Question

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

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

Answer 1

The correct option is C

$4620+441\sqrt{3}sqcm$

Area of segment APB = Area of sector OAPB - Area of $△$OAB
Area of sector OAPB = $\frac{60}{360}\times \pi \times {42}^2$
=$\frac{1}{6}\times \frac{22}{7}\times 42\times 42=924{cm}^2$

Now, since the triangle is isosceles with vertex angle ${60}^{\circ }$, the remaining two angles are also ${60}^{\circ }$ as the sum of three angles should be ${180}^{\circ }$.
Hence, the traingle is equilateral, and
$AB=\text{radius}=42cm$

$\Rightarrow$ We know that area of an equilateral triangle $=\frac{\sqrt{3}}{4}\times r^2$, where $r$ is the radius
Hence, area of $△$OAB = $\frac{\sqrt{3}}{4}\times 42\times 42=441\sqrt{3}{cm}^2$
Hence, area of segment APB = Area of sector OAPB - Area of $△$OAB = 924 - 441 $\sqrt{3}{cm}^2$
Area of major segment = Area of circle - Area of segment APB
= $\frac{22}{7}\times 42\times 42-(924-441\sqrt{3})$ sq.cm.
= $4620+441\sqrt{3}$ sq.cm.