# How to find area of remaining square with shapes inside

Pretty much there is a square with a height and width of $2$", inside there is a perfect circle with radius $1$". Also overlapping the circle and square is a isosceles triangle also with height $2$" and width of the bottom to be $2$". (Truly to see what I'm talking about, you have to click on the image). What is the area of the shaded (gray) region inside of the square not overlapped by the circle or the triangle?

• Please put the picture in the question rather than a link to it. Commented Nov 17, 2015 at 3:06
• The same question is here Commented Nov 17, 2015 at 21:45

Hint: You know the area of the square and the area of the circle, so the question is the area of the funny shapes in the lower left and right corners. In problems like these, you should look for ways to cut up the shapes into something you know. I added two lines and got this

Now your funny shapes are triangles less the bulge of a circle around the secant. That bulge is the difference between a sector of a circle and a triangle. You need to find the coordinates of the intersection of the circle and the downward lines from the top center of the square. Then find the angle the secant from there to the bottom center represents.

It is basically $A(square)-A(triangle)-A(circle)+A(triangle\cap circle)$.

The first three terms are easy so I will do the last term:

First the top angle of the triangle is $\theta=2tan^{-1}({1\over2})$.

We draw to lines from the center of the circle to the intersection points of the triangle and the circle. We also draw a line between the center of the circle to the top point of the triangle.

Now we have two isoceles obtuse triangles on left and right side and an "arc-shaped triangle" in the bottom.

The middle angle for the obtuse triangle is $180^\circ-2({1\over2})(\theta)=180^\circ-\theta$. Hence each obtuse triangle has area ${1\over2}(1\cdot1\cdot \sin(180^\circ-\theta))$

Since we know $\tan({\theta\over2})={1\over2}$, we can get $\sin(\theta)=2\sin({1\over2}\theta)\cos({1\over2}\theta)=2({1\over\sqrt{5}})({2\over\sqrt{5}})={4\over5}$

Hence the obtuse triangles each have area ${1\over2}(1\cdot1\cdot {4\over5})={2\over5}$.

Now we have the bottom "arc-shaped" triangle. The arc length is $2\theta\cdot1=4tan^{-1}({1\over2})$ and hence the shape's area is ${1\over2}(4tan^{-1}({1\over2}))(1)=2tan^{-1}({1\over2})$.

Hence the total area of the last term is ${4\over5}+2tan^{-1}({1\over2})$.