Question

In the given figure, $\Delta ABC$ is an equilateral triangle the length of whose side is equal to 10 cm, and $\Delta DBC$ is right-angled at D and $BD=8cm.$ Find the area of the shaded region. [Take $\sqrt{3}=\mathrm{1.732.}$]

Answer 1

Given, $\Delta ABC$ is an equilateral triangle the length of whose side is equal to 10 cm, and $\Delta DBC$ is right-angled at D
and $BD=8cm.$

From figure:
Area of shaded region $=$ Area of $\Delta ABC-$ Area of $\Delta DBC.....\left(1\right)$

Area of $\Delta ABC:$
Area $=\frac{\sqrt{3}}{4}(side{)}^2=\frac{\sqrt{3}}{4}(10{)}^2=43.30$

So area of $\Delta ABC$ is $43.30c{m}^2$

Area of right $\Delta DBC:$
Area $=\frac{1}{2}\times base\times height...\left(2\right)$

From Pythagoras Theorem:
Hypotenuse${}^2$ = Base${}^2$ + Height${}^2$
$B{C}^2=D{B}^2+Heigh{t}^2$
$100-64=Heigh{t}^2$
$36=Heigh{t}^2$
or Height $=6$
equation $\left(2\right)\Rightarrow$
Area $=\frac{1}{2}\times 8\times 6=24$
So area of $\Delta DBC$ is $24c{m}^2$

Equation (1) implies
Area of shaded region $=43.30-24=19.30$
Therefore, Area of shaded region $=19.3c{m}^2$