Question

In the figure given below, the radius of the circle is 42 cm. The angle in the sector is ${60}^{\circ }$​. What is the area of the major segment?

• A. 5544 sq. cm
• B. $441\sqrt{3}$ sq. cm
• C. $4620+441\sqrt{3}$ sq. cm
• D. 1743 sq. cm

Answer 1

The correct option is C

$4620+441\sqrt{3}$ sq. cm

Area of segment APB = Area of sector OAPB + Area of $△$OAB

Area of sector OAPB = Area of circle - Area of minor sector
$=\pi r^2-\frac{60}{360}\pi r^2=\frac{5}{6}\pi r^2$

$△OAB$ is isosceles as OA = OB = radius. Hence, $\angle OAB=\angle OBA=\theta$
In $△OAB,60+\theta +\theta ={180}^{\circ }$ [Angle Sum Property]
$\Rightarrow \theta ={60}^{\circ }$

Thus, we can say that $△$ OAB is an equilateral triangle.
Area of the $△$ OAB = $\frac{\sqrt{3}}{4}{r}^2$

Hence, area of segment APB = Area of sector OAPB + Area of $△$OAB
$=\frac{5}{6}\pi r^2+\frac{\sqrt{3}}{4}{r}^2$
$=\frac{5}{6}\times \frac{22}{7}\times 42\times 42+\frac{\sqrt{3}}{4}\times 42\times 42$
$=4620+441\sqrt{3}$ sq. cm