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 segment APB?

• A. $924sqcm$
• B. $441\sqrt{3}sqcm$
• C. $924-441\sqrt{3}sqcm$
• D. $1743sqcm$

Answer 1

The correct option is C

$924-441\sqrt{3}sqcm$

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

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 $h=r\sin \left({60}^{\circ }\right)=r\times \frac{\sqrt{3}}{2}=21\sqrt{3}cm$
Hence, area of $△$OAB = $\frac{1}{2}\times 42\times 21\sqrt{3}=441\sqrt{3}sqcm$
Hence, area of segment APB = Area of sector OAPB -Area of $△$OAB $=924-441\sqrt{3}sqcm.$