Shaded region problems

A common geometry question type involves solving for the area of a shaded region. Typically the question will have multiple shapes layered on top of each other, and your task is to calculate the area of the various shapes and add/subtract them in creative ways to figure out just the area of the part that is shaded.

Here’s a simple example:

The diameter of the circle below is exactly half the side length of the square. What is the area of the shaded region if the perimeter of the square is 16?

Before jumping into solving these questions, you should take a moment to think about your approach. In this example, we have a circle laid on top of a square. We can calculate the area of the shaded region by taking the area of the square, and subtracting the area of the circle.

We’re only given one piece of information: the perimeter of the square. We can use this to find the length of a side of the square.

$$\begin{array}{rl}{P}_{\square }& =4s\\ 16& =4s\\ 4& =s\end{array}$$

The length of each side of the square is $4$. We can use this to solve for the area of the square as $A_{\square }=s^2$.

$$\begin{array}{rl}{A}_{\square }& =s^2\\ & =4^2\\ & =16\end{array}$$

The formula for the area of a circle is $A_{\circ }=\pi r^2$. We’re told that the diameter of the circle is exactly half the side length of the square. This means the diameter is $4/2=2$, and since the radius of a circle is half the diameter, the radius is $2/2=1$. We have the radius, so let’s solve for the area:

$$\begin{array}{rl}{A}_{\circ }& =\pi r^2\\ & =\pi (1^2)\\ & =\pi (1)\\ & =\pi \end{array}$$

The area of the shaded region is the square’s area minus the circle’s area. We have all the information needed:

$$\begin{array}{rl}{A}_{\text{shaded}}& =A_{\square }-A_{\circ }\\ & =16-\pi \end{array}$$

This was a simple example, and most GRE questions will be far more complicated. There might be multiple shaded regions that need to be calculated, and there might be multiple clues to help you solve for the area of the various shapes.

You’ll see the words inscribed and circumscribed often in these questions.

Inscribed and circumscribed are essentially opposites of each other.

The figure below is a circle inscribed within a square. It could also be described as a square circumscribed around a circle.

The figure below is a circle circumscribed around a square. It could also be described as a square inscribed within a circle.

Ready to try a more complicated shaded region question?

The drawing below is a circle placed in front of an isosceles right triangle and a square inscribed in the circle. The height of the circle and the triangle are identical, and they are aligned to the left and bottom. What is the area of the shaded region if the circumference of the circle is $12\pi$?

Think about the approach first, then try to solve for the area.

The first step is to determine the approach. The area of the square will be easy to calculate once we find the length of a side. But how do we find the area of the shaded regions the triangle creates? It might not be clear at first, but since the hypotenuse splits the circle exactly in the middle, we can find the shaded region area by taking the area of the triangle and subtracting half the area of the circle.

Now that we have an approach in mind, let’s start solving for the areas of the different pieces. The question tells us that the circumference of the circle is $12\pi$, so let’s start there.

$$\begin{array}{rl}{C}_{\circ }& =2\pi r\\ 12\pi & =2\pi r\\ 6& =r\end{array}$$

Now that we have the circle’s radius, we can calculate the area.

$$\begin{array}{rl}{A}_{\circ }& =\pi r^2\\ & =\pi (6^2)\\ & =\pi (36)\\ & =36\pi \end{array}$$

Let’s work on the area of the triangle next. We’re told the diameter of the circle ($d=2r=2(6)=12$) is the same as the height of the triangle. Since this is an isosceles right triangle, the base will be the same as the height. The area of any triangle is $A=bh/2$.

$$\begin{array}{rl}{A}_{△}& =bh/2\\ & =12(12)/2\\ & =144/2\\ & =72\end{array}$$

As we mentioned earlier, we can find the shaded area of the triangle by subtracting half the area of the circle:

$$\begin{array}{rl}{A}_{\text{shaded}△}& =A_{△}-A_{\circ }/2\\ & =72-36\pi /2\\ & =72-18\pi \end{array}$$

Now let’s work on the area of the square inscribed in the circle. The diagonal of a square is the hypotenuse of a 45-45-90 triangle, so we can use the $x:x:x\sqrt{2}$ ratio. Since we’ve given the hypotenuse, we can manipulate the ratio by dividing each element by $\sqrt{2}$ to make it a little easier to work with, since then the $x$ variable will match up to the length of the hypotenuse. That hypotenuse is the same length as the diameter of the circle, which we’ve already figured out is $12$.

$$\begin{array}{c}x:x:x\sqrt{2}\\ x/\sqrt{2}:x/\sqrt{2}:x\\ 12/\sqrt{2}:12/\sqrt{2}:12\end{array}$$

So the length of each side of the square is $12/\sqrt{2}$. Let’s get the area of the square:

$$\begin{array}{rl}{A}_{\square }& =s^2\\ & =(12/\sqrt{2}{)}^2\\ & =12^2/{\sqrt{2}}^2\\ & =144/2\\ & =72\end{array}$$

Incidentally, there’s a shortcut for that part. The area of a square is typically calculated as $A=s^2$, but it can also be calculated using the rhombus area formula $A=d^2/2$, where $d$ is the diagonal size of the square. Since we already knew the diagonal was $12$, we could have done $A=12^2/2=12\ast 12/2=12\ast 6=72$.

We finally have all the information needed to find the total shaded area.

$$\begin{array}{rl}{A}_{\text{shaded}}& =A_{\text{shaded}△}+A_{\square }\\ & =(72-18\pi )+72\\ & =144-18\pi \end{array}$$

Quite a lot of steps, but very approachable if you go one step at a time!