Area of Composite Figures: Split, Add, Subtract | Grade 8 Math | LevelKite

A composite figure is a figure made up of simpler shapes, such as rectangles and semicircles. To find its area, split it into shapes whose areas you can find, and add their areas. This works because area adds up when the parts do not overlap and together cover the whole figure. If a shape has been cut out of the figure, subtract its area instead.

Every length must be in the same unit, and the area is in the square of that unit, such as square meters ( $\text{m}^2$ ).

The figure below shows an L-shaped patio, $ABCDEF$ . At every corner, the two sides that meet are perpendicular. The figure is not drawn to scale, so use only the facts and lengths stated in the text. The dotted segment $DG$ is a cut you can make: it runs from $D$ to point $G$ on side $AB$ , perpendicular to $AB$ .

Patio

ABCDEF

with

AB = 12

m,

BC = 4

m,

EF = 5

m and

FA = 10

m. The dotted cut

DG

is perpendicular to

AB

.

Worked example

An L-shaped patio

The patio $ABCDEF$ above has $AB = 12$ m, $BC = 4$ m, $EF = 5$ m and $FA = 10$ m. At every corner, the two sides that meet are perpendicular. Find the area of the patio.

Find the two missing sides, CD and DE. At every corner, the two sides that meet are perpendicular, so AB, CD and EF are parallel, and so are BC, DE and FA. Walking around the patio, you go along AB one way and come back along CD and EF, so CD and EF together are as long as AB. In the same way, BC and DE together are as long as FA.
$CD = AB - EF = 12 - 5 = 7 \text{ m}, \qquad DE = FA - BC = 10 - 4 = 6 \text{ m}$
Split the patio along the cut DG, which is perpendicular to AB. It makes two rectangles that do not overlap and together cover the patio. Rectangle AGEF has sides EF = 5 m and FA = 10 m, and rectangle GBCD has sides CD = 7 m and BC = 4 m.
$AGEF: 5 \text{ m} \times 10 \text{ m}, \qquad GBCD: 7 \text{ m} \times 4 \text{ m}$
Find each rectangle's area. A rectangle's sides are perpendicular, so its area is its length times its width.
$5 \times 10 = 50 \text{ m}^2, \qquad 7 \times 4 = 28 \text{ m}^2$
Add the two areas. This is valid because the rectangles do not overlap and together make up the whole patio.
$50 + 28 = 78 \text{ m}^2$
Check by subtracting instead. The patio is a 12 m by 10 m rectangle with a 7 m by 6 m corner missing at the top right, next to CD and DE. Subtract the missing corner from the whole rectangle.
$12 \times 10 - 7 \times 6 = 120 - 42 = 78 \text{ m}^2$

Answer

The patio's area is $78$ square meters.

Some composite figures have a curved edge. A semicircle is half of a circle. Any diameter cuts a circle into two equal halves, so a semicircle's area is half of the circle's area, $\frac{1}{2}\pi r^2$ .

In this lesson, each semicircle stands on one whole side of a rectangle. That side is the semicircle's diameter, $d$ , and the radius is half of it, $r = \frac{d}{2}$ . Both are positive lengths, in the same unit as the rectangle's sides.

Key concept

Add the parts, subtract the cut-outs

To find the area of a composite figure, split it into simple shapes that do not overlap and that together cover the whole figure. The figure's area is the sum of the parts' areas. If a shape is cut out of a larger one, subtract its area from the larger shape's area.

For a rectangle with a semicircle on one whole side, that side is the diameter $d$ , the radius is $r = \frac{d}{2}$ , and the semicircle's area is half of $\pi r^2$ . Add it when the semicircle is outside the rectangle, and subtract it when the semicircle is cut out.

Use one length unit for every measurement. The area is in the square of that unit.

\text{area of figure} = \text{area of rectangle} \pm \tfrac{1}{2}\pi r^2, \qquad r = \tfrac{d}{2} > 0

Pool

ABCD

:

AB = 20

ft,

BC = 10

ft, with a semicircle added outside on side

BC

, its diameter.

Worked example

A semicircle added

A swimming pool has the shape above: rectangle $ABCD$ with $AB = 20$ ft and $BC = 10$ ft, and a semicircle added outside the rectangle on side $BC$ . A cover must fit the whole surface of the pool. Find its area exactly, and then approximately, using $3.14$ for $\pi$ .

Find the rectangle's area. Its sides are perpendicular, so its area is length times width.
$\text{rectangle: } 20 \times 10 = 200 \text{ ft}^2$
Find the semicircle's radius. The semicircle stands on the whole side BC, so BC is its diameter, 10 ft. The radius is half of the diameter.
$r = \frac{d}{2} = \frac{10}{2} = 5 \text{ ft}$
Find the semicircle's area, half of the area of a circle with radius 5 ft. Square the radius before you multiply by π.
$\text{semicircle: } \frac{1}{2}\pi \cdot 5^2 = \frac{1}{2} \cdot 25\pi = 12.5\pi \text{ ft}^2$
Add the two areas, because the semicircle is outside the rectangle and the two parts do not overlap. Keeping π gives the exact area.
$\text{pool: } 200 + 12.5\pi \text{ ft}^2$
Approximate, using 3.14 for π: 12.5 × 3.14 = 39.25. Write ≈ at the step where π is replaced, because 3.14 is a rounded value.
$200 + 12.5\pi \approx 200 + 12.5 \times 3.14 = 200 + 39.25 = 239.25 \text{ ft}^2$

Answer

The cover's area is exactly $200 + 12.5\pi$ square feet, which is about $239.25$ square feet.

Lawn

ABCD

:

AB = 12

m,

BC = 8

m, with a semicircular flower bed cut out on side

AD

.

Worked example

A semicircle cut out

A rectangular lawn $ABCD$ , shown above, has $AB = 12$ m and $BC = 8$ m. A semicircular flower bed is cut out of the lawn along side $AD$ , so $AD$ is the flower bed's diameter. How many square meters of grass are left? Give the exact area and an approximation using $3.14$ for $\pi$ .

Find the area of the whole rectangle, as if nothing were cut out.
$12 \times 8 = 96 \text{ m}^2$
Find the radius. Side AD is opposite side BC in the rectangle, so AD = 8 m. It is the diameter, so the radius is half of it. The radius, 4 m, is less than the 12 m length, so the flower bed fits inside the lawn.
$AD = BC = 8 \text{ m}, \qquad r = \frac{8}{2} = 4 \text{ m}$
Find the flower bed's area, half of the area of a circle with radius 4 m.
$\frac{1}{2}\pi \cdot 4^2 = \frac{1}{2} \cdot 16\pi = 8\pi \text{ m}^2$
Subtract, because the flower bed is cut out of the rectangle and the grass is what is left. Keeping π gives the exact area.
$\text{grass: } 96 - 8\pi \text{ m}^2$
Approximate, using 3.14 for π: 8 × 3.14 = 25.12.
$96 - 8\pi \approx 96 - 8 \times 3.14 = 96 - 25.12 = 70.88 \text{ m}^2$

Answer

The grass covers exactly $96 - 8\pi$ square meters, which is about $70.88$ square meters.

Takeaway. To find the area of a composite figure, split it into simple shapes that do not overlap and that together cover it. Find any missing side from the stated ones, using the sides that are parallel. Add the areas of the parts, and subtract the area of any part that is cut out. For a semicircle on a side of a rectangle, that side is the diameter, the radius is half of it, and the area is $\frac{1}{2}\pi r^2$ . Keep every length in one unit, keep $\pi$ for the exact answer, and write $\approx$ when you use $3.14$ .