A square is inscribed in a circle. Find the total area of the red regions of the circle below. What method for finding the area makes the most sense in this case? Why?

### Composite Shapes

A **composite shape** or a **composite figure** is a shape that is made up of two or more common shapes. Below is an example of a composite shape.

Usually when dealing with a composite shape you are interested in finding its area or perimeter. There are two general methods for finding the area of a composite shape.

**Method 1**: Find the individual areas of each piece of the composite shape. The area of the composite shape will be the sum of the individual areas. This method will be explored in Example A.

**Method 2**: Find the area of a shape larger than the composite shape and the areas of the pieces of the larger shape not included in the composite shape. The area of the composite shape will be the difference between the area of the larger shape and the areas of the pieces of the larger shape not included in the composite shape. This method will be explored in Example B.

Regardless of what method you use, you will often have to think carefully in order to find the dimensions necessary for determining the area or perimeter.

Let's take a look at an example problem using Method 1.

Find the area and perimeter of the composite shape below. To find the area, use Method 1.

The shape has been broken into 1 triangle and 2 rectangles (note that you could have broken it up differently!). First, find all of the missing side lengths.

For the right triangle, you know the hypotenuse is 5 inches because of the tick mark showing it is congruent to the other segment that is 5 inches. The height is 4 inches because the full height of the shape is 10 inches and the height above the triangle is given as 6 inches. Finally, you know the base is 3 inches because of the Pythagorean Theorem (or, because the full base is 15 inches and other portions are 7 inches and 5 inches).

Now you can find the perimeter by finding the sum of all the side lengths.

\begin{align*}P=5+7+6+5+10+12+3=48 \ in\end{align*}

Find the area of each of the three pieces and then the total area.

\begin{align*}A_{Triangle} &=\frac{bh}{2}=\frac{3 \cdot 4}{2}=6 \ in^2 \\
A_{Rectangle \# 1} &=bh=12 \cdot 4=48 \ in^2 \\
A_{Rectangle \# 2} &=bh=5 \cdot 6=30 \ in^2 \\
Total \ Area &=6+48+30=84 \ in^2\end{align*}

Let's take a look at finding the area using Method 2.

Find the area of the composite shape below. To find the area, use Method 2.

Find the area of the full rectangle, the unshaded triangle, and the unshaded rectangle. Subtract to find the area of the composite shape. Just like with Method 1, you will first need to find missing side lengths.

\begin{align*}A_{Large \ Rectangle} &=bh=15 \cdot 10=150 \ in^2 \\
A_{Unshaded \ Rectangle} &=bh=10 \cdot 6=60 \ in^2 \\
A_{Unshaded \ triangle} &=\frac{bh}{2}=\frac{3 \cdot 4}{2}=6 \ in^2 \\
Total \ Area &=150-60-6=84 \ in^2\end{align*}

Compare the two methods for finding the area of the shape given in the problems above. Was one method better than the other?

In each case, finding the area of the composite shape required 4 calculations after filling in the missing side lengths. For this particular shape, one method wasn't necessarily better than the other.

### Examples

#### Example 1

Earlier, you were asked what method for finding the area makes the most sense in this case.

It makes sense to use Method 2 to find the area of the red region. Finding the area of the whole circle and subtracting the area of the square is much simpler than trying to calculate the area of each of the four red pieces directly.

\begin{align*}A_{Circle} &=\pi r^2=\pi 3^2=9 \pi \ in^2 \\ A_{Square} &=\frac{d_1d_2}{2}=\frac{6 \cdot 6}{2}=\frac{36}{2}=18 \ in^2 \\ A_{Red \ Regions} &=9 \pi-18 \ in^2\end{align*}

The figure below is not drawn to scale. Assume all angles that look like right angles are right angles.

#### Example 2

Find the perimeter of the shape.

First find all of the missing lengths. Note that you will need to use the Pythagorean Theorem to find the side length of 10 cm (the hypotenuse of the right triangle).

Now find the perimeter.

\begin{align*}P=17+5+10+11+8+3+17+25=96 \ cm\end{align*}

#### Example 3

Choose a method for finding the area of the shape. Justify your method.

You could choose either method. Method 2 might be slightly easier because you will just need to find the area of the big rectangle and subtract the area of the trapezoid.

#### Example 4

Find the area of the shape using the method you chose in #2.

\begin{align*}Area&=Area_{Rectangle}-Area_{Trapezoid}\\&=(25 \cdot 17)-\left(\frac{(11+17)(8)}{2} \right)\\&=425-112\\&=313 \ cm^2\end{align*}

### Review

The figure below is not drawn to scale. Assume all angles that look like right angles are right angles. Point \begin{align*}O\end{align*} is the center of the partial circle.

1. Find all missing side lengths for the shape.

2. Find the perimeter of the shape.

3. Choose a method for finding the area of the shape. Justify your method.

4. Find the area of the shape using the method you chose in #2.

5. Try to find the area using the other method. If not possible, explain why not.

The figure below is not drawn to scale. Assume all angles that look like right angles are right angles.

6. Find all missing side lengths for the shape.

7. Find the perimeter of the shape.

8. Choose a method for finding the area of the shape. Justify your method.

9. Find the area of the shape using the method you chose in #8.

10. Try to find the area using the other method. If not possible, explain why not.

The figure below is not drawn to scale. Assume all angles that look like right angles are right angles.

11. Find all missing side lengths for the shape.

12. Find the perimeter of the shape.

13. Choose a method for finding the area of the shape. Justify your method.

14. Find the area of the shape using the method you chose in #13.

15. Try to find the area using the other method. If not possible, explain why not.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 1.9.