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In the below figure, from a rectangular region $ A B C D $ with $ A B=20 \mathrm{~cm} $, a right triangle $ A E D $ with $ A E=9 \mathrm{~cm} $ and $ D E=12 \mathrm{~cm} $, is cut off. On the other end, taking $ B C $ as diameter, a semicircle is added on outside the region. Find the area of the shaded region. (Use $ \pi=22 / 7) $."

Academic Mathematics NCERT Class 10

Given:

From a rectangular region \( A B C D \) with \( A B=20 \mathrm{~cm} \), a right triangle \( A E D \) with \( A E=9 \mathrm{~cm} \) and \( D E=12 \mathrm{~cm} \), is cut off. On the other end, taking \( B C \) as diameter, a semicircle is added on outside the region.

To do:

We have to find the area of the shaded region.

Solution:

From the figure,

$AB = 20\ cm, AE = 9\ cm, DE = 12\ cm$ and $\angle AED = 90^o$

In right angled triangle $\mathrm{AED}$, by Pythagoras theorem,

$\mathrm{AD}^{2}=\mathrm{AE}^{2}+\mathrm{DE}^{2}$

$\Rightarrow \mathrm{AD}=\sqrt{9^{2}+12^{2}}$

$=\sqrt{81+144}$

$=\sqrt{225}$

$=15 \mathrm{~cm}$

Therefore,

Area of the shaded portion $=$ Area of rectangle $\mathrm{ABCD}+$ Area of semicircle $-$ Area of $\Delta \mathrm{AED}$

$=(20 \times 15)+\frac{1}{2} \times \frac{22}{7} \times(\frac{15}{2})^{2}-\frac{1}{2} \times 9 \times 12 \mathrm{~cm}^{2}$

$=300+\frac{11}{7} \times \frac{225}{4}-54$

$=300+\frac{2475}{28}-54$

$=246+88.39$

$=334.39 \mathrm{~cm}^{2}$

The area of the shaded region is $334.39 \mathrm{~cm}^{2}$.

Updated on: 10-Oct-2022

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