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

To solve the given problem, we need to analyze the given condition for the side lengths of the triangle:

\(\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}\)

Let's denote the common ratio by \(k\):

a + b = 7k, \quad b + c = 8k, \quad c + a = 9k

Adding these three equations, we get:

(a + b) + (b + c) + (c + a) = 7k + 8k + 9k

2(a + b + c) = 24k \quad \Rightarrow \quad a + b + c = 12k \quad \text{(Equation 1)}

We will express each side in terms of \(k\):

• a = \frac{(7k + 9k - 8k)}{2} = 4k
• b = \frac{(7k + 8k - 9k)}{2} = 3k
• c = \frac{(8k + 9k - 7k)}{2} = 5k

Using the formulae for the radius of the incircle \(r\) and the radius of the circumcircle \(R\):

• The semi-perimeter \(s = \frac{a+b+c}{2} = \frac{12k}{2} = 6k\)
• Triangle area \( \Delta = \sqrt{s(s-a)(s-b)(s-c)} \)
• \( r = \frac{\Delta}{s} \)
• \( R = \frac{abc}{4\Delta} \)

Substituting the values of \(a\), \(b\), and \(c\), we get:

• s - a = 6k - 4k = 2k
• s - b = 6k - 3k = 3k
• s - c = 6k - 5k = k

Therefore, the area \(\Delta\) can be calculated as:

\Delta = \sqrt{6k \cdot 2k \cdot 3k \cdot k} = \sqrt{36k^4} = 6k^2

Now, calculate \(r\) and \(R\):

• r = \frac{\Delta}{s} = \frac{6k^2}{6k} = k
• R = \frac{abc}{4\Delta} = \frac{(4k)(3k)(5k)}{4 \cdot 6k^2} = \frac{60k^3}{24k^2} = \frac{5k}{2}

Finally, the value of \(\frac{R}{r}\) is:

\frac{R}{r} = \frac{\frac{5k}{2}}{k} = \frac{5}{2}

The correct answer is: \(\frac{5}{2}\)