Question

In an equilateral triangle, the ratio of the radii of incircle, circumcircle and excircle is

Answer 1

SOLUTION

TO DETERMINE

In an equilateral triangle, the ratio of the radii of in-circle , circumcircle and ex-circle

EVALUATION

Since the triangle is an equilateral triangle

So every angle = 60°

Let A, B, C be the angles

$\displaystyle \sf{A = B = C = \frac{\pi}{3} }$

Let

R = Radius of the circumcircle

r = Radius of the in-circle

$\sf{r_1 = Radius \: of \: the \: ex circle}$

So

$\displaystyle \sf{ r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}$

$\displaystyle \sf{ \implies \: r = 4R \sin \frac{\pi}{6} \sin \frac{\pi}{6}\sin \frac{\pi}{6}}$

$\displaystyle \sf{ \implies \: r = 4R \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} }$

$\displaystyle \sf{ \implies \: r = \frac{4R}{8} }$

$\displaystyle \sf{ \implies \: r = \frac{R}{2} }$

Again

$\displaystyle \sf{ r_1 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}}$

$\displaystyle \sf{ \implies \: r_1 = 4R \cos \frac{\pi}{6} \cos \frac{\pi}{6} \sin \frac{\pi}{6}}$

$\displaystyle \sf{ \implies \: r_1 = 4R \times \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} \times \frac{1}{2} }$

$\displaystyle \sf{ \implies \: r_1 = \frac{3R}{2} }$

Hence the required ratio of the radii of in-circle , circumcircle and ex-circle

$\displaystyle \sf{ = r: R: r_1 }$

$\displaystyle \sf{ = \frac{R}{2}: R: \frac{3R}{2} }$

$\displaystyle \sf{ = 1: 2: 3 }$

FINAL ANSWER

The required ratio of the radii of in-circle , circumcircle and ex-circle = 1 : 2 : 3

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