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:

- Let’s start off with the equation for the area of a circle:
`A = π × r`

(where^{2}`r`

is the radius) - Now let’s think about the equation for area of a square:
`A = s`

(where^{2}`s`

is the length of one of the sides) - Next we can say that the radius of the inner circle is
`r`

._{1} - After that let’s find the area of the red inner circle relative to
`r`

:_{1}`A`

_{1}= π × r_{1}^{2} - Now let’s find the length of the diagonal of the square (
`d`

) in which the red circle is inscribed relative to`r`

. This will also be the diameter of the outer circle_{1} `s = r`

_{1}+ r_{1}= 2 × r_{1}- The above is true because the red circle is inscribed in the square.
`d`

^{2}= s^{2}+ s^{2}`d`

^{2}= 2 × s^{2}`d = (2 × s`

^{2})^{1/2}`d = s × 2`

^{1/2}`d = 2 × r`

_{1}× 2^{1/2}- Next we should find the area of the black circle (
`A`

), in which the black square is inscribed, relative to_{2}`r`

._{1}- We will use
`r`

to represent the radius of the outer circle:_{2}`r`

_{2}= d / 2`r`

_{2}= (2 × r_{1}× 2^{1/2}) / 2`r`

_{2}= r_{1}× 2^{1/2}

`A`

_{2}= π × r_{2}^{2}`A`

_{2}= π × (r_{1}× 2^{1/2})^{2}`A`

_{2}= π × r_{1}^{2}× 2

- We will use
- Lastly we should find the area of the black outer circle while excluding the area covered by the red inner circle:
- Let’s make the area of the black doughnut shape be represented by
`A`

._{3} `A`

_{3}= A_{2}- A_{1}`A`

_{3}= (π × r_{1}^{2}× 2) - (π × r_{1}^{2})`A`

_{3}= (π × r_{1}^{2}) + (π × r_{1}^{2}) - (π × r_{1}^{2})`A`

_{3}= π × r_{1}^{2}

- Let’s make the area of the black doughnut shape be represented by

So after doing all of the math by using a little geometry and a little algebra we end up with the `A`

(red area) being equal to _{1}`A`

(black area)._{3}

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

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