Question

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

• A. $18\sqrt{3}sq.cm=42.712sq.cm$
• B. $17\sqrt{3}sq.cm=38.712sq.cm$
• C. $15\sqrt{3}sq.cm=22.712sq.cm$
• D. $16\sqrt{3}sq.cm=27.712sq.cm$

Answer 1

The correct option is D $16\sqrt{3}sq.cm=27.712sq.cm$

Given,$△ABC$ and equilateral $△BCD$ with $AD=16cm$ and $BC=8cm$.
$△ABC$ is a right angled triangle
$\therefore {AB}^2+{BC}^2={AD}^2$ .......(Pythagoras Theorem)
${AB}^2+8^2={16}^2{AB}^2={16}^2-8^2=192\therefore AB=\sqrt{192}=8\sqrt{3}cm.$
Now, Area of $△ABC$ = $\frac{1}{2}\times$ base $\times$ height
$\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 8\sqrt{3}\times 8=32\sqrt{3}{cm}^2$
Area of equilateral $△BCD$ =$\frac{\sqrt{3}}{4}{a}^2$ ,where $a$ is side of equilateral triangle.
$=\frac{\sqrt{3}}{4}\times 8^2=16\sqrt{3}{cm}^2$
Area of shaded portion $=$ Area of $△ABC-$ Area of $△BCD$
$=32\sqrt{3}{cm}^2-16\sqrt{3}{cm}^2=16\sqrt{3}{cm}^2=27.712{cm}^2$