Try this beautiful problem from Geometry: Radius of a circle

## Radius of a circle – AMC-8, 2005- Problem 25

A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

- $\frac{5}{\sqrt \pi}$
- $ \frac{2}{\sqrt \pi} $
- $\sqrt \pi$

**Key Concepts**

Geometry

Cube

square

## Check the Answer

But try the problem first…

Answer: $ \frac{2}{\sqrt \pi} $

AMC-8 (2005) Problem 25

Pre College Mathematics

## Try with Hints

First hint

The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Can you now finish the problem ……….

Second Hint

Region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square

The area of the circle -x=Area of the square – x

can you finish the problem……..

Final Step

Given that The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Let the region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square .

Let r be the radius of the circle

Therefore, The area of the circle -x=Area of the square – x

so, \(\pi r^2 – x=4-x\)

\(\Rightarrow \pi r^2=4\)

\(\Rightarrow r^2 = \frac{4}{\pi}\)

\(\Rightarrow r=\frac{2}{\sqrt \pi}\)

## Other useful links

- https://www.cheenta.com/largest-and-smallest-numbers-amc-8-2006-problem-22/
- https://www.youtube.com/watch?v=65RRPvbATsk