Question

Find the area of the shaded region in figure:

Answer 1

By Pythagoras theorem, in $\Delta ADB$
$AB=\sqrt{A{D}^2+B{D}^2}$
$=\sqrt{{12}^2+{16}^2}$
$=\sqrt{144+256}$
$=\sqrt{400}$
$AB=20cm$
$\therefore$ area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
$\{s=\frac{20+48+52}{2}=60cm\}$
$=\sqrt{60(60-20)(60-48)(60-52)}$
$=\sqrt{60\times 40\times 12\times 8}$
$=\sqrt{(12\times 5)\times (5\times 8)\times 12\times 8}$
$=\sqrt{5^2\times 8^2\times {12}^5}=5\times 8\times 12=480c{m}^2$
also area of $\Delta ADB=\frac{1}{2}\left(AD\right)\left(BD\right)$
$=\frac{1}{2}\times 12\times 16=96c{m}^2$
$\therefore$ Shaded area $=ar\left(\Delta ABC\right)-ar\left(\Delta ADB\right)$
$=480-96=384c{m}^2$