In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
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AcademicMathematicsNCERTClass 10

Given:

ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. 

To do:

We have to find the area of the shaded region.

Solution:


Radius of the quadrant of the circle $= 14\ cm$
$\triangle \mathrm{ABC}$ is a right-angled triangle

$\mathrm{AB}=\mathrm{AC}=14 \mathrm{~cm}$

$B C^{2}=A B^{2}+A C^{2}$

$=(14)^{2}+(14)^{2}$

$B C=\sqrt{2 \times14}$

$=14 \sqrt{2} \mathrm{~cm}$

Therefore,

Radius of semicircle $BDC= r$

$=\frac{BC}{2}$

$=\frac{14 \sqrt{2}}{2} \mathrm{~cm}$

$=7 \sqrt{2} \mathrm{~cm}$

Area of the shaded region $=$ Area of semicircle $-$ Area of quadrant $\mathrm{ABC} -$

Area $\triangle \mathrm{ABC}$

$=\frac{\pi r^{2}}{2}-[\frac{90^{\circ}}{360^{\circ}} \times \pi(14)^{2}-\frac{1}{2} \times \mathrm{AC} \times \mathrm{AB}]$

$=\pi(\frac{(7 \sqrt{2})^{2}}{2})-[\frac{\pi(14)^{2}}{4}-\frac{1}{2} \times 14 \times 14]$

$=[\frac{(22 \times 7 \times 7 \times 2)}{(7 \times 2)}]-[\frac{(22 \times 14 \times 14)}{(7 \times 4)}-7 \times 14]$

$=154-(154-98)$

$=98 \mathrm{~cm}^{2}$

raja
Updated on 10-Oct-2022 13:24:29

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