Question

在RT三角形ABC中，角ACB=90度，AB=4,分别以AC，BC为直径作半圆，面积分别记为S1,S2.则S1+S2的值等于？

在RT三角形ABC中，角ACB=90度，AB=4,分别以AC，BC为直径作半圆，面积分别记为S1,S2.则S1+S2的值等于？

Answer 1

三种方法 方法1 r=AC/2，R=BC/2
S1+S2
=π(AC/2)²/2+ π(BC/2)²/2
=π(AC²+BC²)/8
=πAB²/8
=2π
方法2 S1+S2=π(AC/2)²/2+ π(BC/2)²/2
=π(AC²+BC²)/8
=πAB²/8
=2π
方法3 设AC=X BC=Y
S1=0.5×π×(X÷2)²
S2=0.5×π×(Y÷2)²
X²+Y²=4²=16
S1+S2=0.5×π×(X÷2)²+0.5×π×(Y÷2)²
=0.5×π×[（X²+Y²)／4]
=2π

Answer 2

设AC=X BC=Y
S1=0.5×π×(X÷2)²
S2=0.5×π×(Y÷2)²
X²+Y²=4²=16
S1+S2=0.5×π×(X÷2)²+0.5×π×(Y÷2)²
=0.5×π×[（X²+Y²)／4]
=2π

Answer 3

设AC=X BC=Y
S1=0.5×π×(X÷2)²
S2=0.5×π×(Y÷2)²
X²+Y²=4²=16
S1+S2=0.5×π×(X÷2)²+0.5×π×(Y÷2)²
=0.5×π×[（X²+Y²)／4]
=2π

Answer 4

π(½AC)^2+π(½BC)^2
=¼π(AC^2+BC^2)
=¼π(AB^2)
=4π

Answer 5

8π