ABC26GN4639 · Area
Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
The radius of the circumcircle of an equilateral triangle of side 12 cm is
(a)$\frac{4 \sqrt{2}}{3} \mathrm{cm}$
(b)$4 \sqrt{2} \mathrm{cm}$
(c)$\frac{4 \sqrt{3}}{3} \mathrm{cm}$
(d)$4 \sqrt{3} \mathrm{cm}$
Answer
Explanation
Radius of circumcircle $=\frac{a}{\sqrt{3}}=\frac{12}{\sqrt{3}} \mathrm{cm}=4 \sqrt{3} \mathrm{cm}$. 351. Radius of incircle $=\frac{a}{2 \sqrt{3}}=\frac{42}{2 \sqrt{3}} \mathrm{cm}=7 \sqrt{3} \mathrm{cm}$. Area of incircle $=\left(\frac{22}{7} \times 49 \times 3\right) \mathrm{cm}^{2}=462 \mathrm{cm}^{2}$.
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