Question

If $p_1,p_2,p_3$ are respectively the perpendiculars from the vertices of a triangle to the opposite side, then $p_1{p}_2{p}_3$ is equal to

• A. $a^2{b}^2{c}^2$
• B. $2a^2{b}^2{c}^2$
• C. $\frac{4a^2{b}^2{c}^3}{R^2}$
• D. $\frac{a^2{b}^2{c}^2}{8R^3}$

Answer 1

The correct option is D

$\frac{a^2{b}^2{c}^2}{8R^3}$

Explanation for the correct option:

Step 1: Draw a required digram

Where, $\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}$ are the sides of the triangle $ABC$.

We know that the area of the triangle $=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Step 2: Calculate the value of $p_1{p}_2{p}_3$

Area of the triangle ABC

$$\begin{gathered} \therefore A=\frac{1}{2}\times BC\times AD \\ \Rightarrow A=\frac{1}{2}\times a\times p_1 \\ \Rightarrow p_1=\frac{2A}{a} \end{gathered}$$

Similarly,

$\begin{array}{l}\therefore A=\frac{1}{2}\times AC\times BE\\ \Rightarrow A=\frac{1}{2}\times b\times p_2\\ \Rightarrow p_2=\frac{2A}{b}\end{array}$

Similarly,

$\begin{array}{l}\therefore A=\frac{1}{2}\times AC\times CF\\ \Rightarrow A=\frac{1}{2}\times c\times p_3\\ \Rightarrow p_3=\frac{2A}{c}\end{array}$

Now,

$\begin{array}{l}\therefore p_1{p}_2{p}_3=\frac{2A}{a}\times \frac{2A}{b}\times \frac{2A}{c}\\ \Rightarrow p_1{p}_2{p}_3=\frac{8A^3}{abc}\end{array}$

Step 3: Apply the concept $$A={\displaystyle \frac{abc}{4R}}$$ (from $\mathbf{sine}$ rule of the triangle) where $‘R’$is the radius of the circumcircle.
Therefore, putting the value of $‘A’$ in the above expression,

$\begin{array}{c}\Rightarrow p_1{p}_2{p}_3=\frac{8{\left(\frac{abc}{4R}\right)}^3}{abc}\\ =\frac{8a^3{b}^3{c}^3}{64R^{3}abc}\\ =\frac{a^2{b}^2{c}^2}{8R^3}\end{array}$

Hence, the correct option is (D).