Question

An equilateral triangle is inscribed in a circle of radius `6x`. What is the area `A` within the circle but outside the triangle as a function of the length `9x` of the side of the triangle?

Answer 1

Let `y` be a length of a side of a triangle.

If `y=9x`. `S = 36pix^2-(81sqrt(3)x^2)/4~=78.0233x^2`
If `y=6sqrt(3)x` (for inscribed `Delta`), `S = 36pix^2-(36*3*sqrt(3)x^2)/4~=66.3319x^2`

Explanation:

First of all, if a radius of a circle is `r=6x`, the length of a side of an equilateral triangle inscribed into this circle is not `9x`, but `6sqrt(3)x`.

If, however, the length of a side of a triangle is `9x` (which is smaller than `6sqrt(3)x`, the triangle is positioned completely inside the circle and cannot be called inscribed.

To address both cases, let's assume that the length of a side is `y<=6sqrt(3)x` and solve the problem in terms of `y`. Then we can substitute either `y=6sqrt(3)x` or `y=9x` to get both solutions.

The area of an entire circle of a radius `r` is
`S_0=pir^2=36pix^2`

The altitude of an equilateral triangle with a side `y` is
`h=(ysqrt(3))/2`

So, the area of a triangle is
`S_(Delta)=1/2y(ysqrt(3))/2=(y^2sqrt(3))/4`

Therefore, the area inside a circle but outside of a triangle is
`S_0-S_(Delta) = 36pix^2-(y^2sqrt(3))/4`

If `y=9x`, this area is
`S_0-S_(Delta) = 36pix^2-(81sqrt(3)x^2)/4~=78.0233x^2`

If `y=6sqrt(3)x`, this area is
`S_0-S_(Delta) = 36pix^2-(36*3*sqrt(3)x^2)/4~=66.3319x^2`

Choose any answer.