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Question

Mathematics Question on Trigonometric Functions

In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio

A

2:3:52 : 3 : 5

B

1:2:31 : 2 : 3

C

1:3:71 : 3 : 7

D

3:7:93 : 7 : 9

Answer

1:2:31 : 2 : 3

Explanation

Solution

We have, Δ=34a2,s=3a2\Delta=\frac{\sqrt{3}}{4}a^{2}, s=\frac{3a}{2}
Inradius r=Δs=a23r=\frac{\Delta}{s}=\frac{a}{2\sqrt{3}}
Circumradius R=abc4Δ=a33a2=a3R=\frac{abc}{4\Delta}=\frac{a^{3}}{\sqrt{3}a^{2}}=\frac{a}{\sqrt{3}}
and exradii r1=Δsa=3/4a2a/2r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3}/ 4a^{2}}{a / 2}
=32a= \frac{\sqrt{3}}{2}a
\therefore Required ratio =r:R:r1=r: R: r_{1}
=a23:a3:32= \frac{a}{2\sqrt{3}} : \frac{a}{\sqrt{3}} : \frac{\sqrt{3}}{2}
a=1:2:3.a=1 : 2 : 3.