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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 $ )."

Academic Mathematics NCERT Class 10

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 $2.58 \mathrm{~cm}^{2}$.

Updated on: 10-Oct-2022

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