A geometry problem by Naman Sehgal

Let the side of the square be x then its diagonal will be root 2 * x which will be the diameter of the circumcircle and similarly the the diameter of the incircle will be x and hence ratio will be

pie * (root 2 x) (root 2* x)/ pie (x) (x)

which will give u 2

Let S be the side of the square. Then the radius of small circle is $r=\frac{S}{2}$ , and the radius of big circle is $R=\sqrt{\left(\frac{S}{2}\right)^2+\left(\frac{S}{2}\right)^2} = S\sqrt{\frac{1}{2}}$ . Then the ratio of their area is: $\frac{Area_{big}}{Area_{small}}=\frac{\pi R^2}{\pi r^2} = \left(\frac{R}{r}\right)^2 = \left(\frac{S\sqrt{\frac{1}{2}}}{\left(\frac{S}{2}\right)}\right)^2 = \left(2\sqrt{\frac{1}{2}}\right)^2=4\left(\frac{1}{2}\right)=2$

The bigger circle is directer circle to the smatller one that's it has radical2 times radius so area is double

Let $1$ be the diameter of the bigger circle, that is also the diagonal of the square which is a hypotenuse of a right triangle. By pythagorean theorem, the side length of the square is $\dfrac{1}{\sqrt{2}}$ which is also the diameter of the smaller circle. By similar polygons, the ratio of the area of the bigger circle to the area of the smaller circle is

$\dfrac{1^2}{\left(\dfrac{1}{\sqrt{2}}\right)^2}=\dfrac{1}{\dfrac{1}{2}}=$ $\color{#D61F06}\large \boxed{2}$

this is very easy any math reader can solve it ..

The Diameter of the: big Circle is s_/2

Small Circle is s

where in, s is the side of the Square

The ratio of the area will be:

[(s_/2)/2]^2 pi : (s^2) pi

2:1

Giving us, 2...

Larger circle area = pi r^2. Radius of smaller circle is r/sqrt(2) hence its area is (pi r^2)/2. Area of bigger circle/area of smaller circle =2.