close
標題:
發問:
1(a) In Figure 2, BCDE is a rectangle and angle EBA=90 degree. AB=16cm, AE=20cm and BD=15cm.(a)Find BE and ED.(b)Find angle DAC . Correct the answer to 2 decimal places.(c)Find angle ADB. Correct the answer to 2 decimal places.picture:http://postimg.org/image/z33sacnu7/2(a) Show that... 顯示更多 1(a) In Figure 2, BCDE is a rectangle and angle EBA=90 degree. AB=16cm, AE=20cm and BD=15cm. (a)Find BE and ED. (b)Find angle DAC . Correct the answer to 2 decimal places. (c)Find angle ADB. Correct the answer to 2 decimal places. picture: http://postimg.org/image/z33sacnu7/ 2(a) Show that (2t)^2+(t^2-1)^2=(t^2+1)^2 (b)the lengths of a right- angled triangle are x,y and 37, where x and y both integers and x
最佳解答:
1. (a) In ΔABE : AE2 = AB2 + BE2 (Pythagorean theorem) (20 cm)2 = (16 cm)2 + BE2 BE2 = 144 cm2 BE = 12 cm In ΔBED : BD2 = BE2 + ED2 (Pythagorean theorem) (15 cm)2 = (12 cm)2 + ED2 ED2 = 81 cm2 ED = 9 cm (b) In ΔADC : tan∠DAC = DC/AC tan∠DAC = 12/(16 + 9) ∠DAC = 25.64° ∠DAC + ∠ADC + ∠ACD = 180° (∠ sum Δ) 25.64° + ∠ADC + 90° = 180° ∠ADC = 64.36° In ΔBCD : tan∠BDC = BC/CD tan∠BDC = 9/12 ∠BDC = 36.87° ∠ADB = ∠ADC - ∠BDC = 64.36° - 36.87° = 27.49° ===== 2. (a) L.H.S. = (2t)2 + (t2 - 1)2 = 4t2 + [(t2)2 - 2t2 + 1] = (t2)2 + 2(t2) + 1 = (t2 + 1)2 = R.H.S. Hence, (2t)2 + (t2 - 1)2 = (t2 + 1)2 (b) Hypotenuse = 37 By Pythagorean theorem : x2 + y2 = 372 Put x = 2t, y = t2 - 1 : Then t2 + 1 = 37 t2 = 36 For t > 0, then t = 6 Area of the Δ = (1/2) x y = (1/2) [2t] [t2 - 1] = (1/2) [2(6)] [(6)2 - 1] = 210 (square unit) ===== 3. (a) Area of the shaded region = Area of the sector - Area of the triangle = π * (6 cm)2 * (60/360) - (1/2) * (6cm) * (6 cm) * sin60° = 6π - 18 * (√3)/2 cm2 = 6π - 9√3 cm2 (b)(1) The capacity of C = (1/2) * π * (12/2 cm)2 *(20 cm) = 360π cm2 ≈ 1131 cm2 (to 4 sig. fig.) (b)(2) The cross-section of water is identical to the shaded region in figure 3(a). Volume of water in C = [(6π - 9√3) cm2] * (20 cm) = (120π - 180√3) cm3 ≈ 65.22 cm3 (to 4 sig. fig.) (b)(3) [(1/3) * π * (2 cm)2 * (5 cm) * x] + (120π -180√3) ≥ 360π (20/3)πx ≥ 360π - (120π - 180√3) (20/3)πx ≥ 240π + 180√3 x ≥ (240π + 180√3) * (3/20π) x ≥ 50.89 The minimum value of x = 51
其他解答:
I only do 2(a): LHS = (2t)2 + (t2 - 1)2 = 4t2 + (t2)2 - 2t2 + 1 = (t2)2 + 2t2 + 1 = (t2 + 1)2 = RHS
此文章來自奇摩知識+如有不便請留言告知
Math Problem發問:
1(a) In Figure 2, BCDE is a rectangle and angle EBA=90 degree. AB=16cm, AE=20cm and BD=15cm.(a)Find BE and ED.(b)Find angle DAC . Correct the answer to 2 decimal places.(c)Find angle ADB. Correct the answer to 2 decimal places.picture:http://postimg.org/image/z33sacnu7/2(a) Show that... 顯示更多 1(a) In Figure 2, BCDE is a rectangle and angle EBA=90 degree. AB=16cm, AE=20cm and BD=15cm. (a)Find BE and ED. (b)Find angle DAC . Correct the answer to 2 decimal places. (c)Find angle ADB. Correct the answer to 2 decimal places. picture: http://postimg.org/image/z33sacnu7/ 2(a) Show that (2t)^2+(t^2-1)^2=(t^2+1)^2 (b)the lengths of a right- angled triangle are x,y and 37, where x and y both integers and x
最佳解答:
1. (a) In ΔABE : AE2 = AB2 + BE2 (Pythagorean theorem) (20 cm)2 = (16 cm)2 + BE2 BE2 = 144 cm2 BE = 12 cm In ΔBED : BD2 = BE2 + ED2 (Pythagorean theorem) (15 cm)2 = (12 cm)2 + ED2 ED2 = 81 cm2 ED = 9 cm (b) In ΔADC : tan∠DAC = DC/AC tan∠DAC = 12/(16 + 9) ∠DAC = 25.64° ∠DAC + ∠ADC + ∠ACD = 180° (∠ sum Δ) 25.64° + ∠ADC + 90° = 180° ∠ADC = 64.36° In ΔBCD : tan∠BDC = BC/CD tan∠BDC = 9/12 ∠BDC = 36.87° ∠ADB = ∠ADC - ∠BDC = 64.36° - 36.87° = 27.49° ===== 2. (a) L.H.S. = (2t)2 + (t2 - 1)2 = 4t2 + [(t2)2 - 2t2 + 1] = (t2)2 + 2(t2) + 1 = (t2 + 1)2 = R.H.S. Hence, (2t)2 + (t2 - 1)2 = (t2 + 1)2 (b) Hypotenuse = 37 By Pythagorean theorem : x2 + y2 = 372 Put x = 2t, y = t2 - 1 : Then t2 + 1 = 37 t2 = 36 For t > 0, then t = 6 Area of the Δ = (1/2) x y = (1/2) [2t] [t2 - 1] = (1/2) [2(6)] [(6)2 - 1] = 210 (square unit) ===== 3. (a) Area of the shaded region = Area of the sector - Area of the triangle = π * (6 cm)2 * (60/360) - (1/2) * (6cm) * (6 cm) * sin60° = 6π - 18 * (√3)/2 cm2 = 6π - 9√3 cm2 (b)(1) The capacity of C = (1/2) * π * (12/2 cm)2 *(20 cm) = 360π cm2 ≈ 1131 cm2 (to 4 sig. fig.) (b)(2) The cross-section of water is identical to the shaded region in figure 3(a). Volume of water in C = [(6π - 9√3) cm2] * (20 cm) = (120π - 180√3) cm3 ≈ 65.22 cm3 (to 4 sig. fig.) (b)(3) [(1/3) * π * (2 cm)2 * (5 cm) * x] + (120π -180√3) ≥ 360π (20/3)πx ≥ 360π - (120π - 180√3) (20/3)πx ≥ 240π + 180√3 x ≥ (240π + 180√3) * (3/20π) x ≥ 50.89 The minimum value of x = 51
其他解答:
I only do 2(a): LHS = (2t)2 + (t2 - 1)2 = 4t2 + (t2)2 - 2t2 + 1 = (t2)2 + 2t2 + 1 = (t2 + 1)2 = RHS
文章標籤
全站熱搜
留言列表
