SSC CPO SI and ASI 2013 Paper

Question : 114 of 200

Marks: +1, -0

Three circles of equal radius a cm touch each other. The area of the shaded region is

Solution:

(4)

PQ = QR = RP = 2a cm

∠ Q P R=∠ P R Q=∠ P Q R=60°

Area of

\Δ{PQR}= {\√{3}}/{4} ×{\side}^{2}

=\ {√{3}}/{4} ×(2 {a})^{2}=√{3} {a}^{2} {cm}^{2}

Area of three sectors

=3 ×{60}/{360} \×\π{a}^{2}= {\π {a}^{2}}/{2} {cm}^{2}

Area of shaded region = Area ofΔPQR – Area of three sectors

=√{3} a^{2}- {\πa^{2}}/{2}

= ( {2 √{3}-\π}/{2} ) a^{2} c m

Go to Question:

Prev Question

Next Question