SSC CGL Tier II 21 Sep 2014 Paper

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

( √3 =1.732 )

Solution:

(1) In-radius

=a/{2√3}

= {24}/{2√{3}}=4 √{3} {cm}

Area of triangle

= {√{3}}/{4} ×( side )^{2}

= {√{3}}/{4} × 24× 24

=144√{3}

sq.cm.

=144 × 1.732

=249.408 {sq} . {cm}

Area of circle

=\πr^{2}

=\ {22}/{7}× 4 √{3}× 4√{3}

= {1056}/{7}=150.86 {sq.cm} .

Area of remaining part = (249.408 – 150.86) sq.cm.
= 98.548 sq.cm. ≈ 98.55 sq.cm.

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