Properties of Triangles 3 Question 3
3. 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 inradius to the circumradius of the triangle is
(2014 Adv)
(a) $\frac{3 y}{2 x(x+c)}$
(b) $\frac{3 y}{2 c(x+c)}$
(c) $\frac{3 y}{4 x(x+c)}$
(d) $\frac{3 y}{4 c(x+c)}$
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Answer:
Correct Answer: 3. (b)
Solution:
- PLAN (i) $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \quad$ (ii) $R=\frac{a b c}{4 \Delta}, r=\frac{\Delta}{s}$
where, $R, r, \Delta$ denote the circumradius, inradius and area of triangle, respectively.
Let the sides of triangle be $a, b$ and $c$.
$$ \begin{aligned} & \text { Given, } \\ & x=a+b \\ & y=a b \\ & x^{2}-c^{2}=y \\ & \Rightarrow \quad(a+b)^{2}-c^{2}=y \\ & \Rightarrow \quad a^{2}+b^{2}+2 a b-c^{2}=a b \\ & \Rightarrow \quad a^{2}+b^{2}-c^{2}=-a b \\ & \Rightarrow \quad \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}=\cos 120^{\circ} \\ & \Rightarrow \quad \angle C=\frac{2 \pi}{3} \\ & \because \quad R=\frac{a b c}{4 \Delta}, r=\frac{\Delta}{s} \Rightarrow \frac{r}{R}=\frac{4 \Delta^{2}}{s(a b c)} \\ & =\frac{4 \frac{1}{2} a b \sin \frac{2 \pi}{3}{ }^{2}}{\frac{x+c}{2} \cdot y \cdot c} \\ & \therefore \quad \frac{r}{R}=\frac{3 y}{2 c(x+c)} \end{aligned} $$
