Question

The given figure shows a right angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

• A. $3\sqrt{3}c{m}^2$
• B. $4\sqrt{3}c{m}^2$
• C. $5\sqrt{3}c{m}^2$
• D. $6\sqrt{3}c{m}^2$

Answer 1

The correct option is B

$4\sqrt{3}c{m}^2$

Area of shaded porton =Area $\left(\Delta ADB\right)$

Area $\left(\Delta ADB\right)$ =Area $\left(\Delta ABC\right)$ -Area $\left(\Delta BDC\right)--\left(1\right)$

In $\Delta ABC,A{B}^2+B{C}^2=A{C}^2$

$\therefore A{B}^2=A{C}^2-B{C}^2=8^2-4^2=64-16$

$A{B}^2=48$

$\therefore AB=4\sqrt{3}cm$

$\therefore$ Area $\left(\Delta ABC\right)=\frac{1}{2}\times 4\sqrt{3}\times 4=8\sqrt{3}c{m}^2$

Area $\left(\Delta BDC\right)=\frac{\sqrt{3}}{4}(BC{)}^2=\frac{\sqrt{3}}{4}\times 16=4\sqrt{3}c{m}^2$

$\therefore$ Putting values in (1), we get

Area $\left(\Delta ADB\right)=8\sqrt{3}c{m}^2-4\sqrt{3}c{m}^2$

$=4\sqrt{3}c{m}^2$