Question

In an equilateral triangle of side $24cm$, a circle is inscribed touching its sides. The area of the remaining portion of the triangles is ($\sqrt{3}=1.732$)

• A. $98.55sqcm$
• B. $100sqcm$
• C. $101sqcm$
• D. $95sqcm$

Answer 1

The correct option is A $98.55sqcm$

Inradius of an equilateral triangle $=\frac{Side}{2\sqrt{3}}=\frac{24}{2\sqrt{3}}=4\sqrt{3}cm$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times (Side{)}^2$

$=\frac{\sqrt{3}}{4}\times 24\times 24$

$=\sqrt{3}\times 6\times 24$

$=144\sqrt{3}c{m}^2$

$=144\times 1.732$

$=249.41c{m}^2$

Now,

Area of incircle $=\pi r^2$

$=\frac{22}{7}\times 4\sqrt{3}\times 4\sqrt{3}$

$=\frac{22\times 16\times 3}{7}$

$=\frac{1056}{7}$

$=150.86c{m}^2$

Area of remaining part $=$ Area of triangle $-$ Area of incircle

$=249.41-150.86$

$=98.55c{m}^2$