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 $$BD = 8\,cm.$$ Find the area of the shaded region. [Take $$\sqrt{3} = 1.732.$$]

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 $$ BD = 8\,cm.$$

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

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

So area of $$ \Delta ABC $$ is $$ 43.30\,cm^{2}$$

Area of right $$ \Delta DBC: $$
Area $$ = \dfrac12\times base \times height...(2) $$

From Pythagoras Theorem:
Hypotenuse$$^{2}$$ = Base$$^{2}$$ + Height$$^{2}$$
$$ BC^{2} = DB^{2}+Height^{2}$$
$$ 100-64 = Height^{2}$$
$$ 36 = Height^{2} $$
or Height $$ = 6 $$
equation $$ (2)\Rightarrow $$
Area $$ = \dfrac12\times 8\times 6 = 24 $$
So area of $$ \Delta DBC $$ is $$ 24\,cm^{2}$$

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