"
">

Find the area of a shaded region in the below figure, where a circular arc of radius $ 7 \mathrm{~cm} $ has been drawn with vertex $ A $ of an equilateral triangle $ A B C $ of side $ 14 \mathrm{~cm} $ as centre. (Use $ \pi=22 / 7 $ and $ \sqrt{3}=1.73) $"

Academic Mathematics NCERT Class 10

Given:

A circular arc of radius \( 7 \mathrm{~cm} \) has been drawn with vertex \( A \) of an equilateral triangle \( A B C \) of side \( 14 \mathrm{~cm} \) as centre.

To do:

We have to find the area of the shaded region.

Solution:

Radius of the circular arc $r = 7\ cm$

Length of the side of equilateral triangle $ABC (a)= 14\ cm$

Area of the shaded region $=$ Area of the circle $+$ Area of equilateral triangle $-$ 2( Area of the unshaded sector)

$=\pi r^{2}+\frac{\sqrt{3}}{4} a^{2}-2 \times \pi r^{2} \frac{\theta}{360^{\circ}}$

$=\frac{22}{7} \times 7^2+\frac{1.73}{4} \times 14^2-2 \times \frac{22}{7} \times 7^2 \times \frac{60^o}{360^o}$ [Angle in an equilateral triangle is $60^o$]

$=154+84.77-14 \times 22 \times \frac{1}{6}$

$=238.77-51.33$

$=187.44 \mathrm{~m}^{2}$

The area of the shaded region is $187.44\ m^2$.

Updated on: 10-Oct-2022

190 Views

Kickstart Your Career

Get certified by completing the course

Advertisements