Question

In a right-angled isosceles triangle, the ratio of the circumradius and inradius is
A. $$2\left( \sqrt{2} +1\right) \colon 1$$
B. $$\left( \sqrt{2} +1\right) \colon 1$$
C. $$2 \colon 1$$
D. $$\sqrt{2} \colon 1$$

Answer 1

Hint: In this question it is given that we have the ratio of the circumradius and inradius in a right-angled isosceles triangle. So first of all we have to to the diagram to understand it in better way,

So to find the ratio we need to know that,
The circumradius, $$R=\dfrac{abc}{4A }$$......(1)
Where a,b,c are the sides of a triangle and $\triangle$ is the area of the triangle.
And the inradius, $$r=\dfrac{A }{s}$$...........(2)
Where ‘s’ is the semi-perimeter and A is the area.

Complete step-by-step answer:
Here the given triangle is a right-angled isosceles triangle, so let us consider that $$\angle B=90^{\circ}$$ and AB = BC = a.
Therefore, by pythagoras theorem, we can write,
$$\text{hypotenuse} =\sqrt{\left( \text{base} \right)^{2} +\left( \text{perpendicular} \right)^{2} }$$
$$\Rightarrow AC=\sqrt{\left( AB\right)^{2} +\left( BC\right)^{2} }$$
$$\Rightarrow AC=\sqrt{a^{2}+a^{2}}$$
$$\Rightarrow AC=\sqrt{2a^{2}}$$
$$\Rightarrow AC=\sqrt{2} a$$

Now as we know that area of an triangle, $$A=\dfrac{1}{2} \times \text{base} \times \text{height}$$
Therefore, area of $$\triangle ABC$$,
$$A=\dfrac{1}{2} \times BC\times AB$$
  $$=\dfrac{1}{2} \times a\times a$$=$$\dfrac{a^{2}}{2}$$
The semi-perimeter,
$$s=\dfrac{AB+BC+CA}{2}$$
  $$=\dfrac{a+a+\sqrt{2} a}{2}$$
  $$=\dfrac{2a+\sqrt{2} a}{2}$$
Therefore the circumradius,
$$R=\dfrac{abc}{4A }$$
  $$=\dfrac{a\cdot a\cdot \sqrt{2} a}{4\left( \dfrac{a^{2}}{2} \right) }$$
  $$\dfrac{\sqrt{2} a^{3}}{2a^{2}}$$
  $$=\dfrac{a}{\sqrt{2} }$$

And the inradius,
$$r=\dfrac{A }{s}$$
  $$=\dfrac{\left( \dfrac{a^{2}}{2} \right) }{\left( \dfrac{2a+\sqrt{2} a}{2} \right) }$$
  $$=\dfrac{a^{2}}{\left( 2+\sqrt{2} \right) a}$$
  $$=\dfrac{a}{\sqrt{2} \left( \sqrt{2} +1\right) }$$
Now the ratio of circumradius and inradius,
$$\Rightarrow \dfrac{R}{r} =\dfrac{a}{\sqrt{2} }\times \dfrac{\sqrt{2} \left( \sqrt{2} +1\right) }{a}$$
$$\Rightarrow \dfrac{R}{r} =\left( \sqrt{2} +1\right) $$
$$\Rightarrow \dfrac{R}{r} =\dfrac{\left( \sqrt{2} +1\right) }{1}$$
Therefore, $$R\colon r=\left( \sqrt{2} +1\right) \colon 1$$
Hence the correct option is option B.

Note: To solve this type of question you need to know that the pythagorean theorem only applied for a right angle triangle, and the expression we have already mentioned in the solution part. Also when you draw a circumscribed circle in a right-angled triangle then the centre of the circle always lies on the midpoint hypotenuse.