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In fig. a circle is inscribed in a triangle ABC, touches its sides \ AB, BC and AC at points D, E and F respectively.If $AB=12\ cm,\ BC=8\ cm$,Then find the length of AD, BE and CF.
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$-$Area of the $vartiangle ABC$
$=66.33 –30$
$=36.33\ cm^{2}$
Thus, The area of the shaded region is $36.33\ cm^{2}$.
Updated on: 10-Oct-2022
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