Question

In the figure, if the $\angle AOB={60}^{\circ }$ and radius is $12$cm, then find the area of the segment AXB.$(\pi =3.14,\sqrt{3}=1.73)$

• A. $14$
• B. $13.08$
• C. $12$
• D. $3.14$

Answer 1

The correct option is B $13.08$

Area of the segment AXB $=$ Area of the sector OAXB $-$ Area of triangle OAB.
The triangle OAB is an equilateral triangle since the length of sides are equal to the radius of the circle and an angle is ${60}^0$
Area of an equilateral triangle with length of side $a=\frac{\sqrt{3}}{4}{a}^2$
Area of a sector of a circle of radius 'r' and angle $\theta =\frac{\theta }{360}\pi r^2$
Hence, Area of segment AXB $=\frac{60}{360}\times 3.14\times {12}^2-\frac{1.73}{4}\times {12}^2=75.36-62.28=13.08sqcm$