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SSC CGL Tier II 21 Sep 2014 Paper

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Question : 8 of 100

Marks: +1, -0

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)

98.55 sq cm
100 sq cm
101 sq cm
95 sq cm

Solution:

(1) In-radius

=\frac{a}{2\sqrt{3}}

=\frac{24}{2\sqrt{3}}=4\sqrt{3}\,\text{cm}

Area of triangle

=\frac{\sqrt{3}}{4}\times(\text{side})^{2}

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

=144\sqrt{3}

sq.cm.

=144\times 1.732

=249.408\,\text{sq. cm}

Area of circle

=\pi r^{2}

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

=\frac{1056}{7}=150.86\,\text{sq.cm}.

Area of remaining part = (249.408 – 150.86) sq.cm.

= 98.548 sq.cm. ≈ 98.55 sq.cm.

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