有一個直角三角形,底係a,高係b 斜係c 面積係24 求c (a同b都係
整數)
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2010-10-02 7:19 am
回答 (2)
2010-10-02 8:05 am
✔ 最佳答案
................................................................................設 a =< b :a^2 + b^2 = c^2......(1)ab/2 = 24.....(2)由 (2) :ab = 48ab = (1*48) , (2*24) , (3*16) , (4*12) 或 (6*8) , 代入(1) :1^2 + 48^2 = 2305 = c^2 , c 非整數 ,2^2 + 24^2 = 580 = c^2 , c 非整數 ,3^2 + 16^2 = 265 = c^2 , c 非整數 ,4^2 + 12^2 = 160 = c^2 , c 非整數 ,6^2 + 8^2 = 100 = c^2 , c = 10
2010-10-03 3:46 am
ab/2=24
ab=48
The numbers of a and b possible are (1,48),(2,24),(3,16),(4,12),(6,8)
1^2+48^2=sqrt2305
2^2+24^2=2sqrt145
3^2+16^2=sqrt265
4^2+12^2=sqrt160
6^2+8^2=10(possible)
ab=48
The numbers of a and b possible are (1,48),(2,24),(3,16),(4,12),(6,8)
1^2+48^2=sqrt2305
2^2+24^2=2sqrt145
3^2+16^2=sqrt265
4^2+12^2=sqrt160
6^2+8^2=10(possible)
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