Mensuration Practice Test 5

Area of the shaded region of the semicircles
= $\frac{\mathit{\pi }\left({\mathit{r}}^2\right)}{2}+\frac{\mathit{\pi }{\left({\mathit{r}}_1\right)}^2}{2}$
= $\frac{\mathit{\pi }{\left(6\right)}^2}{2}+\frac{\mathit{\pi }{\left(\right)}^2}{2}$
= $\frac{36\mathit{\pi }}{2}+\frac{16\mathit{\pi }}{2}$
= 18π + 8π
= 26π
Area of the unshaded region = $\frac{\mathit{\pi }\left({\mathit{r}}^2\right)}{2}-\frac{\mathit{\pi }{\left({\mathit{r}}_1\right)}^2}{2}$
= $\mathit{\pi }\left(\frac{{\left(6\right)}^2}{2}-\frac{{\left(4\right)}^2}{2}\right)$
= $\mathit{\pi }\left(\frac{36}{2}-\frac{16}{2}\right)$
=π (18 - 8)
= 10π
Ratio of the shaded to unshaded region is:
10π : 26π
10 : 26
5 : 13