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

The Correct Option is A

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): \(\frac 52\)