Question

ΔABC is an isosceles right angled triangle. If r is its in radius and R its circum radius, then what is \(\frac{R}{r}\) equal to ?

Answer 1

Let ABC be the right ∠d isosceles triangle, where AB = BC = x and ∠ABC = 90º.

∴ AC = \(\sqrt{AB^2+BC^2}\) = \(\sqrt{x^2+x^2}\) = \(x\sqrt2\)

s = Semi-perimeter of ΔABC = \(\frac{x+x+x\sqrt2}{2}\) = \(\frac{2x+x\sqrt2}{2}\) = \(\frac{x\sqrt2(\sqrt2+1)}{2}\)

A = Area of ΔABC = \(\frac{1}{2}\) x \(x\) x \(x\) x = \(\frac{x^2}{2}\)

∴ In-radius of ΔABC (r) = \(\frac{\text{Area}}{\text{Semi-perimeter}}\) = \(\frac{\frac{x^2}{2}}{\frac{x\sqrt2(1+\sqrt2)}{2}}\) = \(\frac{x}{\sqrt2(1+\sqrt2)}\)

Circum radius of ΔABC (R) = \(\frac{abc}{4\times\text{Area}}\) = \(\frac{x\times{x}\times{x}\sqrt2}{4\times\frac{x^2}{2}}\) = \(\frac{x\sqrt2}{2}\) = \(\frac{x}{\sqrt2}\)

∴ Required ratio = \(\frac{R}{r}\) = \(\frac{\frac{x}{\sqrt2}}{\frac{x}{\sqrt2(1+\sqrt2)}}\) = 1 + √2.