Solve Geometry Applications: Circles and Irregular Figures

By the end of this section, you will be able to: use the properties of circles, and find the area of irregular figures.

In this section, we’ll continue working with geometry applications, adding a few new formulas to our collection.

Use the properties of circles

Recall the properties of circles:

Properties of circles.

$r$ is the length of the radius.
$d$ is the length of the diameter, and $d = 2r$ .
Circumference is the perimeter of a circle. The formula for circumference is $C = 2\pi r$ .
The formula for the area of a circle is $A = \pi r^2$ .

Remember, we approximate $\pi$ with $3.14$ or $\tfrac{22}{7}$ depending on whether the radius of the circle is given as a decimal or a fraction. If you use the $\pi$ key on a calculator, your answers will be slightly different from the answers shown here, since that key uses more decimal places.

Example. A circular sandbox has a radius of $2.5$ feet. Find (a) the circumference and (b) the area of the sandbox.

(a) Let $c =$ circumference. Substituting into $C = 2\pi r$ :

C = 2\pi(2.5) \qquad\Rightarrow\qquad C \approx 2(3.14)(2.5) \qquad\Rightarrow\qquad C \approx 15.7 \text{ ft}

Checking: if we draw a square around the circle, its sides would be $5$ ft (twice the radius), so its perimeter would be $20$ ft — slightly more than the circle’s circumference, which makes sense. The circumference of the sandbox is $15.7$ feet.

(b) Let $A =$ area. Substituting into $A = \pi r^2$ :

A = \pi(2.5)^2 \qquad\Rightarrow\qquad A \approx (3.14)(2.5)^2 \qquad\Rightarrow\qquad A \approx 19.625 \text{ sq. ft}

Checking: the square around the circle has area $25$ sq ft, slightly more than the circle’s area, which makes sense. The area of the sandbox is $19.625$ square feet.

A circular mirror has radius of 5 inches. Find the circumference. Use 3.14 for $\pi$ .

A circular mirror has radius of 5 inches. Find the area. Use 3.14 for $\pi$ .

We usually see the formula for circumference in terms of the radius $r$ . But since the diameter of a circle is two times the radius, we can also write the formula in terms of $d$ . Using the commutative property, $C = 2\pi r = \pi \cdot 2r$ , and substituting $d = 2r$ gives $C = \pi d$ . We use this form when we’re given the length of the diameter instead of the radius.

Example. A circular table has a diameter of four feet. What is the circumference of the table?

Let $c =$ the circumference. Substituting into $C = \pi d$ :

C = \pi(4) \qquad\Rightarrow\qquad C \approx (3.14)(4) \qquad\Rightarrow\qquad C \approx 12.56 \text{ feet}

Checking: a square around the circle would have side $4$ and perimeter $16$ ; it makes sense that the circumference, $12.56$ , is a little less than $16$ . The circumference of the table is $12.56$ feet.

Find the circumference of a circular fire pit whose diameter is 5.5 feet. Use 3.14 for $\pi$ .

If the diameter of a circular trampoline is 12 feet, what is its circumference? Use 3.14 for $\pi$ .

Example. Find the diameter of a circle with a circumference of $47.1$ centimeters.

Let $d =$ the diameter. Substituting into $C = \pi d$ with $C = 47.1$ :

47.1 \approx 3.14d

Dividing both sides by $3.14$ :

\frac{47.1}{3.14} \approx \frac{3.14d}{3.14} \qquad\Rightarrow\qquad 15 \approx d

Checking: $47.1 \overset{?}{=} (3.14)(15)$ , and $47.1 = 47.1$ . The diameter of the circle is approximately $15$ centimeters.

Find the diameter of a circle with circumference of 94.2 centimeters. Use 3.14 for $\pi$ .

Find the diameter of a circle with circumference of 345.4 feet. Use 3.14 for $\pi$ .

Find the area of irregular figures

So far, we have found area for rectangles, triangles, trapezoids, and circles. An irregular figure is a figure that is not a standard geometric shape — its area cannot be calculated using any single standard area formula. But some irregular figures are made up of two or more standard geometric shapes. To find the area of one of these irregular figures, we can split it into figures whose formulas we know, and then add the areas of the figures.

Example. Find the area of an L-shaped figure: a wide rectangle $12$ units long and $4$ units tall along the top, with a narrower rectangular tab hanging down $10$ units total (so $6$ more units below the top rectangle) and $2$ units wide, attached to the right side.

The figure is irregular, but we can split it into two rectangles: a blue rectangle with width $12$ and length $4$ across the top, and a red rectangle attached below it. The right side of the whole figure is $10$ units, and the blue rectangle’s right side is $4$ units, so the red rectangle’s length is $10 - 4 = 6$ units, with width $2$ .

A_{\text{figure}} = A_{\text{rectangle}} + A_{\text{rectangle}} = bh + bh = 12 \cdot 4 + 2 \cdot 6 = 48 + 12 = 60

The area of the figure is $60$ square units. (There’s more than one way to split an irregular figure into rectangles — try splitting this one a different way and check that you still get the same total area.)

An irregular figure looks like a narrow column 3 units wide and 6 units tall, with a wider arm attached across its top: the whole figure is 8 units wide at the top, and that top arm is 2 units tall. Find the total (shaded) area.

An irregular figure is shaped like a 14-by-10 rectangle with a rectangular notch cut out of the bottom-left corner. The notch is 6 units wide, and the un-notched top strip is 5 units tall (so the notch itself is $10 - 5 = 5$ units tall). Find the total (shaded) area.

Example. Find the area of a shaded region made of a rectangle $8$ units long and $4$ units tall, topped by a right triangle whose vertical leg is $3$ units (the difference between the total right-side height of $7$ and the rectangle’s height of $4$ ) and whose horizontal leg (base) is $3$ units (the difference between the rectangle’s length of $8$ and the $5$ -unit top edge of the rectangle).

We break this irregular figure into a triangle and a rectangle, and the area of the figure is the sum of their areas. The rectangle has length $8$ and width $4$ . Since both vertical sides of the rectangle are $4$ , the vertical leg of the triangle is $7 - 4 = 3$ . Since the rectangle’s length is $8$ , the base of the triangle is $8 - 5 = 3$ .

A_{\text{figure}} = A_{\text{rectangle}} + A_{\text{triangle}} = lw + \tfrac{1}{2}bh = 8 \cdot 4 + \tfrac{1}{2} \cdot 3 \cdot 3 = 32 + 4.5

A_{\text{figure}} = 36.5 \text{ sq. units}

An irregular figure is a rectangle 8 units long and 4 units tall, with a triangular wedge attached at the top-right corner whose base and height are both 3 units. Find the total (shaded) area.

An irregular figure is a rectangle 12 units long and 5 units tall, topped by a triangle whose base is $2.5 + 2.5 = 5$ units and whose height is 4 units. Find the total area (rectangle area $12 \cdot 5$ , plus triangle area $\tfrac{1}{2}(5)(4)$ ).

Example. A high school track is shaped like a rectangle with a semicircle (half a circle) on each end. The rectangle has length $105$ meters and width $68$ meters. Find the area enclosed by the track, rounded to the nearest hundredth.

We break the figure into a rectangle and two semicircles. The rectangle has length $105$ m and width $68$ m. The semicircles have a diameter of $68$ m, so each has radius $34$ m.

A_{\text{figure}} = A_{\text{rectangle}} + A_{\text{semicircles}} = bh + 2\left(\tfrac{1}{2}\pi \cdot r^2\right)

A_{\text{figure}} \approx 105 \cdot 68 + 2\left(\tfrac{1}{2} \cdot 3.14 \cdot 34^2\right) \qquad\Rightarrow\qquad A_{\text{figure}} \approx 7140 + 3629.84

A_{\text{figure}} \approx 10{,}769.84 \text{ square meters}

A shaded figure is a rectangle 15 units long and 9 units wide, with a semicircular bite (diameter equal to the rectangle's width, 9 units, so radius 4.5) cut out of one end. Find the shaded area, rounded to the nearest tenth. Use 3.14 for $\pi$ .

A shaded figure is a trapezoid with parallel top and bottom sides of 5.2 units and 3.3 units and height 6.5 units, topped by a semicircle whose diameter is 5.2 units (radius 2.6). Find the total (shaded) area, rounded to the nearest hundredth. Use 3.14 for $\pi$ .

Key terms

radius — the distance from the center of a circle to any point on the circle. diameter — the distance across a circle through its center, equal to twice the radius. circumference — the perimeter (distance around) a circle, $C = 2\pi r$ or $C = \pi d$ . irregular figure — a figure that is not a standard geometric shape, whose area can often be found by splitting it into rectangles, triangles, trapezoids, and circles (or semicircles) and adding their areas.

This section is adapted from Prealgebra 2e, Section 9.5: Solve Geometry Applications: Circles and Irregular Figures by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: described the irregular-figure diagrams and the track diagram in prose instead of hotlinking images; omitted the Be Prepared quiz, Media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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