POW Answer – Circle, Square, Circle

Last week’s problem involved using your geometry/trigonometry skills to find areas of shapes. We start off with a large (black) circle which has a (black) square inscribed in it which in turn has a (red) circle inscribed in it:

1. Let’s start off with the equation for the area of a circle: A = π × r2 (where r is the radius)
2. Now let’s think about the equation for area of a square: A = s2 (where s is the length of one of the sides)
3. Next we can say that the radius of the inner circle is r1.
4. After that let’s find the area of the red inner circle relative to r1: A1 = π × r12
5. Now let’s find the length of the diagonal of the square (d) in which the red circle is inscribed relative to r1. This will also be the diameter of the outer circle
1. s = r1 + r1 = 2 × r1
2. The above is true because the red circle is inscribed in the square.
3. d2 = s2 + s2
4. d2 = 2 × s2
5. d = (2 × s2)1/2
6. d = s × 21/2
7. d = 2 × r1 × 21/2
6. Next we should find the area of the black circle (A2), in which the black square is inscribed, relative to r1.
   1. We will use r2 to represent the radius of the outer circle:
      1. r2 = d / 2
      2. r2 = (2 × r1 × 21/2) / 2
      3. r2 = r1 × 21/2
   2. A2 = π × r22
   3. A2 = π × (r1 × 21/2)2
   4. A2 = π × r12 × 2
7. Lastly we should find the area of the black outer circle while excluding the area covered by the red inner circle:
   1. Let’s make the area of the black doughnut shape be represented by A3.
   2. A3 = A2 - A1
   3. A3 = (π × r12 × 2) - (π × r12)
   4. A3 = (π × r12) + (π × r12) - (π × r12)
   5. A3 = π × r12

So after doing all of the math by using a little geometry and a little algebra we end up with the A1 (red area) being equal to A3 (black area).

Even though the image makes it look like there is more red than black, there really isn’t. Interesting stuff, huh? 😎