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$

A)\frac{y}{{13}} \\

B)\frac{c}{{\sqrt 3 }} \\

C)\frac{c}{3} \\

D)\frac{3}{2}y \\

$

Answer

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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.

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.