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