Question

If r₁ = 2, r₂ = 3 and r₃ = 6 and r = 1, prove that a = 3, b = 4 and c = 5.

Answer 1

We take

a = 3, b = 4 and c = 5

We know that

s = (a+b+c)/2 = (3+4+5)/2 = 12/2 = 6

s-a = 6 - 3 = 3

s-b = 6 - 4 = 2

s-c = 6- 5 = 1

Now

$\Delta=\sqrt{s(s-a)(s-b)(s-c)}\\\\\Delta=\sqrt{6(3)(2)(1)}\\\\\Delta=\sqrt{36}\\\\\Delta=6$

Now

$r_{1} = \frac{\Delta}{s-a}=\frac{6}{3}=2\\  \\r_{2} = \frac{\Delta}{s-b}=\frac{6}{2}=3\\\\r_{3} = \frac{\Delta}{s-c}=\frac{6}{1}=6\\\\r= \frac{\Delta}{s}=\frac{6}{6}=1$

Which is required.

Clearly for r₁ = 2, r₂ = 3 and r₃ = 6 and r = 1,

there must be a = 3, b = 4 and c = 5