Question

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

• A. $\frac{3y}{2x(x+c)}$
• B. $\frac{3y}{2c(x+c)}$
• C. $\frac{3y}{4x(x+c)}$
• D. $\frac{3y}{4c(x+c)}$

Answer 1

Answer: (B)

$\frac{3y}{2c(x+c)}$

We know, in-radius $=\frac{area}{semi-perimeter}=\frac{△}{s}$

And, circum-radius $=\frac{abc}{4△}$

Ratio $=\frac{4{△}^2}{abcs}$

$=\frac{8{△}^2}{yc(x+c)}$

Now, $△=\frac{1}{2}ab.sinC$

$cosC=\frac{a^2+b^2-c^2}{2ab}=\frac{x^2-2y-c^2}{2y}=\frac{-1}{2}$

$\Rightarrow C={120}^{\circ }$

So, $sinC=\frac{\sqrt{3}}{2}$

$△=\frac{1}{2}y.\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}y$

Substituting this in the ratio expression gives Ratio $=\frac{3y}{2c(x+c)}$

Hence, Option (B)