Question

If ${{p}_{1}},{{p}_{2}},{{p}_{3}}$ are altitudes of a triangle $ABC$ from the vertices $A,B,C$ and $\Delta $ is the area of the triangle, then \[p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}\] is equal to
A. $\dfrac{a+b+c}{\Delta }$
B. $\dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{4{{\Delta }^{2}}}$
C. $\dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{{{\Delta }^{2}}}$
D. None of these.

Answer 1

Hint: To solve this question, we will use the formula of area of right angled triangle and calculate area from each of the vertices$A,B,C$ of the triangle and then we will derive three equation for ${{p}_{1}},{{p}_{2}},{{p}_{3}}$. We will then reciprocate the equations ,square them and add them to get the value of \[p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}\].

Formula Used: Area of right-angled triangle: $\Delta =\dfrac{1}{2}\times b\times h$ where $b$ is base and $h$ is height

Complete step by step solution: We are given a triangle $ABC$ in which ${{p}_{1}},{{p}_{2}},{{p}_{3}}$ are the altitudes of a triangle $ABC$ from the vertices $A,B,C$ and $\Delta $ is the area of the triangle. And we have to calculate the value of \[p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}\].
We will use the area of the right angled triangle and calculate the area of the triangle from each of the vertices $A,B,~$ of the triangle $ABC$.
Now,
Area of triangle with ${{p}_{1}}$ as altitude and $a$ as a base.
$\Delta =\dfrac{1}{2}a{{p}_{1}}$
We will derive the equation for ${{p}_{1}}$.
$\dfrac{2\Delta }{a}={{p}_{1}}$…. (i).
Area of triangle with ${{p}_{2}}$ as altitude and $b$ as a base.
$\Delta =\dfrac{1}{2}b{{p}_{2}}$
We will derive the equation for ${{p}_{2}}$.

$\dfrac{2\Delta }{b}={{p}_{2}}$…. (ii)
Area of triangle with ${{p}_{3}}$ as altitude and $c$ as a base.
$\Delta =\dfrac{1}{2}c{{p}_{3}}$
We will derive the equation for ${{p}_{3}}$.
$\dfrac{2\Delta }{c}={{p}_{3}}$…. (iii)
We will now reciprocate all the equations (i), (ii) and (iii).
\[\begin{align}
  & \dfrac{1}{\dfrac{2\Delta }{a}}=\dfrac{1}{{{p}_{1}}} \\
 & \dfrac{a}{2\Delta }=\dfrac{1}{{{p}_{1}}}....(iv) \\
 & \dfrac{1}{\dfrac{2\Delta }{b}}=\dfrac{1}{{{p}_{2}}} \\
 & \dfrac{b}{2\Delta }=\dfrac{1}{{{p}_{2}}}....(v)
\end{align}\]
\[\begin{align}
  & \dfrac{1}{\dfrac{2\Delta }{c}}=\dfrac{1}{{{p}_{3}}} \\
 & \dfrac{c}{2\Delta }=\dfrac{1}{{{p}_{3}}}....(vi)
\end{align}\]
We will now square the equations (iv), (v) and (vi) and add them.
\[\begin{align}
  & {{\left( \dfrac{a}{2\Delta } \right)}^{2}}+{{\left( \dfrac{b}{2\Delta } \right)}^{2}}+{{\left( \dfrac{c}{2\Delta } \right)}^{2}}=\dfrac{1}{p_{1}^{2}}+\dfrac{1}{p_{2}^{2}}+\dfrac{1}{p_{2}^{3}} \\
 & \dfrac{{{a}^{2}}}{4{{\Delta }^{2}}}+\dfrac{{{b}^{2}}}{4{{\Delta }^{2}}}+\dfrac{{{c}^{2}}}{4{{\Delta }^{2}}}=\dfrac{1}{p_{1}^{2}}+\dfrac{1}{p_{2}^{2}}+\dfrac{1}{p_{2}^{3}} \\
 & \dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{4{{\Delta }^{2}}}=\dfrac{1}{p_{1}^{2}}+\dfrac{1}{p_{2}^{2}}+\dfrac{1}{p_{2}^{3}} \\
 & \dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{4{{\Delta }^{2}}}=p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}
\end{align}\]

The value of \[p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}\] when ${{p}_{1}},{{p}_{2}},{{p}_{3}}$, are altitudes of a triangle $ABC$ from the vertices $A,B,C$ and $\Delta $ is the area of the triangle. Hence the correct option is (C).

Note: The relationship between the three sides of a right angled triangle is stated by Pythagoras theorem.