### Solved Examples

**Question 1: **Find the area of a circular sector whose chord is the side of the square inscribed in a circle with a 5 cm radius.

** Solution: **

Given: radius, r = 5cms.

Central angle = 90° (angle in a square)

Area of sector, A = $\frac{\pi r^2 \theta^o}{360}$

A = $\frac{\pi .5^2.90}{360}$ = $\frac{25 \pi}{4}$ = 19.63 $cm^2$

Area of the given circular sector whose chord is the side of a square is 19.63 $cm^2$.

**Question 2: **The area of a circular sector of 60° is 6$\pi cm^2$. Find the radius and the length of the circle to which the sector belongs.

** Solution: **

Given, Area of the circular sector = 6$\pi cm^2$ and $\theta$ = 60°.

Using the formula, Area of sector, A = $\frac{\pi r^2 \theta^o}{360}$

6$\pi$ = $\frac{\pi r^2 60^o}{360}$

$\frac{\pi r^2}{6}$ = 6$\pi$

$r^2$ = 36

r = $\sqrt{36}$

r = 6 cms

Thus the radius of the circle is 6 cms.

The length of the circle to which the sector belongs to is the circumference which is given by 2$\pi$r.

Thus, the length of the circle = 2 $\times \pi \times$ 6 = 12$\pi$ = 37.7 cm.

**Question 3: **Calculate the area of a circular sector whose chord is the side of an
inscribed equilateral triangle in a circle with a 3 cm radius.

** Solution: **

Given, radius = 3 and $\theta$ = 120° (The angle in an Equilateral
triangle is 60° and that becomes the inscribed angle. Also, the
intercepted angle is twice that of the inscribed angle).

Area of sector, A = $\frac{\pi r^2 \theta}{360}$

A = $\frac{\pi 3^2 120}{360}$ = 3$\pi$ = 9.43 $cm^2$

The
area of the sector whose chord is the side of an inscribed equilateral
triangle in a circle with a 3 cm radius is 9.43 $cm^2$.

**Question 4: **A circle with radius r = 6 and a sector with subtended angle measuring
45° is given, find the area of the sector and the arc length.

** Solution: **

Given, radius = 6 and $\theta$ = 45°

The arc length is given by, L = $\theta$$\frac{\pi r}{180}$

L = 45 $\frac{\pi (6)}{180}$

L = $\frac{3}{2}$$\pi$

The area of Sector of circle, A = $\frac{\pi r^2 \theta^o}{360}$

= $\frac{\pi 6^2 45}{360}$

= $\frac{\pi 36}{8}$

A = $\frac{9 \pi}{2}$

Thus the area A of the given sector = $\frac{9 \pi}{2}$ and the arc-length L = $\frac{3 \pi}{2}$.

**Question 5: **A circular sector with radius r = 5 and a corresponding arc length of 10p is given, find the area of the sector.

** Solution: **

Given, the radius, r = 5 and arc length = 10$\pi$

Arc length of a circle in radians, L = $\theta$ r.

10$\theta$ = ($\theta$)(5)

$\theta$ = 10 $\frac{\pi}{5}$ = 2$\pi$

Thus, the central angle is 2$\pi$.

Then the area of the circular sector is, A = $\frac{r^2 \theta}{2}$ = $\frac{(5)^2 (2 \pi)}{2}$ = 25$\pi$

The area of the given circular sector is 25$\pi$.