Area of Composite Figures

A composite figure (or composite shape) is a figure made up of two or more simple geometric shapes, like rectangles, triangles, or circles. To find the total area of a composite figure, you don't need a special formula. Instead, you break the shape down into simpler pieces that you already know how to measure.

The 3-Step Method

1. Decompose the Figure: Split the composite shape into simpler, non-overlapping figures (like squares, rectangles, triangles, or semicircles).
2. Calculate Each Area: Use standard area formulas to find the area of each individual piece.
3. Combine: Add the areas together to get the total area. (If there is a "hole" cut out of the shape, you subtract the area of the hole instead).

Quick Formula Review

Before we solve examples, let's review the area formulas for basic shapes:

• Rectangle: $A = l \times w$ (length $\times$ width)
• Triangle: $A = \frac{1}{2} b h$ (base $\times$ height)
• Circle: $A = \pi r^2$ (radius squared $\times \pi$)
• Semicircle: $A = \frac{1}{2} \pi r^2$ (half of a circle)

Example 1: An L-Shaped Figure

Imagine an L-shaped figure. You can easily split an L-shape into two separate rectangles. Let's say after drawing a line to split the shape, we have:

• Rectangle 1 (vertical): $5\text{ cm}$ by $2\text{ cm}$
• Rectangle 2 (horizontal): $6\text{ cm}$ by $3\text{ cm}$

Step 1: Find the area of Rectangle 1 $A_1 = 5 \times 2 = 10\text{ cm}^2$

Step 2: Find the area of Rectangle 2 $A_2 = 6 \times 3 = 18\text{ cm}^2$

Step 3: Add them together $\text{Total Area} = 10 + 18 = 28\text{ cm}^2$

Example 2: Rectangle and a Semicircle

Suppose you have a shape that looks like a rectangular window with a rounded top (a semicircle). The rectangle has a width of $4\text{ m}$ and a height of $6\text{ m}$. The semicircle sits exactly on top of the $4\text{ m}$ width.

Step 1: Find the area of the rectangle $A_{\text{rect}} = 4 \times 6 = 24\text{ m}^2$

Step 2: Find the area of the semicircle The diameter of the semicircle is the width of the rectangle ($4\text{ m}$), which means the radius $r$ is exactly half of that ($2\text{ m}$). $A_{\text{semi}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2^2) = \frac{1}{2} \pi (4) = 2\pi \approx 6.28\text{ m}^2$

Step 3: Add the areas $\text{Total Area} = 24 + 6.28 = 30.28\text{ m}^2$

By breaking complex shapes into familiar pieces, finding the area becomes a simple puzzle of adding and subtracting!