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中三數學................... 淨做B) http://photo.xuite.net/machunkit/1427765/1.jpg 更新: 答案係18
最佳解答:
(b) MR =√(25^2-20^2) =√225 =15 cm PR=2MR=30 cm QR =√(PR^2-PQ^2) =√(30^2-24^2) =√324 =18 cm
其他解答:
(b) MR =√(25^2-20^2) =√225 =15 cm PR=2MR=30 cm QR =√(PR^2-PQ^2) =√(30^2-24^2) =√324 =18 cm|||||you will get the answer 15 cm in part a) i.e. MR = 15cm b) PM = MR (長方形特性/rectanglar properties) 所以 PM = 15cm, PR = 30cm 在三角形 PQR 裡, PR^2 = PQ^2 + QR^2 (畢氏定理) 30^2 = 24^2 + QR^2 900 = 576 + QR^2 QR^2 = 324 QR = 18 所以QR 是18 cm|||||you will get the answer 15 cm in part a) i.e. MR = 15cm b) PM = MR (長方形特性/rectanglar properties) 所以 PM = 15cm, PR = 30cm 在三角形 PQR 裡, PR^2 = PQ^2 + QR^2 (畢氏定理) 30^2 = 24^2 + QR^2 900 = 576 + QR^2 QR^2 = 324 QR = 18 所以QR 是18 cm
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中三數學................發問:
中三數學................... 淨做B) http://photo.xuite.net/machunkit/1427765/1.jpg 更新: 答案係18
最佳解答:
(b) MR =√(25^2-20^2) =√225 =15 cm PR=2MR=30 cm QR =√(PR^2-PQ^2) =√(30^2-24^2) =√324 =18 cm
其他解答:
(b) MR =√(25^2-20^2) =√225 =15 cm PR=2MR=30 cm QR =√(PR^2-PQ^2) =√(30^2-24^2) =√324 =18 cm|||||you will get the answer 15 cm in part a) i.e. MR = 15cm b) PM = MR (長方形特性/rectanglar properties) 所以 PM = 15cm, PR = 30cm 在三角形 PQR 裡, PR^2 = PQ^2 + QR^2 (畢氏定理) 30^2 = 24^2 + QR^2 900 = 576 + QR^2 QR^2 = 324 QR = 18 所以QR 是18 cm|||||you will get the answer 15 cm in part a) i.e. MR = 15cm b) PM = MR (長方形特性/rectanglar properties) 所以 PM = 15cm, PR = 30cm 在三角形 PQR 裡, PR^2 = PQ^2 + QR^2 (畢氏定理) 30^2 = 24^2 + QR^2 900 = 576 + QR^2 QR^2 = 324 QR = 18 所以QR 是18 cm
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