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# If ${{p}_{1}},{{p}_{2}},{{p}_{3}}$ are respectively the perpendicular from the vertices of a triangle to the opposite sides, then find the value of ${{p}_{1}}{{p}_{2}}{{p}_{3}}$.

Last updated date: 01st Mar 2024
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Hint: Find the area of triangle by taking three different sets of base and height. Then express ${{p}_{1}}$,${{p}_{2}}$ and ${{p}_{3}}$ in terms of area. After getting the values of ${{p}_{1}}$,${{p}_{2}}$ and ${{p}_{3}}$ multiply them and do the necessary simplification to get the required result.

Let us consider a triangle $\vartriangle ABC$ with sides a, b and c.
${{p}_{1}},{{p}_{2}},{{p}_{3}}$ are the perpendiculars drawn from the vertices A, B and C respectively.
We know the area of triangle $=\dfrac{1}{2}\times base\times height$
Again let us denote the area of triangle ABC as ‘A’.
Here the area can be expressed in three different ways as we have three different sets of base and height.
So, the area of triangle ABC
\begin{align} & A=\dfrac{1}{2}\times BC\times AD \\ & \Rightarrow A=\dfrac{1}{2}\times a\times {{p}_{1}} \\ \end{align}
So, ${{p}_{1}}$ can be written as
$\Rightarrow {{p}_{1}}=\dfrac{2A}{a}$
Again,
\begin{align} & A=\dfrac{1}{2}\times AC\times BE \\ & \Rightarrow A=\dfrac{1}{2}\times b\times {{p}_{2}} \\ \end{align}
So, ${{p}_{2}}$ can be written as
$\Rightarrow {{p}_{2}}=\dfrac{2A}{b}$
Again,
\begin{align} & A=\dfrac{1}{2}\times AC\times CF \\ & \Rightarrow A=\dfrac{1}{2}\times c\times {{p}_{3}} \\ \end{align}
So, ${{p}_{3}}$ can be written as
$\Rightarrow {{p}_{3}}=\dfrac{2A}{c}$
Hence,
\begin{align} & {{p}_{1}}{{p}_{2}}{{p}_{3}}=\dfrac{2A}{a}\times \dfrac{2A}{b}\times \dfrac{2A}{c} \\ & \Rightarrow {{p}_{1}}{{p}_{2}}{{p}_{3}}=\dfrac{8{{A}^{3}}}{abc} \\ \end{align}
Again as we know $A=\dfrac{abc}{4R}$ (from sine rule of triangle) where ‘R’ is the radius of the circumcircle.
So, putting the value of ‘A’ in the above expression, we get
\begin{align} & \Rightarrow {{p}_{1}}{{p}_{2}}{{p}_{3}}=\dfrac{8{{\left( \dfrac{abc}{4R} \right)}^{3}}}{abc} \\ & \Rightarrow {{p}_{1}}{{p}_{2}}{{p}_{3}}=\dfrac{8{{a}^{3}}{{b}^{3}}{{c}^{3}}}{64{{R}^{3}}abc} \\ & \Rightarrow {{p}_{1}}{{p}_{2}}{{p}_{3}}=\dfrac{{{a}^{2}}{{b}^{2}}{{c}^{2}}}{8{{R}^{3}}} \\ \end{align}
This is the required solution of the given question.

Note: For the above triangle $\vartriangle ABC$ with sides a, b, c and area ‘A’, the measure of circumradius ‘R’ is $R=\dfrac{abc}{4A}$.
Which can be written as
$\Rightarrow A=\dfrac{abc}{4R}$
This value of area is used in the above solution to get the result in a simplified form.