Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let a,b and c be the length of sides of a triangle ABC such that <img alt="\mathrm{\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

Let a,b and c be the length of sides of a triangle ABC such that \mathrm{\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}}

If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of \frac{\mathrm{R}}{\mathrm{r}} is equal to:

Option: 1

\frac{5}{2}


Option: 2

2


Option: 3

\frac{3}{2}


Option: 4

1


Answers (1)

best_answer

\mathrm{ Let \: \frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k}
\mathrm{ So \: \, a+b=7 k,\quad b+c=8 k, \quad c+a=9 k}

Solving this we get \mathrm{ a=4 k, b=3 k, c=5 k}  and calculating semi-perimeter\mathrm{ s=\frac{a+b+c}{2} \Rightarrow s=6 k}
\mathrm{\text{we know that} \: \: \frac{r}{R}=4 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \sin \frac{C}{2}}
\mathrm{ \Rightarrow \frac{r}{R}=4 \sqrt{\frac{(s-b)(s-c)}{b c}} \times \sqrt{\frac{(s-a)(s-c)}{a c}} \times \sqrt{\frac{(s-a)(s-b)}{a b}}}
\mathrm{ \Rightarrow \frac{r}{R}=4 \cdot \frac{(s-a)(s-b)(s-c)}{a b c}}
\mathrm{ \Rightarrow \frac{r}{R}=4 \cdot \frac{(2 k)(3 k)(k)}{(4 k)(3 k)(5 k)} \Rightarrow \frac{r}{R}=\frac{2}{5}}

\mathrm{ \text{So the value of}\: \: \frac{R}{r}=\frac{5}{2}}

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question