题目内容
已知:如图,以Rt△ABC的三边为斜边分别向外作等腰直角三角形,若斜边AB=5,则图中阴影部分的面积为______.
在Rt△ABC中,AB2=AC2+BC2,AB=5,
S阴影=S△AHC+S△BFC+S△AEB=
×(
)2+
×(
)2+
×(
)2,
=
(AC2+BC2+AB2),
=
AB2,
=
×52
=
.
故答案为
.
S阴影=S△AHC+S△BFC+S△AEB=
| 1 |
| 2 |
| AC | ||
|
| 1 |
| 2 |
| BC | ||
|
| 1 |
| 2 |
| AB | ||
|
=
| 1 |
| 4 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=
| 25 |
| 2 |
故答案为
| 25 |
| 2 |
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