Question

Find the area of the shaded region:)

Answer 1

Answer:

first get the area of whole triangle

then get area of the non shaded portion

then subtract both areas

that will be the area if the shaded part

Answer 2

To find:

$\sf{The \ area \ of \ the \ shaded \ region. }$

Solution:

$\sf{By \ heron's \ formula}$

$\sf{In \ \triangle ABC,}$

$\sf{s=\dfrac{20+52+48}{2}}$

$\sf{s=60}$

$\sf{A(\triangle ABC)=\sqrt{60(60-20)(60-48)(60-52)}}$

$\sf{\therefore{A(\triangle \ ABC)=\sqrt{60\times40\times12\times8}}}$

$\sf{\therefore{A(\triangle ABC)=\sqrt{230400}}}$

$\sf{\therefore{A(\triangle ABC)=480 \ cm^{2}}}$

$\sf{In \ \triangle AOB,}$

$\sf{OA=\sqrt{20^{2}+16^{2}}}$

$\sf{By \ pythagoras \ theorem}$

$\sf{\therefore{OA=\sqrt{656}}}$

$\sf{\therefore{OA=25.6 \ cm}}$

$\sf{A(\triangle \ AOB)=\dfrac{1}{2}\times16\times25.6}$

$\sf{\therefore{A(\triangle \ AOB)=204.8 \ cm^{2}}}$

$\sf{Now,}$

$\sf{Area \ of \ shaded \ region=A(\triangle \ ABC)-A(\triangle \ AOB)}$

$\sf{\therefore{Area \ of \ shaded \ region=480-204.8}}$

$\sf{\therefore{Area \ of \ shaded \ region=275.2 \ cm^{2}}}$

$\sf\purple{\tt{The \ area \ of \ shaded \ region \ is \ 275.2 \ cm^{2}}}$