Question

Find the area of the shaded region in figure.

Answer 1

In right angle triangle $ADB$

${AB}^2={AD}^2+{BD}^2={12}^2+{16}^2$ (Pythagoras theorem)

$\Rightarrow AB=\sqrt{144+256}=\sqrt{400}=20cm$

In $\Delta ABC$, we can see that

${AB}^2+{BC}^2={20}^2+{48}^2=400+2304=2704$

$\Rightarrow {AB}^2+{BC}^2=2704={52}^2={\left(AC\right)}^2$

$\therefore$ $\Delta ABC$ is right angled at $B$

Area of $\Delta ADB=\frac{1}{2}\times 12\times 16=96{cm}^2$

Area of $\Delta ABC=\frac{1}{2}\times 20\times 48=480{cm}^2$

$\therefore$ Area of shaded region $=$ Area of $\Delta ABC$ $-$ Area of $\Delta ADB$

$=480-96$

$=384{cm}^2$