Search any question & find its solution
Question:
Answered & Verified by Expert
Let $p, q$ and $r$ be the altitudes of a triangle with area $S$ and perimeter $2 t .$ Then, the value of $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ is
Options:
Solution:
1122 Upvotes
Verified Answer
The correct answer is:
$\frac{t}{S}$
According to question, $S=\frac{1}{2} a \cdot p$
$$
\begin{array}{c}
=\frac{1}{2} b \cdot q=\frac{1}{2} c \cdot r \\
\therefore \quad \frac{1}{p}=\frac{a}{2 S}, \frac{1}{q}=\frac{b}{2 S}, \frac{1}{r}=\frac{c}{2 S} \\
\therefore \quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2 S}(a+b+c)=\frac{2 t}{2 S}=\frac{t}{S}
\end{array}
$$
$$
\begin{array}{c}
=\frac{1}{2} b \cdot q=\frac{1}{2} c \cdot r \\
\therefore \quad \frac{1}{p}=\frac{a}{2 S}, \frac{1}{q}=\frac{b}{2 S}, \frac{1}{r}=\frac{c}{2 S} \\
\therefore \quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2 S}(a+b+c)=\frac{2 t}{2 S}=\frac{t}{S}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.