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AMC 8 Geometry Math Olympiad USA Math Olympiad

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?

radius of the circle
  • $\frac{5}{\sqrt \pi}$
  • $ \frac{2}{\sqrt \pi} $
  • $\sqrt \pi$

Key Concepts


Geometry

Cube

square

Check the Answer


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 ……….

Shaded region of the figure

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……..

Shaded region of the figure

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, \(\pi r^2 – x=4-x\)

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

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

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

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