题目内容
如图,Rt△ABC的周长为(5+3
)cm,以AB、AC为边向外作正方形ABPQ和正方形ACMN.若这两个正方形的面积之和为25 cm2,则△ABC的面积是______ cm2.
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5 |
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如图,a2=c2+b2=25,则a=5.
又∵Rt△ABC的周长为(5+3
)cm,
∴a+b+c=5+3
,
∴b+c=3
(cm).
∴△ABC的面积=
bc=
[(c+b)2-(c2+b2)]÷2=
[(3
)2-25]÷2=5(cm2).
故答案是:5.
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又∵Rt△ABC的周长为(5+3
5 |
∴a+b+c=5+3
5 |
∴b+c=3
5 |
∴△ABC的面积=
1 |
2 |
1 |
2 |
1 |
2 |
5 |
故答案是:5.
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