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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.
"
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 $\triangle \mathrm{ABC}$
$=\frac{\pi r^{2}}{2}-[\frac{90^{\circ}}{360^{\circ}} \times \pi(14)^{2}-\frac{1}{2} \times \mathrm{AC} \times \mathrm{AB}]$
$=\pi(\frac{(7 \sqrt{2})^{2}}{2})-[\frac{\pi(14)^{2}}{4}-\frac{1}{2} \times 14 \times 14]$
$=[\frac{(22 \times 7 \times 7 \times 2)}{(7 \times 2)}]-[\frac{(22 \times 14 \times 14)}{(7 \times 4)}-7 \times 14]$
$=154-(154-98)$
$=98 \mathrm{~cm}^{2}$