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 $\frac Rr$ is equal to :

• A. $\frac 52$
• B. $2$
• C. $\frac 32$
• D. $1$

Answer 1

Answer: (A)

$\frac 52$

Answer 2

Let $\frac {a+b}{7}=\frac {b+c}{8}=\frac {c+a}{9}= λ$
Then we can write,
$a+b = 7λ$ ....... (1)
$b+c = 8λ$ ....... (2)
$c+a = 9λ$ ....... (3)

On adding eq(1), (2) and (3), we get
$a+b+c = 12λ$
On solving,
$a = 4λ,\ b= 3λ$ and $c = 5λ$
$s = \frac {4λ+3λ+5λ}{2}$
$s = 6λ$

$Δ = \sqrt {s(s−a)(s−b)(s−c)}$
$Δ = \sqrt {(6λ)(2λ)(3λ)(λ)}$
$Δ= 6λ^2$

$R = \frac {abc}{4Δ}$

$R = \frac {(4λ)(3λ)(5λ)}{4(6λ^2)}$

$R = \frac 52 λ$

$r =\frac Δs$

$r= \frac {6λ^2}{6λ} = λ$
Now,
$\frac Rr = \frac {\frac {52}λ}{λ}$

$\frac Rr = \frac 52$

So, the correct option is (A): $\frac 52$