Question

In a triangle the sum of two sides is $x$ and their product is $y$ such that $(x+z)(x-z)=y$ where $z$ is the third side of the triangle. The inradius of the triangle is :

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

Answer 1

Answer: (B)

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

Sum of two sides $=x$

Product of two sides $=y$ such that $(x+z)(x-z)=y$

Inadius of a triangle is $r=\frac{△}{s}$ where $△=\sqrt{s(s-a)(s-b)(s-c)}$

and $s=\frac{a+b+c}{2}\Rightarrow \frac{x+z}{2}$

$△=\sqrt{\left(\frac{x+z}{2}\right)\left(\frac{x-z}{2}\right)\left[(s-a)(s-b)\right]}$

$=\frac{1}{2}\sqrt{y[s^2-(a+b)s+ab]}$ ....... $[(x+z)(x-z)=y]$

$=\frac{1}{2}\sqrt{y[{\left(\frac{x+z}{2}\right)}^2-x\left(\frac{x+z}{2}\right)+y]}$

$=\frac{\sqrt{y}}{2}\sqrt{\frac{x^2+z^2+2xz}{4}-\frac{x^2+xz}{2}+y}$

$=\frac{\sqrt{y}}{2}\sqrt{\frac{x^2+z^2+2xz-2x^2-2xz+4y}{4}}$

$=\frac{\sqrt{y}}{4}\sqrt{(z^2-x^2)+4y}=\frac{\sqrt{y}}{4}\times \sqrt{3y}=\frac{y\sqrt{3}}{4}$

$\therefore r=\frac{△}{5}=\frac{\frac{y\sqrt{3}}{4}}{\frac{x+z}{2}}=\frac{\sqrt{3}y}{2(x+z)}$

Hence, $\frac{\sqrt{3}y}{2(x+z)}$ is the correct answer.