# Prove that \[{{r}_{1}}{{r}_{2}}{{r}_{3}}=r{{s}^{2}}\], where \[{{r}_{1}},{{r}_{2}}\]and \[{{r}_{3}}\]is the radius of the exterior circle on side A, B and C.

Last updated date: 17th Mar 2023

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Answer

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Hint: Use heron’s formula to solve. Multiply the radius of these circles considered to prove it. Find the radius of \[{{r}_{1}},{{r}_{2}}\]and \[{{r}_{3}}\]. Prove the LHS by multiplying \[{{r}_{1}},{{r}_{2}}\]and \[{{r}_{3}}\]..

Complete step-by-step answer:

Let us consider that \[{{r}_{1}},{{r}_{2}}\]and \[{{r}_{3}}\]are radius of the circle, opposite to a, b and c of the triangle ABC.

Let the triangle be the area of the triangle ABC.

S is the sum of the length of the triangle ABC.

\[\therefore s=\dfrac{a+b+c}{2}\]

Radius of the triangle, \[r=\dfrac{area}{sum-opposite side}\]

\[\therefore \]radius of \[{{r}_{1}}=\dfrac{\Delta }{s-a}\]

Similarly, \[{{r}_{2}}=\dfrac{\Delta }{s-b}\]and \[{{r}_{3}}=\dfrac{\Delta }{s-c}\]

\[{{r}_{1}}{{r}_{2}}{{r}_{3}}=\left( \dfrac{\Delta }{s-a} \right)\left( \dfrac{\Delta }{s-b} \right)\left( \dfrac{\Delta }{s-c} \right)\]

Where radius \[{{r}_{1}}\]is opposite to the side ‘a’ of \[\vartriangle ABC\]

radius \[{{r}_{2}}\]is opposite to the side ‘b’ of \[\vartriangle ABC\]

radius \[{{r}_{3}}\]is opposite to the side ‘c’ of \[\vartriangle ABC\]

\[\therefore {{r}_{1}}{{r}_{2}}{{r}_{3}}=\dfrac{{{\Delta }^{3}}}{\left( s-a \right)\left( s-b \right)\left( s-c \right)}-(1)\]

By using, Heron’s formula, we can take the area of the triangle, when the length of all three sides of triangle are known:

\[\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}-(2)\]

The figure shows the geometrical significance of \[\left( s-a \right),\left( s-b \right)\]and \[\left( s-c \right)\].

In equation (1) multiply the numerator & denominator by ‘s’.

\[{{r}_{1}}{{r}_{2}}{{r}_{3}}=\dfrac{{{\Delta }^{3}}\times s}{s\times \left( s-a \right)\left( s-b \right)\left( s-c \right)}-(3)\]

We know, \[\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]

Squaring equation (2) on both sides

\[{{\Delta }^{2}}=s\left( s-a \right)\left( s-b \right)\left( s-c \right)-(4)\]

Substitute the value of (4) in (3)

\[=\dfrac{s{{\Delta }^{3}}}{{{\Delta }^{2}}}=s\Delta -(5)\]

Multiply numerator and denominator by ‘s’ in equation (5)

\[={{s}^{2}}\dfrac{\Delta }{s}\]

We know the radius in circle, r = a / (sum of lengths of triangle) = \[\dfrac{a}{s}\]

By substituting the same, we get

\[{{r}_{1}}{{r}_{2}}{{r}_{3}}=r{{s}^{2}}\].

Note: Remember to use heron’s formula to simplify equation (1).

Complete step-by-step answer:

Let us consider that \[{{r}_{1}},{{r}_{2}}\]and \[{{r}_{3}}\]are radius of the circle, opposite to a, b and c of the triangle ABC.

Let the triangle be the area of the triangle ABC.

S is the sum of the length of the triangle ABC.

\[\therefore s=\dfrac{a+b+c}{2}\]

Radius of the triangle, \[r=\dfrac{area}{sum-opposite side}\]

\[\therefore \]radius of \[{{r}_{1}}=\dfrac{\Delta }{s-a}\]

Similarly, \[{{r}_{2}}=\dfrac{\Delta }{s-b}\]and \[{{r}_{3}}=\dfrac{\Delta }{s-c}\]

\[{{r}_{1}}{{r}_{2}}{{r}_{3}}=\left( \dfrac{\Delta }{s-a} \right)\left( \dfrac{\Delta }{s-b} \right)\left( \dfrac{\Delta }{s-c} \right)\]

Where radius \[{{r}_{1}}\]is opposite to the side ‘a’ of \[\vartriangle ABC\]

radius \[{{r}_{2}}\]is opposite to the side ‘b’ of \[\vartriangle ABC\]

radius \[{{r}_{3}}\]is opposite to the side ‘c’ of \[\vartriangle ABC\]

\[\therefore {{r}_{1}}{{r}_{2}}{{r}_{3}}=\dfrac{{{\Delta }^{3}}}{\left( s-a \right)\left( s-b \right)\left( s-c \right)}-(1)\]

By using, Heron’s formula, we can take the area of the triangle, when the length of all three sides of triangle are known:

\[\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}-(2)\]

The figure shows the geometrical significance of \[\left( s-a \right),\left( s-b \right)\]and \[\left( s-c \right)\].

In equation (1) multiply the numerator & denominator by ‘s’.

\[{{r}_{1}}{{r}_{2}}{{r}_{3}}=\dfrac{{{\Delta }^{3}}\times s}{s\times \left( s-a \right)\left( s-b \right)\left( s-c \right)}-(3)\]

We know, \[\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]

Squaring equation (2) on both sides

\[{{\Delta }^{2}}=s\left( s-a \right)\left( s-b \right)\left( s-c \right)-(4)\]

Substitute the value of (4) in (3)

\[=\dfrac{s{{\Delta }^{3}}}{{{\Delta }^{2}}}=s\Delta -(5)\]

Multiply numerator and denominator by ‘s’ in equation (5)

\[={{s}^{2}}\dfrac{\Delta }{s}\]

We know the radius in circle, r = a / (sum of lengths of triangle) = \[\dfrac{a}{s}\]

By substituting the same, we get

\[{{r}_{1}}{{r}_{2}}{{r}_{3}}=r{{s}^{2}}\].

Note: Remember to use heron’s formula to simplify equation (1).

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