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. $5544sqcm$
• 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 = $\frac{60}{360}\times \pi \times {42}^2=924{cm}^2$

Now area of $△$OAB = $\frac{1}{2}\times base\times height$
Now the triangle is isosceles and vertex angle is ${60}^{\circ }$. Hence, the remaining two angles are equal and the sum of three angles is ${180}^{\circ }$.
Hence, base = radius = 42 cm

Now $h=rsin({60}^{\circ }$) = $r\times \frac{\sqrt{3}}{2}=21\sqrt{3}cm$
Hence, area of $△$OAB = $\frac{1}{2}\times 42\times 21\sqrt{3}=441\sqrt{3}{cm}^2$
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
= $\frac{22}{7}\times 42\times 42-(924-441\sqrt{3})sq.cm.$
= $4620+441\sqrt{3}sq.cm.$