In the given figure, $\triangle A B C$ is an equilateral triangle the length of whose side is equal to $10 \mathrm{~cm}$, and $\triangle D B C$ is right-angled at $D$ and $B D=8 \mathrm{~cm}$. Find the area of the shaded region. Take $\sqrt{3}=1.732$.
Given:
Side of equilateral triangle ABC = 10 cm
BD = 8 cm
Area of equilateral $\Delta A B C=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$ (where $a=10 \mathrm{~cm}$ )
Area of equilateral $\triangle A B C=\frac{\sqrt{3}}{4} \times 10^{2}$
$=25 \sqrt{3}$
$=25 \times 1.732$
$=43.30 \mathrm{~cm}^{2}$
In the right $\Delta B D C$, we have:
$B C^{2}=B D^{2}+C D^{2}$
$\Rightarrow 10^{2}=8^{2}+C D^{2}$
$\Rightarrow C D^{2}=10^{2}-8^{2}$
$\Rightarrow C D^{2}=36$
$\Rightarrow C D=6$
Area of triangle $\triangle B C D=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 8 \times 6$
$=24 \mathrm{~cm}^{2}$
Area of the shaded region = Area of $\Delta A B C-$ Area of $\Delta B D C$
$=43.30-24$
$=19.3 \mathrm{~cm}^{2}$