Math Core

Real floor plans, windows, gardens and logos are rarely a single rectangle or circle. A composite figure is built from simpler shapes, and its area comes from combining the areas of those shapes. The skill is in choosing the pieces.

Two strategies: add or subtract

Because area is additive, you can find the area of a composite figure in two ways:

Add: split the figure into non-overlapping pieces and add their areas.
Subtract: surround the figure with a larger simple shape, then subtract the pieces that aren't part of the figure.

Both give the same answer. Often one of them is much shorter, so it's worth a moment to look for the easiest decomposition before you start computing.

Area of a composite figure

A_{\text{figure}} = A_1 + A_2 + \cdots \qquad \text{or} \qquad A_{\text{figure}} = A_{\text{outer}} - A_{\text{removed}}

The pieces you add must not overlap, and together they must cover the whole figure exactly once.

An L-shaped floor. The dashed segment splits it into two rectangles.

Worked example: An L-shaped floor, two ways

Find the area of the L-shaped floor above.

Adding. The dashed segment splits the floor into a left rectangle that is 4 m wide and 8 m tall, and a right rectangle that is $10 - 4 = 6$ m wide and 4 m tall:

A = 4 \cdot 8 + 6 \cdot 4 = 32 + 24 = 56 \text{ m}^2

Subtracting. The floor fits in a 10 m by 8 m rectangle. The missing upper-right corner is $6$ m wide and $8 - 4 = 4$ m tall:

A = 10 \cdot 8 - 6 \cdot 4 = 80 - 24 = 56 \text{ m}^2

Notice that some side lengths weren't labeled. Finding them is part of the problem: opposite sides of each rectangle are equal, so a missing horizontal length is a difference of two labeled horizontal lengths, and the same goes for vertical ones.

Figures with curves

Circles and parts of circles show up constantly in composite figures. Keep these on hand:

Piece	Area
Circle of radius $r$	$\pi r^2$
Semicircle	$\frac{1}{2}\pi r^2$
Quarter circle	$\frac{1}{4}\pi r^2$
Sector with central angle $\theta^\circ$	$\frac{\theta}{360}\pi r^2$

Unless a problem asks for a decimal, leave answers in terms of $\pi$ . An exact answer like $20 + 2\pi$ is more precise than any rounded decimal.

A window made of a rectangle topped by a semicircle.

Worked example: A window

The window above is a 4 ft by 5 ft rectangle topped by a semicircle. Find its area, exactly and to the nearest tenth.

The semicircle's diameter is the top of the rectangle, 4 ft, so its radius is 2 ft.

\begin{aligned} A &= (4)(5) + \tfrac{1}{2}\pi(2)^2 \\ &= 20 + 2\pi \\ &\approx 26.3 \text{ ft}^2 \end{aligned}

Common mistake

The dimension shown on a circle is often the diameter, not the radius. In the window, $\tfrac{1}{2}\pi(4)^2 = 8\pi$ would be four times too large. Always halve a diameter before squaring.

Shaded regions

A shaded region problem shows one shape with another removed from it. These almost always call for subtraction.

A circle inscribed in a square with 10 cm sides. The region to find is the part of the square outside the circle.

Worked example: Outside the circle, inside the square

A circle is inscribed in a square with 10 cm sides. Find the area of the region inside the square but outside the circle.

The circle touches all four sides, so its diameter equals the side of the square, 10 cm. The radius is 5 cm.

A = 10^2 - \pi(5)^2 = 100 - 25\pi \approx 21.5 \text{ cm}^2

Tip

Estimate before you trust a calculation. The circle fills most of the square, so the leftover area should be a small fraction of 100. An answer near 21 passes the check. An answer like 78.5 means you found the circle instead of the region outside it.

Putting it to work

Composite area problems often end with a real decision: how much paint, sod or tile to buy. Materials come in whole units, so round up at the end, even when the decimal is small.

A field shaped like a rectangle with a semicircle on each end.

Worked example: Seeding a field

The field above is a 40 m by 20 m rectangle with a semicircle on each short end. One bag of grass seed covers 150 square meters. How many bags are needed?

The two semicircles have diameter 20 m, so together they make one full circle of radius 10 m.

\begin{aligned} A &= 40 \cdot 20 + \pi(10)^2 \\ &= 800 + 100\pi \\ &\approx 1114.2 \text{ m}^2 \end{aligned}

Bags: $1114.2 \div 150 \approx 7.43$ . Seven bags would leave part of the field bare, so buy 8 bags.

Practice

Practice 1

A rectangular room is 10 ft by 8 ft, but a 4 ft by 3 ft closet is cut out of one corner. What is the area of the room, not counting the closet, in square feet?