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In figure, O is the centre of a circle such that diameter $AB=13\ cm$ and $AC=12\ cm$. $BC$ is joined. Find the area of the shaded region. $( Take\ \pi \ =\ 3.14)$
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Given: O is the center of a circle such that diameter $AB=13\ cm$ and $AC =12\ cm$. $BC$ is joined.
To do: To find the area of the shaded region.
Solution:
Diameter, $AB = 13\ cm$
$\therefore$ Radius of the circle, $r =\frac{13}{2}$
$=\ 6.5\ cm$
$\angle ACB$ is the angle in the semi-circle.
$\therefore \ \angle ACB=90^{o}$
Now, in $\vartriangle ACB$, using Pythagoras theorem, we have
$AB^{2} =AC^{2}+BC^{2}$
$( 13)^{2}=( 12)^{2}+( BC)^{2}$
$( BC)^{2}=( 13)^{2} –( 12)^{2} =169–144=25$
$BC=\sqrt{25}$
$=5\ cm$
Now, Area of shaded region $=$ Area of semi-circle
$=66.33 –30$
$=36.33\ cm^{2}$
Thus, The area of the shaded region is $36.33\ cm^{2}$.
Updated on: 10-Oct-2022
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