Shaded region problems

A common geometry question type asks you to find the area of a shaded region. Usually, the diagram has multiple shapes layered on top of each other. Your job is to find the areas of the relevant shapes and then add or subtract them to isolate just the shaded part.

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 you start calculating, decide on an approach. Here, a circle sits inside a square, and the shaded region is the part of the square outside the circle. So you can find the shaded area by taking the area of the square and subtracting the area of the circle.

You’re given only one piece of information: the perimeter of the square. Use that to find the side length.

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

Each side of the square is $4$, so the area of the square is $A_{\square }=s^2$.

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

The area of a circle is $A_{\circ }=\pi r^2$. You’re told the diameter of the circle is half the side length of the square. Since the side length is $4$, the diameter is $4/2=2$. The radius is half the diameter, so $r=2/2=1$.

Now compute the circle’s area:

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

The shaded area is the square’s area minus the circle’s area:

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

This was a simple example. On the GRE, shaded-region questions are often more complex: you might need to find multiple shaded areas, and you might need to combine several clues to determine missing lengths.

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

Inscribed and circumscribed are essentially opposites.

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.

Start by choosing a strategy. The square’s area will be straightforward once you know its side length. The triangle’s shaded portion is less obvious, but notice that the hypotenuse splits the circle into two equal halves. That means you can find the triangle’s shaded area by taking the area of the triangle and subtracting half the area of the circle.

Now compute each piece.

The circumference of the circle is $12\pi$, so solve for the radius.

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

With $r=6$, the circle’s area is:

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

Next, find the area of the triangle. The diameter of the circle is $d=2r=2(6)=12$, and you’re told this equals the height of the triangle. Because the triangle is an isosceles right triangle, its base equals its height, so $b=12$ and $h=12$. Use $A=bh/2$.

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

Now subtract half the circle’s area to get the shaded part created by the triangle:

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

Now find the area of the square inscribed in the circle. The diagonal of the square is the diameter of the circle, so the diagonal is $12$. A square’s diagonal is the hypotenuse of a 45-45-90 triangle, which has side ratio $x:x:x\sqrt{2}$. If the hypotenuse is $12$, then each leg (the side length of the square) is $12/\sqrt{2}$.

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

So the side length is $12/\sqrt{2}$, and the square’s area is:

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

Incidentally, there’s a shortcut here. The area of a square is usually written as $A=s^2$, but you can also use the rhombus area formula $A=d^2/2$, where $d$ is the diagonal of the square. Since the diagonal is $12$, you could compute $A=12^2/2=12\ast 12/2=12\ast 6=72$.

Finally, add the two shaded parts:

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

There are several steps, but the process stays manageable if you (1) choose a clear add/subtract plan and (2) compute one shape at a time.