Question

If $P_1{P}_2,P_3$ are the perpendiculars of a triangle $ABC$ from the vertices $A,B,C$ to sides opposites to $\angle A,\angle B,\angle C$ respectively and $2S$ be the perimeter of triangle ABC $,\Delta$ is the area of the triangle.

In the $\Delta ABC$ $P_1,P_2,P_3$,are in A.P. then

• A. $a,b,c\in A.P.$
• B. $a,b,c\in G.P.$
• C. $a,b,c\in H.P.$
• D. None of these

Answer 1

The correct option is C $a,b,c\in H.P.$

Let the length of the sides be $a,b,c$.

Hence, Area $=△$ $=\frac{a.P_1}{2}$ $=\frac{b.P_2}{2}$ $=\frac{c.P_3}{2}$
Hence
$P_1=\frac{2△}{a}$
$P_2=\frac{2△}{b}$
$P_3=\frac{2△}{c}$

Since
$P_1,P_2,P_3$ are A.P

Hence
$\frac{2△}{a},\frac{2△}{b},\frac{2△}{c}$ are in A.P

Hence
$a,b,c$ are in H.P.