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# In the figure below, $A B C$ is an equilateral triangle of side $8 \mathrm{~cm} . A, B$ and $C$ are the centres of circular arcs of radius $4 \mathrm{~cm}$. Find the area of the shaded region correct upto 2 decimal places. (Take $\pi=3.142$ and $\sqrt{3}=1.732$ )."

Given:

$A B C$ is an equilateral triangle of side $8 \mathrm{~cm} . A, B$ and $C$ are the centres of circular arcs of radius $4 \mathrm{~cm}$.

To do:

We have to find the area of the shaded region correct upto 2 decimal places.

Solution:

Length of each side of $\triangle ABC = 8\ cm$

Area of the triangle $=\frac{\sqrt{3}}{4} a^{2}$

$=\frac{\sqrt{3}}{4}(8)^{2}$

$=\frac{1.732 \times 64}{4}$

$=1.732 \times 16$

$=27.712 \mathrm{~cm}^{2}$

Angle of each sector $=60^{\circ}$ Area of three sectors $=3 \times \pi r^{2} \times \frac{60^{\circ}}{360^{\circ}}$

$=3 \times 3.142 \times 4 \times 4 \times \frac{1}{6}$

$=1.571 \times 16$

$=25.136 \mathrm{~cm}^{2}$

Therefore,

Area of the shaded region $=$ Area of $\Delta \mathrm{ABC}-$ Areas of three sectors

$=27.712-25.136$

$=2.576 \mathrm{~cm}^{2}$

The area of the shaded region correct upto 2 decimal places is $2.58 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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