How do you find the area of the gray region in the problem.
Pretty much the isosceles triangle is 2" tall and 2" wide at the bottom.
The circle has a radius of 1"
The square is 2" tall and 2" wide.
My question is: How do you find the area of the shaded (gray) region of the square not getting overlapped by the circle or triangle.
The area of the square surrounding the triangle and the circle is 4. The area of the triangle is $2*2*\frac{\sqrt{3}}{4}$=$\sqrt{3}$. The area of the circle is $\pi$ because it has a diameter 2 and radius 1. The common region of the circle and the triangle is a sector of $60^\circ$ and the area is $\frac{60}{360}*\pi$ which equals to $\frac{\pi}{6}$. The area of the region not in the gray is $\sqrt{3}+\pi-\frac{\pi}{6}$. This needs to be subtracted from the area of the square, which is 4, to calculate the gray region, getting an answer of $\boxed{4-\sqrt{3}-\frac{5\pi}{6}}$. Hope this helps!
Let $S$ be the square $C$ the circle and $T$ the triangle. What you look for is: $$ |S \setminus (C\cup T)| = |S| - |C\cup T| = |S| - (|C| + |T| - |C\cap T|). $$
Let's consider $C\cap T$. $T$ is an isosceles triangle, and the angle on the top vertex is $$ 2\alpha = \arctan \frac 1 2. $$ The region $C\cap T$ is the union of a circle sector of amplitude $4\alpha$ with two isosceles triangles with angle $\alpha$ at the bases and equal sides of length $1$. The circle sector has area $2\alpha$ and the triangles together have area $2\cos\alpha \sin\alpha$. So $$ |C \cap T| = 2\alpha + 2\sin\alpha\cos \alpha = 2\alpha +\frac{2\tan \alpha}{1+\tan^2 \alpha} = 2\arctan\frac 1 2 + \frac 2 5. $$ and the grey area has measure $$ 4 - (\pi + 2 - (2\arctan \frac 1 2 + \frac 2 5))= \frac{12}{5} - \pi + 2\arctan \frac 1 2. $$
Since you asked how rather than compute, I'll just describe the method:
