Question

In the given figure, ∆ABC is an equilateral triangle the length of whose side is equal to 10 cm, and ∆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:
Side of equilateral triangle ABC = 10 cm
BD = 8 cm
Area of $\mathrm{equilateral}∆ABC=\frac{\sqrt{3}}{4}{\mathrm{a}}^2$ (where a = 10 cm)
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{equilateral}△ABC=\frac{\sqrt{3}}{4}\times {10}^2 \\ =25\sqrt{3} \\ =25\times 1.732 \\ =43.30{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{right}∆BDC,\mathrm{we}\mathrm{have}: \\ B{C}^2=B{D}^2+C{D}^2 \\ \Rightarrow {10}^2=8^2+C{D}^2 \\ \Rightarrow C{D}^2={10}^2-8^2 \\ \Rightarrow C{D}^2=36 \\ \Rightarrow CD=6 \end{gathered}$$

Area of triangle $△$BCD = $\frac{1}{2}\times b\times h$

$$\begin{gathered} =\frac{1}{2}\times 8\times 6 \\ =24{\mathrm{cm}}^2 \end{gathered}$$

Area of the shaded region = $\mathrm{Area}\mathrm{of}∆ABC-\mathrm{Area}\mathrm{of}∆BDC$
= 43.30 $-$ 24
= 19.3 cm²