JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Properties of Triangle

In a triangle the sum of two sides is

x

and the product of the same two sides is

y

. If

x^{2} - c^{2} = y

, where

c

is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

Answer: (B)

Solution

Let the two sides be

a

and

b

with

a + b = x

and

ab = y

.

Given

x^{2} - c^{2} = y

, we have

(a + b)^{2} - c^{2} = ab

.

Expanding:

a^{2} + b^{2} + 2 ab - c^{2} = ab

, so

a^{2} + b^{2} - c^{2} = - ab

.

By the cosine rule:

cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = - \frac{1}{2}

, giving

C = 12 0^{\circ}

.

The semi-perimeter is

s = \frac{x + c}{2}

.

In-radius:

r = \frac{Δ}{s} = \frac{\frac{1}{2} ab s i n C}{\frac{x + c}{2}} = \frac{y s i n 12 0 ^{\circ}}{x + c} = \frac{y 3}{2 ( x + c )}

.

Circum-radius:

R = \frac{c}{2 s i n C} = \frac{c}{2 \cdot \frac{3}{2}} = \frac{c}{3}

.

Thus

\frac{r}{R} = \frac{y 3}{2 ( x + c )} \cdot \frac{3}{c} = \frac{3 y}{2 c ( x + c )}

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option