Question

The area of an equilateral triangle is $49\sqrt{3}c{m}^2$. Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in the figure. Find the area of the portion in the triangle not included in the circles.

Answer 1

GIVEN:

Area of $△ABC=49\sqrt{3}$

$\theta =60$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times sid{e}^2$

$49\sqrt{3}=\frac{\sqrt{3}}{4}\times a^2$

$a=7\times 2=14cm$

Radius of the circle half the length of the side of the $△ABC$ (GIVEN)

Radius of the circle $=\frac{1}{2}\times 14=7cm$

Area of sector $=\frac{\theta }{360}\times \pi r^2$

$=\frac{60}{360}\times \frac{22}{7}\times 7^2=\frac{77}{3}c{m}^2$

Required area = Area of $△ABC$ - 3 ( area of a sector of angle 60° in a circle of radius 7 cm)

Required area $=49\sqrt{3}-3\times \frac{77}{3}$

$=7.77c{m}^2$

Hence, the area of the triangle not included in the circle is $7.77c{m}^2$