Question

The area of the segment in a quarter of a circle of radius 8 inches is
$inche{s}^2$

• A. $16\pi +32$
• B. $16\pi -8$
• C. $16\pi -32$

Answer 1

The correct option is C $16\pi -32$

Area of a segment = area of the sector - area of the triangle

Area of the sector $=\frac{\theta }{{360}^{\circ }}\times areaofthecircle$

Here, $\theta ={90}^{\circ }$ since it is a quadrant in a circle and $r=8inches$.

Area of the sector $=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times 8^2$

Area of the sector $=16\pi inche{s}^2$

The triangle is a right-angled triangle since it is formed in a quarter circle.

Area of the triangle $=\frac{1}{2}\times b\times h$
$b=h=8inches$

Area of the triangle $=\frac{1}{2}\times 8\times 8$

Area of the triangle $=32inche{s}^2$

Area of the segment $=16\pi -32inche{s}^2$