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

#1**+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

#1**+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