Question 42

The ratio of circumradius and inradius of an equilateral triangle is

Solution

Circumradius of a triangle = $$R = \frac{abc}{4\triangle}$$

Inradius = $$r=\frac{2\triangle}{a+b+c}$$

Let the side of the equilateral triangle = $$a$$ cm

Also, area of equilateral triangle = $$\triangle = \frac{\sqrt3}{4}a^2$$

=> Ratio of circumradius and inradius

= $$(\frac{a^3}{4\triangle})\div(\frac{2\triangle}{3a})$$

= $$\frac{3a^4}{8\triangle^2} = \frac{3a^4}{8a^4 \times \frac{3}{16}}$$

= $$\frac{16}{8}=\frac{2}{1}$$

=> Ans - (C)


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