat sleeps by day and flies by night.
2. The tourists have decided to travel to Australia by sea rather than by air.
3. Usha usually goes to college in her scooter, but sometimes she goes by bus.
4. It was only by chance that he passed the examination. He hadn’t worked at all.
5. The officer did not come to work in his car on Monday. His car had broken down and he had to come by taxi.
6. We decided not to go by bus. We went in my bike instead.
7. The journey takes 20 minutes by bus and about 50 minutes on foot.
8. They went for a ride on/in a motorbike.
9. I traveled to Delhi by train.
10. How long does it take to get to Sri Lanka by boat?

CHSE Odisha Class 11 English Grammar Prepositions

(Adjective + Preposition) Combinations
Certain adjectives are always used in combination with particular prepositions. Here are some common examples of these adjective + preposition combinations.

of
1. Children are afraid of snakes.
2. We are proud of being Indians.
3. Cats are fond of milk.
Some more examples :

frightened of           jealous of             conscious of         tolerant of

scared of                 envious of             capable of            independent of

full of                       suspicious of        short of                typical of

ashamed of              aware of               critical of              aware of

tolerant of

about/with
1. We are excited about our sister’s wedding tomorrow.
2. My younger sister was delighted with the present I gave her.
Some more examples:
worried/upset/nervous/happy/annoyed/furious etc. + about.
pleased/satisfied/disappointed/bored/angry etc. with
at/by/with
1. My friend is very good at mathematics.
2. Everybody was shocked at/by what they heard.

Some more examples.
bad at        brilliant at
clever at     surprised at/by
hopeless    at astonished at/by
excellent    at amazed at/by

of/to
1. It is so nice / kind of you to visit us.
2. My mother is very nice / kind to the poor.
Some more examples.
silly/stupid/generous/good + of somebody (to do something)
cruel/friendly/polite/generous/good + to somebody.

Activity – 13

Fill in the blanks with suitable prepositions.
1. The boy’s quite capable __________ solving the problem.
2. Are you worried __________ your examination?
3. The children were very disappointed __________ the magic show.
4. Minu is still upset __________ what you said to her on last Sunday.
5. One should not be cruel __________ animals.
6. We were amused __________ the way he spoke.
7. My elder brother is not aware __________ his responsibility.
8. The beggar is not ashamed __________ what other people think of them.

Answer:
1. The boy’s quite capable of solving the problem.
2. Are you worried about your examination?
3. The children were very disappointed with the magic show.
4. Minu is still upset about what you said to her on last Sunday.
5. One should not be cruel to animals.
6. We were amused at the way he spoke.
7. My elder brother is not aware of his responsibility.
8. The beggar is not ashamed of what other people think of them.

CHSE Odisha Class 11 English Grammar Prepositions

Activity – 14

Write sentences using the adjectives in brackets followed by suitable prepositions. The first two have been done for you.
1. Can I help you with your luggage?
(kind) It’s very kind of you.
2. I went out in the rain without an umbrella.
(silly) It was silly of me to go out in the rain without an umbrella.
3. My friend didn’t thank me for the present.
(not polite) ____________________________
4. The couple had an argument and now they refuse to speak to one another.
(childish) ____________________________
5. Uma offered to carry the luggage to the station.
(nice) ____________________________
6. Mohan shouted at his friends in his birthday party.
(not nice) ____________________________
7. I needed money, and Shyam gave me some.
(generous) ____________________________

Answer:
3. My friend did not thank me for the present.
(not polite) It’s not polite of him.
4. The couple had an argument and now they refuse to speak to one another.
(childish) It’s childish of them.
5. Uma offered to carry the luggage to the station.
(nice) It’s nice of her.
6. Mohan shouted at his friends in his birthday party.
(not nice) It’s not nice of Mohan.
7. I needed money and Shyam gave me some.
(generous) It’s generous of Shyam.

Activity – 15

Put each of the adjectives followed by an appropriate preposition in the right blank.
[rude, brilliant, annoyed, typical, bored, furious]
1. Why do you always get so ___________ little things.
2. He isn’t happy at school. He says he’s ___________ the courses he’s doing.
3. The teacher was ___________ us for making so much noise in the class.
4. Why are you always so ___________ your friends? Can’t you be ___________ them?
5. We’re not surprised he changed his mind at the last moment. That’s ___________ him.
6. He is ___________ telling jokes.

Answer:
1. Why do you always get so worried about little things.
2. He isn’t happy at school. He says he’s annoyed with the courses he’s doing.
3. The teacher was furious at us for making so much noise in the class.
4. Why are you always so rude to your friends? Can’t you be nice to them?
5. We’re not surprised he changed his mind at the last moment. That’s typical of him.
6. He is bored with telling jokes.

CHSE Odisha Class 11 English Grammar Prepositions

SECTION – 6

(Verb + Preposition) Combinations
Certain verbs are followed by particular prepositions. Here are some common examples
of these verb + preposition combinations.
at
1. (a) Don’t point that knife at the child. It’s dangerous.
(b) He spoke in such a manner that everyone laughed at him.
Some more verbs used with ‘at’

aim at              jump at             stare at            glance at

wonder at       hint at               peck at              jeer at

for
2. (a) The students are waiting for their teacher to come.
(b) One should always hope for the best.
Some more verbs used with ‘for’

apologize for         cry for                send for

beg for                   feel for               strive for

care for                   forgive for         sue for

charge for               long for             wish for

compensate for      mourn for          search for

from
3. (a) The speaker digressed from the topic and spoke a lot of nonsense.
(b) Students will benefit from the new book.
Some more verbs used with ‘from’

hide from              derive from           escape from          save from

abstain from         desist from            exclude from         separate from

borrow from         differ from             prevent from         subtract from

emerge from        digress from           recover from         refrain from

in
4. (a) Most people believe in astrology.
(b) My friend did well in English but failed in Mathematics.
Some more verbs with ‘in’

abound in            excel in            involved in           succeed in

admit of               boast of           dispose of             suspect of

approve of           smell of           dream of               think of

 

CHSE Odisha Class 11 English Grammar Prepositions

of
5. (a) The students are tired of doing the same task every day.
(b) Many smokers die of lung cancer.
Some more verbs used with ‘of’

accuse of                beware of              complain of              deprive of

admit of                 boast of                  dispose of                 suspect of

approve of             smell of                  dream of                   think of

on
6. (a) Success in life depends on hard work.
(b) My brother spends a lot on books.
Some more verbs used with ‘on’

base on               impose on            reflect on              comment on

insist on              rely on                  congratulate on    intrude on

resolve on          decide on              operate on            feast on

to
7. (a) None listened to the Chief Guests lecture.
(b) Students contributed a lot to the Indian Freedom Struggle.
Some more verbs with ‘to’

add to          attend to             conform to          submit to

adhere to    belong to             consent to           surrender to

amount to   commit to            introduce to         yield to

aspire to      confine to            object to              speak to

with
8. (a) Our neighbor always quarrels with children.
(b) Gandhi never compromised with falsehood.
Some more verbs with ‘with’

coincide with      fill with               part with               unite with

comply with        grapple with      supply with           deal with

cope with            interfere with    sympathize with    overwhelm with

disagree with      meddle with      threaten with         reconcile with

Activity – 16

Complete the sentences using appropriate prepositions.
1. The old man accused the boy ___________ trying to steal his bag.
2. My friend invited me ___________ his sister’s marriage.
3. The local people have warned the tourists ___________ swimming in that part of the river.
4. Will you please remind us ___________ the party next Monday?
5. He always borrows money ___________ his friends and never cares to return it.
6. Our teacher congratulated us ___________ our good performance in the examination.
7. Pradip blames other people ___________ his own mistakes.
8. Our Independence Day coincides ___________ the birthday of Sri Aurobindo.
9. The principal approved ___________ the stand taken by the students on the dowry system.
10. We are involved ___________ a project on the uplift of the poor.

CHSE Odisha Class 11 English Grammar Prepositions

Answer:
1. The old man accused the boy of trying to steal his bag.
2. My friend invited me to his sister’s marriage.
3. The local people have warned the tourists of swimming in that part of the river.
4. Will you please remind us of the party next Monday?
5. He always borrows money from his friends and never cares to return it.
6. Our teacher congratulated us on our good performance in the examination.
7. Pradip blames other people for his own mistakes.
8. Our Independence Day coincides with the birthday of Sri Aurobindo.
9. The principal approved of the stand taken by the students on the dowry system.
10. We are involved in a project on the uplift of the poor.

Activity – 17

Complete the sentences, using one of the following verbs in the correct form with appropriate prepositions.
write, speak, glance, listen, talk, explain, ask, shout
1. Please ___________ him. ! He may have something interesting to tell you.
2. He didn’t have his watch. He ___________ my watch to see what the time was.
3. I had an argument with Sikha and now we’re not ___________ each another.
4. Please don’t ___________ the children. Be nice to them.
5. That old woman’s a bit lonely. She needs somebody to ___________.
6. Can you ___________ me how this machine works?
7. Don’t forget to ___________ me while you’re away.
8. His son is always ___________ him ___________ money.

Answer:
1. Please listen to him. ! He may have something interesting to tell you.
2. He didn’t have his watch. He glanced at my watch to see what the time was.
3. I had an argument with Sikha and now we’re not talking to each another.
4. Please don’t shout at the children. Be nice to them.
5. That old woman’s a bit lonely. She needs somebody to speak to.
6. Can you explain to me how this machine works?
7. Don’t forget to write to me while you’re away.
8. His son is always asking him for money.

Activity – 18

Insert the correct prepositions.
1. The parents searched everywhere ___________ their baby but couldn’t find it.
2. ‘Are you going to Calcutta tomorrow ?’
‘I hope so. It depends ___________ the weather.’
3. I don’t mind buying this book. But who is going to pay ___________ it?
4. These two brothers are suspected ___________ stealing a car.
5. You can rely ___________ your grandfather, who has never refused you anything.
6. That handbag belongs ___________ the old man standing in the comer.
7. Warm clothes protect us ___________ cold.
8. Mother asked me to fill the bucket ___________ water.

Answer:
1. The parents searched everywhere for their baby but couldn’t find it.
2. ‘Are you going to Calcutta tomorrow ?’
‘I hope so. It depends on the weather.’
3. I don’t mind buying this book. But who is going to pay for it?
4. These two brothers are suspected of stealing a car.
5. You can rely on your grandfather, who has never refused you anything.
6. That handbag belongs to the old man standing in the comer.
7. Warm clothes protect us from cold.
8. Mother asked me to fill the bucket with water.

CHSE Odisha Class 11 English Grammar Prepositions

Activity – 19

Look at this paragraph from a letter and put in these verbs with suitable prepositions.
applied, agree, care, ask, caring, decided, concentrate suffering, pay

I’m working at a private factory now. I ___________ a caterer’s job last August and started in November. I don’t earn much money, and I even had to___________ my uniform out of my own money. Perhaps I should ___________ a pay rise. But I don’t really ___________ the money. The work is the important thing. Of course, it’s very hard work ___________ the patients, and at the moment I’m ___________ backache. But I knew it would be like this even when I ___________ a career in a hospital. I just try to forget all the problems and ___________ the job. I think it’s a worthwhile thing to do: I hope you ___________ me.

Answer:
I’m working at a private factory now. I decided about a caterer’s job last August and started in November. I don’t earn much money, and I even had to pay for my uniform out of my own money. Perhaps I should ask for a pay rise. But I don’t really care for the money. The work is the important thing. Of course, it’s very hard work caring for the patients, and at the moment I’m suffering from backache. But I knew it would be like this even when I applied for a career in a hospital. I just try to forget all the problems and concentrate on the job. I think it’s a worthwhile thing to do; I hope you agree with me.

CHSE Odisha Class 11 English Grammar The Imperative

Odisha State Board CHSE Odisha Class 11 Invitation to English 4 Solutions Grammar The Imperative Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar The Imperative

Look at the sentences below.
1. (a) Hurry up!
(b) Stop!
2. (a) Be careful!
(b) Mind your language!
3. Borrow the book from your teacher, read the chapter and make notes on the main points.
All the sentences above are imperative. Sentences 1 and 2 are used for orders, advice, and warnings but Sentence 3 is used to give direction. There is no special form of the verb imperative in English.

Look at the following sentences.
4. Come in, make yourselves at home.
5. Please start, don’t wait for me.

These above Sentences 4 and 5 are used for ‘invitations’.
6. Push.
7. Keep off the grass.

These sentences are used for signboards and notices.
To suggest doing something together we use Let’s / Let us.
8. Let’s hire a taxi.
9. Let us go out.

CHSE Odisha Class 11 English Grammar The Imperative

Activity – 1

Match the sentences in Column A with the imperatives in Column B.
Match the sentences in Column A with the imperatives in Column B.
Answer:
Activity 1

Activity – 2

Here is a recipe for making an omelette. Put the following verbs in the correct spaces.
( break, beat, heat, add, turn, make, put, pour)
1. First, two eggs and _____________ them in a bowl.
2. _____________ them up and _____________ salt and pepper to taste.
3. _____________ a little butter or oil in a frying pan.
4. Then _____________ the mixture in, and _____________ sure that the omelette doesn’t burn.
5. _____________ it over when the omelette is fried.
6. Serve it hot.
Answer:
1. First, two eggs and break them in a bowl.
2. Heat them up and add salt and pepper to taste.
3. Pour a little butter or oil in a frying pan.
4. Then pour the mixture in and make sure that the omelette doesn’t burn.
5. Beat it over when the omelette is fried.
6. Serve it hot.

CHSE Odisha Class 11 English Grammar The Imperative

Activity – 3

Match each of the imperatives in column A with their functions in column B.
Activity - 3
Answer:
1. Enjoy yourself! —Making a friendly remark.
2. Come in and make yourself at home. — Inviting.
3. Get a Number 7 bus, that’s direct. — Making a suggestion.
4. Turn right at the traffic lights. — Giving direction.
5. Open your book at page 20. — Giving an instruction.
6. Mind the floor, it’s slippery. — Warning.

Activity – 4

Write suggestions that will match the statements below. Use clues to form sentences beginning with Let.
1. Tomorrow is my sister’s birthday. (buy / her / present)
___________________________
2. 1 can’t wait for the train. (take / taxi)
___________________________
3. I want to see a film. (go / cinema)
___________________________
4. It’s hard to believe. (forget / it)
___________________________
5. It’s very cold in here. (light/fire)
___________________________
Answer:
1. Tomorrow is my sister’s birthday.
Let’s buy her a present.
2. I can’t wait for the train.
Let’s take a taxi.
3. I want to see a film.
Let’s go to see the cinema.
4. It’s hard to believe.
Let’s forget it.
5. It’s very cold in here.
Let’s light a fire.

CHSE Odisha Class 11 English Grammar The Imperative

Activity – 5

Fill in each blank choosing the right verb from the following list.
(turn, ask, go, cross, walk, continue, take.)
Stranger: Excuse me, could you tell me how to get to Lewis Road, please?
Young boy: Yes, certainly. _________ along this road till you come to the traffic lights. Then, _________ over and _________ going till you come to the roundabout. _________ right there, then _________ as far as the next roundabout. _________ the first turning to the left and then the first turning to the right. Lewis Road is the second one on the left. _________ someone if you get lost, but it’s really quite easy to find.
Stranger: Thank you very much.
Answer:
Stranger: Excuse me, could you tell me how to get to Lewis Road, please?
Young boy: Yes, certainly. Go along this road till you come to the traffic lights. Then, take over and continue going till you come to the roundabout. Turn right there, then walk as far as the next roundabout. Cross the first turning to the left and then the first turning to the right. Lewis Road is the second one on the left. Ask someone if you get lost, but it’s really quite easy to find.
Stranger: Thank you very much.