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 ${{p}_{1}}{{p}_{2}}{{p}_{3}}$.
Answer
Verified
436.8k+ views
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.
Complete step by step answer:
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.
Complete step by step answer:
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.
Recently Updated Pages
Class 10 Question and Answer - Your Ultimate Solutions Guide
Master Class 10 Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 10 General Knowledge: Engaging Questions & Answers for Success
Master Class 10 Social Science: Engaging Questions & Answers for Success
Master Class 10 English: Engaging Questions & Answers for Success
Trending doubts
What is Commercial Farming ? What are its types ? Explain them with Examples
List out three methods of soil conservation
Complete the following word chain of verbs Write eat class 10 english CBSE
Compare and contrast a weekly market and a shopping class 10 social science CBSE
Imagine that you have the opportunity to interview class 10 english CBSE
On the outline map of India mark the following appropriately class 10 social science. CBSE