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F.6 Geometric sequence
發問:
http://i729.photobucket.com/albums/ww291/kaka1307/0540.jpg ans: ai) root 5 a cm aii) 5/3a cm aiii)5(root5)a/9 cm b) (5^n/9^n-1)a^2 cm^2 ci) 1.9 cm^2 cii)43 cm^2 ciii)45 cm^2 Please show clear steps.thank you!!
最佳解答:
31. (a) (i) A1B = 3a ′[2/(1 + 2)] cm = 2a cm B1B = 3a ′[1/(1 + 2)] cm = a cm By Pythagorean theorem: A1B12 = A1B2 + B1B2 A1B12 = [(2a)2 + a2] cm2 A1B12 = 5a2 cm2 A1B1 = (√5)a cm (a) (ii) A2B1 = (√5)a ′ [2/(1 + 2)] cm = 2(√5)a/3 cm B2B1 = (√5)a ′ [1/(1 + 2)] cm = (√5)a/3 cm By Pythagorean theorem: A2B22 = A2B12 + B2B12 A2B22 = [(2(√5)a/3)2 + ((√5)a/3)2] cm2 A2B22 = 25a2/9 cm2 A2B2 = 5a/3 cm (a) (iii) A3B2 = (5a/3) ′ [2/(1 + 2)] cm = 10a/9 cm B3B2 = (5a/3) ′ [1/(1 + 2)] cm = 5a/9 cm By Pythagorean theorem: A3B32 = A3B22 + B3B22 A3B32 = [10a/9)2 + (5a/9)2]cm2 A3B32 = 125a2/81 cm2 A3B3 = 5(√5)a/9 cm ===== (b) Area of square A1B1-C1D1 = [(√5)a]2 cm2 = 5a2 cm2 Area of square A2B2C2D2 = (5a/3)2 cm2 = 52a2/9 cm2 Area of square A3B3C3D3 = [5(√5)a/9]2 cm2 = 53a2/92 cm2 It can be deduced that : Area of square AnBnCnDn = 5^n a^2 / 9^(n - 1) cm2 ===== (c) (i) CD = 3a cm = 6 cm Hence, a = 2 Area of square A5B5C5D5 = (5)^5 (2)^2 / 9^4 cm2 = 1.9 cm2 (to 1 sig. fig.) (c) (ii) Areas (in cm2) of squares A1B1C1D1,A2B2C2D2, A3B3C3D3...... form a geometric sequence. T(1) = a = 5(2)2 = 20 r = [5^n a^2 / 9^(n - 1)] / 5^(n - 1) a^2 / 9^(n - 2) = 5/9 The total area of the squares A1B1C1D1,A2B2C2D2, ......, A5B5C5D5 = a(1 - r^5) / (1 - r) = 20[1 - (5/9)^5] / [1 - (5/9)] cm2 = 43 cm2 (to2 sig. fig.) (c) (iii) The total area of the squares A1B1C1D1,A2B2C2D2, A3B3C3D3 = a / (1 - r) = 20 / [1 - (5/9)] cm2 = 45 cm2 2011-10-15 22:02:38 補充： The first line of (c) (iii) should be : The total area of the squares A1B1C1D1,A2B2C2D2, A3B3C3D3 ......
F.6 Geometric sequence
發問:
http://i729.photobucket.com/albums/ww291/kaka1307/0540.jpg ans: ai) root 5 a cm aii) 5/3a cm aiii)5(root5)a/9 cm b) (5^n/9^n-1)a^2 cm^2 ci) 1.9 cm^2 cii)43 cm^2 ciii)45 cm^2 Please show clear steps.thank you!!
最佳解答:
31. (a) (i) A1B = 3a ′[2/(1 + 2)] cm = 2a cm B1B = 3a ′[1/(1 + 2)] cm = a cm By Pythagorean theorem: A1B12 = A1B2 + B1B2 A1B12 = [(2a)2 + a2] cm2 A1B12 = 5a2 cm2 A1B1 = (√5)a cm (a) (ii) A2B1 = (√5)a ′ [2/(1 + 2)] cm = 2(√5)a/3 cm B2B1 = (√5)a ′ [1/(1 + 2)] cm = (√5)a/3 cm By Pythagorean theorem: A2B22 = A2B12 + B2B12 A2B22 = [(2(√5)a/3)2 + ((√5)a/3)2] cm2 A2B22 = 25a2/9 cm2 A2B2 = 5a/3 cm (a) (iii) A3B2 = (5a/3) ′ [2/(1 + 2)] cm = 10a/9 cm B3B2 = (5a/3) ′ [1/(1 + 2)] cm = 5a/9 cm By Pythagorean theorem: A3B32 = A3B22 + B3B22 A3B32 = [10a/9)2 + (5a/9)2]cm2 A3B32 = 125a2/81 cm2 A3B3 = 5(√5)a/9 cm ===== (b) Area of square A1B1-C1D1 = [(√5)a]2 cm2 = 5a2 cm2 Area of square A2B2C2D2 = (5a/3)2 cm2 = 52a2/9 cm2 Area of square A3B3C3D3 = [5(√5)a/9]2 cm2 = 53a2/92 cm2 It can be deduced that : Area of square AnBnCnDn = 5^n a^2 / 9^(n - 1) cm2 ===== (c) (i) CD = 3a cm = 6 cm Hence, a = 2 Area of square A5B5C5D5 = (5)^5 (2)^2 / 9^4 cm2 = 1.9 cm2 (to 1 sig. fig.) (c) (ii) Areas (in cm2) of squares A1B1C1D1,A2B2C2D2, A3B3C3D3...... form a geometric sequence. T(1) = a = 5(2)2 = 20 r = [5^n a^2 / 9^(n - 1)] / 5^(n - 1) a^2 / 9^(n - 2) = 5/9 The total area of the squares A1B1C1D1,A2B2C2D2, ......, A5B5C5D5 = a(1 - r^5) / (1 - r) = 20[1 - (5/9)^5] / [1 - (5/9)] cm2 = 43 cm2 (to2 sig. fig.) (c) (iii) The total area of the squares A1B1C1D1,A2B2C2D2, A3B3C3D3 = a / (1 - r) = 20 / [1 - (5/9)] cm2 = 45 cm2 2011-10-15 22:02:38 補充： The first line of (c) (iii) should be : The total area of the squares A1B1C1D1,A2B2C2D2, A3B3C3D3 ......
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