一個直角三角型,低長14,(最長果條邊斜長25),仲有一條點求?
回答 (8)
2008-01-05 7:32 am
✔ 最佳答案
用畢氐定理的逆定理設個條為a
14^2+a^2=25^2
196+a^2=625
a^2=429
a=開方429
個條邊係開方429
2008-01-04 23:33:26 補充：
大概係20.7123.....
參考: 自己
2010-05-25 11:13 pm
逆定理???
2008-01-06 1:34 am
25x25 = 14x14 + yxy
yxy = 625 - 196 = 429
y = 20.7
yxy = 625 - 196 = 429
y = 20.7
2008-01-05 7:33 am
25的二次方－１４的二次方等於４２９，再開方，就係答案啦
2008-01-05 7:29 am
另外果條係=root(25^2-14^2)
=root429
=root429
2008-01-05 7:29 am
斜二次方=(低乘高)二次方
2008-01-05 7:29 am
用畢史定理
設要求個條邊為x
25^2=14^2+x^2
625=196+x^2
x^2=625-196
x=429^1/2
大概等於20.7123
設要求個條邊為x
25^2=14^2+x^2
625=196+x^2
x^2=625-196
x=429^1/2
大概等於20.7123
2008-01-05 7:29 am
let the remaining side = y
25x25 = 14x14 + yxy
yxy = 625 - 196 = 429
y = 20.71231518.....
25x25 = 14x14 + yxy
yxy = 625 - 196 = 429
y = 20.71231518.....
收錄日期: 2021-04-14 18:03:55
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