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# What is the ratio of the area of the smaller circle to the area of the larger square?

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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

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+7266
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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
#1
+7266
+2

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