Search any question & find its solution
Question:
Answered & Verified by Expert
In $\triangle \mathrm{ABC}$, if $\mathrm{r}=1, \mathrm{R}=4$ and $\Delta=8$, then $\frac{1}{\mathrm{ab}}+\frac{1}{\mathrm{bc}}+\frac{1}{\mathrm{ca}}=$
Options:
Solution:
1194 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{8}$
Given,
$\begin{aligned} & \text { In } \triangle A B C, \\ & r=1, \quad R=4 ; \Delta=8 \\ & \text { since } \left.\quad r=\frac{\Delta}{s} \right\rvert\, \text { since } a b c=4 R \Delta \\ & \left.\Rightarrow \quad 1=\frac{\Delta}{s} \right\rvert\, \quad=4.4 .8 \\ & \Rightarrow \quad \Delta=s=8 \mid=128 \\ & \text { Now } \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{(a+b+c)}{a b c}=\frac{2 s}{128} \\ & =\frac{2 \times 8}{128}=\frac{1}{8}\end{aligned}$
$\begin{aligned} & \text { In } \triangle A B C, \\ & r=1, \quad R=4 ; \Delta=8 \\ & \text { since } \left.\quad r=\frac{\Delta}{s} \right\rvert\, \text { since } a b c=4 R \Delta \\ & \left.\Rightarrow \quad 1=\frac{\Delta}{s} \right\rvert\, \quad=4.4 .8 \\ & \Rightarrow \quad \Delta=s=8 \mid=128 \\ & \text { Now } \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{(a+b+c)}{a b c}=\frac{2 s}{128} \\ & =\frac{2 \times 8}{128}=\frac{1}{8}\end{aligned}$
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.