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Find the area of the shaded region in figure.
"
To do:
We have to find the area of the shaded region in the figure.
Solution:
From the figure,
In $\triangle ABC$,
$AC = 52\ cm, BC = 48\ cm$
In right angle triangle $ADC$,
$\angle D = 90^o, AD = 12\ cm, BD = 16\ cm$
By Pythagoras theorem,
$AB^2=AD^2 + BD^2$
$=(12)^2 + (16)^2$
$= 144 + 256$
$= 400$
$= (20)^2$
$\Rightarrow AB = 20\ cm$
Area of the $\triangle \mathrm{ABD}=\frac{1}{2} \times \mathrm{AD} \times \mathrm{BD}$
$=\frac{1}{2} \times 12 \times 16$
$=96 \mathrm{~cm}^{2}$
Area of $\triangle \mathrm{ABC}$ with sides $52 \mathrm{~cm}, 48 \mathrm{cm}, 20 \mathrm{~cm}$ is,
$s=\frac{a+b+c}{2}$
$=\frac{52+48+20}{2}$
$=\frac{120}{2}$
$=60$
Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{60(60-52)(60-48)(62-20)}$
$=\sqrt{60 \times 8 \times 12 \times 40}$
$=\sqrt{10 \times 2 \times 3 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 10 \times 2 \times 2}$
$=10 \times 2 \times 2 \times 2 \times 2 \times 3$
$=480 \mathrm{~cm}^{2}$
Area of the shaded portion $=480-96$
$=384 \mathrm{~cm}^{2}$