Tiêu đề Grade 11 - Answer sheet.pdf ✅

Page 1 9.2 × 8.4 < 72 × 84 × 2.1 < 42  (2.03) 2 < 8 × (10.05) 2 > 100  √102 + 1 < 10 × log2 16 = 4  ms< s;=re – m q.Kl 10. 2 11. 1 4 12. 2x − y 13. 216 14. 1̅. 9741 15. 2̅. 8602 16. 1 293 17. 0.9030 18. q 2 p3 19. 3 20. 9 16 21. 32 22. −8 23. 2 24. x = 30 25. 1 7 9 26. 0.04 27. 2 28. 9 29. x = −6 30. log 100 = 2 31. 34. i ) 29 ii) 1.42 37. i) 2 ii) 8.5 iii) 5.631 40. i) x = 36 ii) 4 iii) 3.917 43. i) 257 16x 2 ii) x = 12√2 iii) 0.00009436 45. i) 9 16 ii) ,>q 120 + 3 iii) 11.1 (m
Page 3 w) i. 24 cm2 ii. 528 cm3 wd) i. πr 2h = 540 ii. h = 540 πr 2 iii. h = 31.73 cm 105. . 106. . 107. 108. . 109. . 110. . 111. . 112. . w) i. 2πrh + 2πr 2 2πr + 2r + 2πr 2 6πr 2 ii. r = 2 wd) a 3 = r 3h 16 22 cm 8 cm a) i. 4a ii. 2 3 πa 3 cm iii. 2πa 3 cm3 b) i. ;yvQfõ mßudj = l =i,dkfha mßudj ksid 2x 2 = 2πa 3 x 2 = πa 3 ii. 7223 iii. 85 cm i. >kldNfha mßudj = 10.86 × 10.86 × 0.85 cm2 f.d a, 3 mßudj = 3 × 4πr 3 3 r = q 3 × 4πa 3 3 = 10. 862 × 0.85 a 3 = 10.862 ×0.85 4π ii. a 3 = 10.862 ×0.85 12.55 a 3 = 2 ,>q 10.86 + ,>q 0.95 − ,>q 12.55 = 2 × 1.0359 + 1̅. 9294 − 1.0986 = 2.0718 + 1̅. 9294 − 1.0986 = 2.0012 − 1.0986 = m%' ,' 0.9026 = 7.991 iii. a 3 = 8 a = 2 3 a = 2 cm i. j3⁄4.M,h = 20 × 14 = 280m2 ii. πr 2 = 22 7 × 7 × 7 = 154 cm2 iii. 20 m 7 m 1 m 4 m 14 m 7 m 8 m X 5 m iv. 4 × 4 = 16 m2 v. 154 m2 + 16 m2 = 170 m2 280 − 170 = 110 m2 ( ;K ìu i`oyd ) vi. x 2 = 5 2 + 8 2 = 25 + 64 x = √89 iii. h = 24 × (1.75) 3 49 log h = log 24 + log 1.75 − log 49 = 1.3802 + 3 × 0.2430 − 1.6902 = 1.3802 + 0.7290 − 1.6902 = 2. 1092 − 1.6902 = antilog 0.4190 = 2.624 cm i. c, mßudj = πr 2h = 22 7 × 7 × 7 × 10 = 1540 cm2 ii. 4 3 πr 3 = 4 3 πa 3 18 × 4 3 πa 3 = π × 7 × 7 × h 24 a 3 = 49 h h = 24a 3 49 cm i. ùÿrefõ c, mßudj = 1 3 πr 2h = 1 3 π × r 2 × 12 = 4 πr 2 ii. a × a × b = 4 πr 2 b × a 2 = 4 πr 2 a 2 = 4 πr 2 b iii. a 2 = 4 πr 2 b = 12.56 ×9.572 18 log a 2 = log 12.56 + log 9. 572 − log 18 log a 2 = 1.0990 + 2 × 0.9809 − 1.2553 log a 2 = 1.0990 + 1.9618 − 1.2553 = 3.0508 − 1.2553 = antilog 1.8055 a 2 = 63.90 = 64 a = 8 cm i. wkq,df.a fldgi ;%sfldaKhl s' lu,df.a fldgi ;%mSishuls' ii. wkq,df.a = 1 2 × 80 × 80 = 3200 m2 iii. lu,df.a fldgfia j'M,h = 1 2 × 80 (100 + 20) = 40 × 120 = 4800 m2 wkq,d ( lu,d 3200 ( 4800 32 ( 48 2 ( 3 80 m 32 m 3200 m2 4800 48 m m2 100 m iv. v. 20m
Page 4 C 28 m 28 m 50 m A D B f.d a,fha mßudj = 4 3 π(2) 3 fl a;=fõ m;=f,a wrh r iy ,ïN Wi h kï" r h = 4 3 fl a;=fõ mßudj = 1 3 π(r) 2 ( 4 3 r) ∴ 4 3 π(2) 3 = 1 3 π(r) 2 ( 4 3 r) 8 × 3 = r 3 ∴ r = 2 × √3 3 log r = log 2 + 1 2 log 3 = 0.3010 + 1 3 (0.4771) = 0.3010 + 0.1590 = 0.4600 r = 2.884 r = 2.88 cm 113. . 114. . 115. . 116. . 117. . 118. i. 8cm ii. 6cm 119. 2 120. . 121. . 122. . 123. . 124. 3a a i. is, sk avrfha mßudj = πr 2h = 22 7 × a 2 × 2a = 44a 2 7 ii. m % siaufha mßudj = a 2b a 2b = 44 a 2 7 b = 44 a 2 7a2 b = 22 × 2a 2 7a2 b = 2πa i. fl a;=fõ mßudj = 1 3 πr 2h = 1 3 × π × a 2 × 3a = πa 3 ii. f.d a,fha mßudj = 4 3 πr 3 = 4 3 π (a/2) 3 = 4 3 π × a 3 / 8 = πa 3 6 f.d a, .Kk = πa 3 ÷ πa 3 6 = πa 3 × 6 πa 3 = 6 iii. f.d a,fha mßudj = 4 3 × πr 3 = 4 3 × 22 7 × ( a 2 ) 3 = 4 3 × 22 7 × a 3 8 = 11a3 21 is, skavrfha mßudj = πa 2 × 6a = 6πa 3 w3⁄4O f.da,fha mßudj = 1 2 × 4 3 π × a 3 = 2 3 πa 3 = 6 πa 3 − 2 3 πa 3 = 18πa 3 − 2πa 3 3 = 16πa 3 3 = 4 × 4 3 πa 3 = 4 × >k f,d ay f.da,hl mßudj i. m;=f,a mßñ;sh = 28 × 4 + ( 1 4 × 2 × 22 7 × 28) = 112 + 44 = 156m ii. m;=f,ys j3⁄4.M,h = iup;=ri% fldgfia j'M + jD;a;dldr fldgfia j'M = 28 × 28 + ( 1 4 × 22 7 × 28 × 28) = 784 + 616 = 1400m2 iii. w ̈;a ;% sfld aKfha j'M = mdofha j'M 1 2 × 56 × AD = 1400 AD = 1400 28 = 50 m iv. m;=, c, frdaOkh lsßu ioyd jehjk uqo, = 2800 × 500 = re' 1 400 000 i. B Ndckfha Odß;dj = úkdä 22 oS c,h msfrk mßudj B Ndckfha Odß;dj = 22 × 6 cm3 = 132 cm3 1 3 πr 2h = 132 1 3 × 22 7 r 2 × 14 = 132 r 2 = 9 r = ±3 ∴ wrh = 3 cm ii. A Ndckh u.sk a c,h iïmQ3⁄4Kfhk a bj;a l sÍug hk ld,h t kï" ∴ t × 6 = 1 2 × 4 3 πr 3 t = 1 6 × 1 2 × 4 3 × 22 7 × r 3 = 22 63 r 3 ;;amr i. π(6) 2 × 7 = 22 7 × 36 × 7 = 792 cm2 ii. 4 3 πr 3 = 44 + 792 4 3 × 22 7 × r 3 = 836 r 3 = 836 ×21 88 r 3 = 199.5 r = √199.5 3 i. 1 2 × 2 × 22 7 × 7 = 22m ii. 1 2 × 22 7 × 7 × 7 = 77m2 iii. AD = 77 ×3 14 = 16.5 m iv. mr;r .Kk = 14 1.4 = 10 ∴ 22 10 = 2.2 m ùÿre f.da,fha mßudj = 4 3 π(21) 3 is, sk avrdldr ùÿre ;eáhl mßudj = πr 2 ( r 9 ) ∴ 4x 3 (21) 3 = 240πr 2 ( r 9 ) r = ( 213 20 ) r = r √20 3 lgr = lg 21 − 1 3 lg 20 = 1.3222 − 1 3 (1.3010) = 1.3222 − 0.4336 r = antilog 0.8886 = 7.737 = 7.74 123. 22 7 × r × r × 7 = 88 r = 2m fyd a 2