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^{2}
(wherer
is the radius) - Now let’s think about the equation for area of a square:
A = s^{2}
(wheres
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 tor_{1}
. This will also be the diameter of the outer circle 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_{2}
), in which the black square is inscribed, relative tor_{1}
.- We will use
r_{2}
to represent the radius of the outer circle: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_{1}
(red area) being equal to A_{3}
(black area).
Even though the image makes it look like there is more red than black, there really isn’t. Interesting stuff, huh?