question_answer

In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle is \[(\sqrt{3}=1.732)\] [SSC CGL Tier II. 2014]

A) 98.55 sq cm

B) 100 sq cm

C) 101 sq cm

D) 95 sq cm

Correct Answer: A

Solution :

[a] Radius of incircle\[=\frac{a}{2\sqrt{3}}=\frac{24}{2\sqrt{3}}=\frac{12}{\sqrt{3}}\]                         Area of triangle \[\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{1.732}{4}\times {{(24)}^{2}}\] \[=\frac{1.732\times 576}{4}\] \[=249.408\] Area of circle \[=\pi {{r}^{2}}=\frac{22}{7}\times {{\left( \frac{12}{\sqrt{3}} \right)}^{2}}\] \[=\frac{22}{7}\times \frac{144}{3}=\frac{3168}{21}=150.857\] \[\therefore \]Area of remaining portion \[=249.408-150.857=98.55\,\,c{{m}^{2}}\]