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Radius of the Circle | AMC-8, 2005 | Problem 25

Try this beautiful problem from Geometry:Radius of the Circle from AMC-8(2005). You may use sequential hints to solve the problem

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

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

AMC-8 (2005) Problem 25

Pre College Mathematics

Try with Hints

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 ……….

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……..

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, $\pi r^2 – x=4-x$

$\Rightarrow \pi r^2=4$

$\Rightarrow r^2 = \frac{4}{\pi}$

$\Rightarrow r=\frac{2}{\sqrt \pi}$