Question

In the figure given below, the radius of the circle is 84 cm. The angle in the sector is ${60}^{\circ }$​. What is the area of the minor segment APB?

• A. $3696-1764\sqrt{3}{cm}^2$
• B. $3696+1764\sqrt{3}{cm}^2$
• C. $18480+1764\sqrt{3}{cm}^2$
• D. $3696{cm}^2$

Answer 1

The correct option is A $3696-1764\sqrt{3}{cm}^2$

Area of segment APB = Area of sector OAPB - Area of $△$OAB
Area of sector OAPB = $\frac{60}{360}\times \pi \times {84}^2=3696{cm}^2$

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 = 84 cm

Now $h=rsin({60}^{\circ }$) = $r\times \frac{\sqrt{3}}{2}=84\times \frac{\sqrt{3}}{2}=42\sqrt{3}cm$
Hence, area of $△$OAB = $\frac{1}{2}\times 84\times 42\sqrt{3}=1764\sqrt{3}{cm}^2$
Hence, area of segment APB = Area of sector OAPB - Area of $△$OAB $=3696-1764\sqrt{3}{cm}^2$