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標題:
F.4 Maths trigonometry
發問:
1.PQRS is a trapezium, which PS//QR,PS=12cm, PR=25cm and RS=17cm. a)find the area of triangle PRS. b)find the area of triangle PQS. Thank you!! 更新: re:nmnmm http://postimg.org/image/rgga1qgnb/ 更新 2: re 土扁 : 為甚麼有相同的altitude就表示它們是congrent?? 我就是想不到b part 怎樣證朋它們是congrent才在這裹問,thx
最佳解答:
1. (a) In ΔPRS : s = (12 + 17 + 25)/2 cm = 27 cm By Heron's formula : Area of ΔPRS = √[27 x (27 - 12) x (27 - 17) x (27 - 25)] cm2 = √(27 x 15 x 10 x 2) cm2 = 90 cm2 (b) Compare ΔPRS and ΔPQS. They have the same base (PS) and the same altitude (distance between PS andQR). Hence, ΔPQS = ΔPRS = 90 cm2 2014-05-03 19:26:30 補充： It is NOT necessary to prove that they are congruent, but it is only necessary to prove that their areas are equal. Area of Δ = (1/2) x base x altitude Area of ΔPRS = (1/2) x PS x (equal altitude) Area of ΔPQS = (1/2) x PS x (equal altitude) Hence, ΔPQS = ΔPRS = 90 cm2
其他解答:
F.4 Maths trigonometry
發問:
1.PQRS is a trapezium, which PS//QR,PS=12cm, PR=25cm and RS=17cm. a)find the area of triangle PRS. b)find the area of triangle PQS. Thank you!! 更新: re:nmnmm http://postimg.org/image/rgga1qgnb/ 更新 2: re 土扁 : 為甚麼有相同的altitude就表示它們是congrent?? 我就是想不到b part 怎樣證朋它們是congrent才在這裹問,thx
最佳解答:
1. (a) In ΔPRS : s = (12 + 17 + 25)/2 cm = 27 cm By Heron's formula : Area of ΔPRS = √[27 x (27 - 12) x (27 - 17) x (27 - 25)] cm2 = √(27 x 15 x 10 x 2) cm2 = 90 cm2 (b) Compare ΔPRS and ΔPQS. They have the same base (PS) and the same altitude (distance between PS andQR). Hence, ΔPQS = ΔPRS = 90 cm2 2014-05-03 19:26:30 補充： It is NOT necessary to prove that they are congruent, but it is only necessary to prove that their areas are equal. Area of Δ = (1/2) x base x altitude Area of ΔPRS = (1/2) x PS x (equal altitude) Area of ΔPQS = (1/2) x PS x (equal altitude) Hence, ΔPQS = ΔPRS = 90 cm2
其他解答:
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can u draw a fig of trapezium-pqrs ans (a) 解答 (b) 解答
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