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A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

bbelt711  Aug 17, 2017

Best Answer 

 #1
avatar+4480 
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Let's call the side length of the larger square  " s "  .

 

Then...

 

diameter of the larger circle  =  s  

 

Let's draw a diameter of the larger circle from the corner of the smaller square to the opposite corner of the smaller square. This creates a 45 - 45 - 90 triangle, where the side across from the 90° angle is  "s"  . So....

 

side across from the  45°  angle  =   \(\frac{s}{\sqrt2}\)

 

diameter of the smaller circle       =  \(\frac{s}{\sqrt2}\)

 

radius of the smaller circle            =  \(\frac12\,*\,\text{diameter}=\frac12\,*\,\frac{s}{\sqrt2}=\frac{s}{2\sqrt2}\)

 

\(\frac{\text{area of smaller circle}}{\text{area of larger square}}=\frac{\pi(\frac{s}{2\sqrt2})^2}{s^2}=\frac{\pi(\frac{s^2}{4*2})}{s^2}=\frac{\pi s^2}{8}\,*\,\frac1{s^2}=\frac{\pi}{8}\)

hectictar  Aug 17, 2017
edited by hectictar  Aug 17, 2017
edited by hectictar  Aug 17, 2017
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1+0 Answers

 #1
avatar+4480 
+2
Best Answer

Let's call the side length of the larger square  " s "  .

 

Then...

 

diameter of the larger circle  =  s  

 

Let's draw a diameter of the larger circle from the corner of the smaller square to the opposite corner of the smaller square. This creates a 45 - 45 - 90 triangle, where the side across from the 90° angle is  "s"  . So....

 

side across from the  45°  angle  =   \(\frac{s}{\sqrt2}\)

 

diameter of the smaller circle       =  \(\frac{s}{\sqrt2}\)

 

radius of the smaller circle            =  \(\frac12\,*\,\text{diameter}=\frac12\,*\,\frac{s}{\sqrt2}=\frac{s}{2\sqrt2}\)

 

\(\frac{\text{area of smaller circle}}{\text{area of larger square}}=\frac{\pi(\frac{s}{2\sqrt2})^2}{s^2}=\frac{\pi(\frac{s^2}{4*2})}{s^2}=\frac{\pi s^2}{8}\,*\,\frac1{s^2}=\frac{\pi}{8}\)

hectictar  Aug 17, 2017
edited by hectictar  Aug 17, 2017
edited by hectictar  Aug 17, 2017

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