From a thin metallic piece, in the shape of a trapezium $ A B C D $, in which $ A B \| C D $ and $ \angle B C D=90^{\circ} $, a quarter circle BEFC is removed. Given $ A B=B C=3.5 $ $ \mathrm{cm} $ and $ D E=2 \mathrm{~cm} $, calculate the area of the remaining piece of the metal sheet."

Given:

\( A B \| C D \) and \( \angle B C D=90^{\circ} \)

\( A B=B C=3.5 \) \( \mathrm{cm} \) and \( D E=2 \mathrm{~cm} \)

To do: 

We have to find the area of the remaining piece of the metal sheet.

Solution:

$ABCD$ is a trapezium.

$DC = DE + EC$

$= DE + BC$

$= 2 + 3.5$

$= 5.5\ cm$

Area of the trapezium $ABCD = \frac{1}{2}(AB + CD) \times BC$

$=\frac{1}{2}(3.5+5.5) \times 3.5$

$=\frac{1}{2} \times 9 \times 3.5$

$=4.5 \times 3.5$

$=15.75 \mathrm{~cm}^{2}$

Radius of the quadrant $CEFB (r)=3.5 \mathrm{~cm}$

Area of the quadrant $=\frac{1}{4} \times \pi r^{2}$

$=\frac{1}{4} \times \frac{22}{7}(3.5)^{2}$

$=\frac{1}{4} \times \frac{22}{7} \times 12.25$

$=\frac{134.75}{14}$

$=9.6\ cm^2$

Therefore,

Area of the remaining piece of the metal sheet $=15.75-9.625$

$=6.125 \mathrm{~cm}^{2}$

The area of the remaining piece of the metal sheet is $6.125 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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