A square is inscribed in a circle with radius r. What is the ratio of the area of the square to the area of the circle?  

Expert Answers
mathsworkmusic eNotes educator| Certified Educator

The formula for the area of a circle is

`A_c = pi r^2`

if the radius of the circle is given by `r```

A square inscribed in such a circle has its four corners touching the inside of the circle (this is what 'inscribed' means). This implies that the distance from corner to corner on the square (the diagonal) is equal to `2r` .

To find the length of the sides of the square, consider that the square can be divided into two identical isosceles triangles (triangles with two sides equal in length) whose hypotenuse (long side) has length `2r` .

Using Pythagoras' equation (that the square of the hypotenuse of a triangle is equal to the sum of the squares of the other two sides), we then have that the length `L_s` of the sides of the square (which we are seeing as the length of the two equal sides of the isosceles triangles which the square is made up of) satisfies

`L_s^2 + L_s^2 = (2r)^2`  that is

`2L_s^2 = 4r^2`  `implies`

`L_s^2 = 2r^2`  `implies`

`L_s = sqrt(2)r`

The area `A_s` of the square is given simply by multiplying the lengths of its two (equal sides), that is

`A_s = L_s^2 = 2r^2`

Remembering that the area of the circle in which the square is inscribed is

`A_c = pir^2`

then we have that the ratio of the area of the square to the area of the circle is given by

`R_(sc) = A_s/A_c = (2r^2)/(pir^2) = 2/pi`    since the `r^2` term on each of the top and bottom of the ratio cancel.


mathscihelp eNotes educator| Certified Educator

What is quite beautiful about this solution is that the final ratio is independent of the size of the radius r. Regardless of how large or small our circle is, the ratio of the area of the inscribed square over the area of the circle, `R_sc` , will always be equal to `2/pi` !

I have provided a link to the "Math is Fun" website which shows similar very interesting mathematical relationships. `<br> `