Question

Find the area of the shaded region in figure.

Answer 1

In $\Delta$ ABC, AC =52 cm, BC= 48 cm and in right $\Delta$ ADC, $\angle D={90}^0$
AD= 12 cm, BD = 16 cm
$\therefore A{B}^2=A{D}^2+B{D}^2$ (Pythagoras Theorem)

$(12{)}^2+(16{)}^2=144+256=400=(20{)}^2$

$\therefore AB=20cm$

=$\frac{1}{2}\times 12\times 16=96c{m}^2$ and area of $\Delta$ ABC with sides 52 cm 48 cm, 20 cm

$\therefore s=\frac{a+b+c}{2}=\frac{52+48+20}{2}=\frac{120}{2}=60$

$\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}$

= $\sqrt{60(60-52)(60-48)(62-20)}$

= $\sqrt{60\times 8\times 12\times 40}$

= $\sqrt{10\times 2\times 3\times 2\times 2\times 2\times 2\times 2\times 3\times 10\times 2\times 2}$

=$\sqrt{10\times 2\times 2\times 2\times 2\times 3}=480c{m}^2$

$\therefore$ Area of shaded portion = $480-96=384c{m}^2$