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Find the area of the shaded region in the below figure, if $ A C=24 \mathrm{~cm}, B C=10 \mathrm{~cm} $ and $ O $ is the centre of the circle. (Use $ \pi=3.14) $"
Given:
\( A C=24 \mathrm{~cm}, B C=10 \mathrm{~cm} \) and \( O \) is the centre of the circle.
To do:
We have to find the area of the shaded region.
Solution:
In right angled triangle $ABC$, by Pythagoras theorem,
$AB^2=AC^2 + BC^2$
$= (24)^2+ (10)^2$
$= 576 + 100$
$= 676$
$= (26)^2$
$\Rightarrow AB = 26\ cm$
Diameter of the circle $= 26\ cm$
Radius of the circle $=\frac{26}{2}$
$=13\ cm$
Area of the shaded region $=$ Area of the semicircle $-$ Area of right angled triangle $ABC$
$=\frac{1}{2} \pi r^{2}-\frac{1}{2} AC \times BC$
$=\frac{1}{2}(3.14) \times 13^2-\frac{1}{2} \times 24 \times 10$
$=265.33-120$
$=145.33\ cm^2$
The area of the shaded region is $145.33\ cm^2$.
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