Question

Let $a , b$ and $c$ be the length of sides of a triangle $ABC$ such that $\frac{ a + b }{7}=\frac{ b + c }{8}=\frac{ c + a }{9}$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\frac{ R }{ r }$ is equal to

• A. $\frac{5}{2}$
• B. 2
• C. $\frac{3}{2}$
• D. 1

Answer 1

Answer: (A)

$\frac{ a + b }{7}=\frac{ b + c }{8}=\frac{ c + a }{9}=\lambda$
$a + b =7 \lambda, b + c =8 \lambda, a + c =9 \lambda$
$\Rightarrow a + b + c =12 \lambda$
Now $a =4 \lambda, b =3 \lambda, c =5 \lambda$
$\because c ^{2}= b ^{2}+ a ^{2}$
$\angle C =90^{\circ}$
$\Delta=\frac{1}{2} ab \sin C =\frac{1}{2} ab$
$\frac{ R }{ r }=\frac{ c }{2 \sin C } \times \frac{ s }{\Delta}=\frac{ c }{2} \times \frac{6 \lambda}{\frac{1}{2} ab }=\frac{ c }{ ab } \times 6 \lambda=\frac{5}{2}$