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 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
$=\frac{60}{360}\times \pi r^2$

$=\frac{60}{360}\times \pi \times {42}^2$

$=\frac{1}{6}\times \frac{22}{7}\times 42\times 42=924sq.cm$

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 we can conclude that it is an equilateral triangle : base = radius = 42 cm

Now Area of an equilateral triangle
$=\frac{\sqrt{3}}{4}\times s^2=\frac{\sqrt{3}}{4}\times {42}^2$

Hence, Area of $△$OAB
$=441\sqrt{3}sq.cm$

Hence, Area of segment APB = Area of sector OAPB - Area of $△$OAB
$=924-441\sqrt{3}$ sq. cm