Question

# In a triangle, sum of lengths of two sides is $x$ and the product of length of same two sides is $y$.If${x^2} - {y^2} = {c^2}$ where $c$ the length of third side of triangle then circumradius of triangle is :$A)\frac{y}{{13}} \\ B)\frac{c}{{\sqrt 3 }} \\ C)\frac{c}{3} \\ D)\frac{3}{2}y \\$

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Let us consider a triangle$\Delta abc$, where $a,b,c$ are the sides of triangle
Given $c$ is the length of the third side of the triangle, then $a,b$ be the other two sides of the triangle.
Given sum of lengths of two sides = $x$
$a + b = x \to (1)$
And also given product of length of same two sides = $y$
$ab = y \to (2)$
Given condition
$\Rightarrow$${x^2} - {y^2} = {c^2}$
Let us substitute the $x$ and $y$ values in the above equation
$\Rightarrow {(a + b)^2} - {c^2} = ab \\ \Rightarrow {a^2} + {b^2} + 2ab - {c^2} = ab \\ \Rightarrow {a^2} + {b^2} - {c^2} = ab - 2ab \\ \Rightarrow {a^2} + {b^2} - {c^2} = - ab \\$
Let us divide with $2ab$on both sides we get
$\Rightarrow \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \dfrac{{ - ab}}{{2ab}} \\ \Rightarrow \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \dfrac{{ - 1}}{2} \to (3) \\$
Apply the cosine rule formula where $COSC = \dfrac{{{a^2} + {b^2} - c2}}{{2ab}}$
From the cosine rule we can rewrite the equation $'3'as$
$COSC = \dfrac{{ - 1}}{2} \\ \angle C = \dfrac{{2\pi }}{3} \\$
We know that circumradius of triangle is $R = \dfrac{C}{{2SINC}}$
On substituting the value we get $R = \dfrac{C}{{\sqrt 3 }}$
Therefore circumradius of triangle is $R = \dfrac{C}{{\sqrt 3 }}$
( B) is the correct option
NOTE: Make a note that after substituting the value in the given condition we have divided the equation with $2ab$ on both sides, where we directly get the required value of circumradius after applying cosine rule.