Question

The window is shown in the shape of a semi-circle with radius $4$ ft. The distance from $S$ to $T$ is $2$ ft and the measure of arc $AB={45}^{\circ }$. Area of the glass in the region $ABCD$ is

• A. 4.7 sq. ft
• B. 4.9 sq. ft
• C. 4.5 sq. ft
• D. 5 sq. ft

Answer 1

The correct option is A

4.7 sq. ft

Let $r_1$ be the radius of the larger semi-circle and $r_2$ be the radius of the smaller semi-circle.

So, $r_1=4$ ft and $r_2=4-2=2$ ft

Required area $=$ Area of the sector $OAB$ $-$ Area of the sector $ODC$
$=\pi r_1^2\times \frac{\theta }{{360}^{\circ }}-\pi r_2^2\times \frac{\theta }{{360}^{\circ }}$
$=\pi r_1^2\times \frac{{45}^{\circ }}{{360}^{\circ }}-\pi r_2^2\times \frac{{45}^{\circ }}{{360}^{\circ }}$
Taking common terms and solving we get,
$=\frac{\pi }{8}(4^2-2^2)$
$=\frac{\pi }{8}\times 12=4.7$ sq. ft