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?
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 = \(\frac12\,*\,\text{diameter}=\frac12\,*\,\frac{s}{\sqrt2}=\frac{s}{2\sqrt2}\)
\(\frac{\text{area of smaller circle}}{\text{area of larger square}}=\frac{\pi(\frac{s}{2\sqrt2})^2}{s^2}=\frac{\pi(\frac{s^2}{4*2})}{s^2}=\frac{\pi s^2}{8}\,*\,\frac1{s^2}=\frac{\pi}{8}\)
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 = \(\frac12\,*\,\text{diameter}=\frac12\,*\,\frac{s}{\sqrt2}=\frac{s}{2\sqrt2}\)
\(\frac{\text{area of smaller circle}}{\text{area of larger square}}=\frac{\pi(\frac{s}{2\sqrt2})^2}{s^2}=\frac{\pi(\frac{s^2}{4*2})}{s^2}=\frac{\pi s^2}{8}\,*\,\frac1{s^2}=\frac{\pi}{8}\)