Question

A sector with acute central angle q is cut from a circle of diameter 14 cm. The area (in $c{m}^2$ ) of the circle circumscribing the sector is

• A. $\frac{22}{7}se{c}^2\frac{\theta }{2}$
• B. $\frac{77}{2}se{c}^2\theta$
• C. $\frac{7}{2}co{s}^2\frac{\theta }{2}$
• D. $\frac{77}{2}se{c}^2\frac{\theta }{2}$

Answer 1

The correct option is D $\frac{77}{2}se{c}^2\frac{\theta }{2}$

Suppose the vertices of the sector are labelled $A,B$ and $C$ with angle BAC $=\theta$.

Given diameter $=14cm$ $⟹$ Radius $\left(R\right)=7cm$

Then $|AB|=|AC|=R=7$.

From $A$, draw the bisector $m$ of $\theta$ in the sector.

This line $m$ goes through the centre of the circumscribing circle and intersects that circle at point $S$(opposite to vertex $A$).

The length $AS=2r$, where r is the radius that we need to determine.

Draw from $S$ the line segment $SB$.

Since $AS$ goes through the centre of the circumscribing circle and $B$ is a point on that same circle,

$△ASB$ is a right triangle and $\angle SBA={90}^o$.

Furthermore, angle $BAS=\theta /2$

So $|AS|=\frac{|AB|}{\cos \left(\theta /2\right)}$

Since $|AS|=2r$ and $|AB|=R=7$, we have

$2r=\frac{R}{cos\left(\theta /2\right)}=\frac{7}{cos\left(\theta /2\right)}$

$⟹r=\frac{1}{2}\times \frac{7}{cos\left(\theta /2\right)}=\frac{7}{2}\sec \left(\theta /2\right)$

$\therefore$ Area of the circle circumscribing the sector $ABC=\pi r^2=\frac{22}{7}\times \frac{7^2}{4}se{c}^2\left(\frac{\theta }{2}\right)=\frac{77}{2}se{c}^2\frac{\theta }{2}$.

Hence, option D is correct.