Topics
9. Modeling in Two and Three dimensions
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United States of America
Grades 9-12

9.01 Area and perimeter

Lesson
Practice

Introduction

When solving problems in the real world, we can often approximate them with geometric problems by matching the properties of 2D figures, such as the perimeter or area, to the context.

Perimeter

Perimeter is a term for the distance around the boundary of a two-dimensional shape. To calculate the perimeter of any polygon we simply add up all the lengths of the sides.

The perimeter of a circle has a special name, the circumference.

Exploration

Let's explore the formula for the circumference of a circle:

  • Change the diameter of the circle by dragging the blue dot.

  • Using the slider, roll the circle out along the line.

Loading interactive...

  1. How many diameters fit around the circumference of a circle?

  2. Did this change as the diameter changed?
  3. How does this relate to the formula for the circumference of a circle?

Just over 3 diameters will fit around the circumference of a circle, or to be more precise approximately 3.14. This leads to the following formula for the circumference of a circle:

\displaystyle C=\pi d
\bm{C}
circumference
\bm{d}
diameter

and because the diameter is twice the radius, we can also write the formula as

\displaystyle C=2\pi r
\bm{C}
circumference
\bm{r}
radius

Examples

Example 1

Simy is building a fence around her vegetable patch to keep her dog from digging up the potatoes. The vegetable patch is a 24 \text{ ft} by 9 \text{ ft} rectangle. She will create the fence with wire netting supported by posts placed every 3 \text{ ft}, and with a single gate placed between two of the posts.

Estimate the total cost to fence off the vegetable patch if wire netting costs \$ 0.60 per foot, the posts cost \$ 17.50 each, and the gate costs \$75.

Worked Solution
Create a strategy

Both the amount of wire netting and number of posts depend on the perimeter, P, of the vegetable patch. So, we can first find the perimeter of the patch using P=2l+2w, where l and w are the length and width of the rectangle respectively.

The amount of wire netting required is the perimeter minus 3 \text{ ft} for the gap with the gate.

The number of posts required will be the perimeter divided by three, for the 3 \text{ ft} gaps between each post.

Apply the idea

Finding the perimeter:

\displaystyle P\displaystyle =\displaystyle 2l+2wPerimeter formula
\displaystyle =\displaystyle 2 \cdot 24+2 \cdot 9Substitute l=24 and w=9
\displaystyle =\displaystyle 66 \text{ ft}Evaluate the multiplication and addition

The vegetable patch has a perimeter of 66 \text{ ft}, so we require P-3=63 \text{ ft} of wire netting, and \dfrac{P}{3}=22 posts.

Total cost:

\displaystyle C\displaystyle =\displaystyle \$0.60 \cdot \text{feet of wire} + \$17.50 \cdot \text{ number of posts} + \text{cost of gate}Set up equation
\displaystyle =\displaystyle \$0.60 \cdot 63+ \$17.50 \cdot 22 +\$75Substitute 63 feet of wire, 22 posts, and \$ 75 gate
\displaystyle =\displaystyle \$497.80Evaluate multiplication and addition

The total cost of the fence should be approximately \$497.80.

Example 2

The wheel of Kirara's bicycle has a diameter of 26 \text{ in}. Determine the number of feet she would travel if the wheels made 240 complete revolutions.

Worked Solution
Create a strategy

On each revolution Kirara will travel approximately the circumference of the wheel. We can multiply the circumference of the wheel by the number of revolutions to find the distance traveled in inches and then convert to feet.

Apply the idea
\displaystyle \text{Distance}\displaystyle =\displaystyle \pi d \times 240Circumference multiplied by the number of revolutions
\displaystyle =\displaystyle \pi \cdot 26 \cdot 240Substitute d=26
\displaystyle =\displaystyle 6240 \pi Evaluate the multiplication
\displaystyle =\displaystyle 6240 \pi \div 12 Convert inches to feet
\displaystyle =\displaystyle 520 \pi \text{ ft}Evaluate the division

Kirara traveled 520 \pi \text{ ft}, which is approximately 1\,634 \text{ ft}.

Reflect and check

What might affect the accuracy of the estimated distance? For example, if the tire is flat and reduced the diameter of the wheel by half an inch, by how much would the distance calculated above overestimate the actual distance traveled?

Idea summary

Perimeter is a term for the distance around the boundary of a two-dimensional shape. To calculate the perimeter of any polygon we simply add up all the lengths of the sides.

The perimeter of a circle is called the circumference and can be found using the formula:

\displaystyle C=\pi d
\bm{C}
circumference
\bm{d}
diameter

and because the diameter is twice the radius, we can also write the formula as

\displaystyle C=2\pi r
\bm{C}
circumference
\bm{r}
radius

Area

Area is the measure of the space enclosed by the boundary of a two-dimensional shape. We have previously encountered several formulas for simple shapes that can be used in a wide variety of real-world problems and can also be used to build or approximate the area of more complex figures.

The formula for the area of a rectangle is:

\displaystyle A = bh
\bm{b}
base
\bm{h}
height

The formula for the area of a triangle is:

\displaystyle A = \dfrac{1}{2}bh
\bm{b}
base
\bm{h}
height

Exploration

Let's explore the formula for the area of a circle:

  • Use the first slider to unwrap the circumference of the circle.

  • Use the second slider to select the number of slices to divide the circle into.

  • Use the third slider to rearrange the slices.

Loading interactive...

  1. Once rearranged, what figure do the slices approximate as the number of slices increases?

  2. Explain how the width of the shape relates to the circumference of the circle.

  3. Explain how the area of this figure relates to the area of the circle.

By decomposing a circle into equal sectors and rearranging them into a parallelogram, we see how the formula for the area of a circle is:

\displaystyle A = \pi r^2
\bm{r}
radius

Examples

Example 3

Find the area of the following figure to two decimal places:

Worked Solution
Create a strategy

We can use the area formula for a circle, A=\pi r^2, and multiply by \frac{3}{4} to find area of the given figure.

Apply the idea
\displaystyle A\displaystyle =\displaystyle \frac{3}{4} \cdot \pi r^2The proportion of the circle times the area of the circle
\displaystyle =\displaystyle \frac{3}{4} \cdot \pi \cdot4^2Substitute r=4
\displaystyle =\displaystyle 12 \pi\text{ cm}^2Evaluate the multiplication
\displaystyle =\displaystyle 37.70 \text{ cm}^2Approximate to 2 decimal places

Example 4

An area of floor measuring 2\,280\text{ cm}^2 is to be paved with identical tiles in the shape of parallelograms. Each tile measures 12\text{ cm} along the base, and has a perpendicular height of 5\text{ cm}.

How many tiles are needed to cover the whole area?

Worked Solution
Create a strategy

Find the area of a single tile and divide the floor area by this value.

Apply the idea

Area single tile:

\displaystyle \text{Area}\displaystyle =\displaystyle bhFormula for area of a parallelogram
\displaystyle =\displaystyle 5 \cdot 12Substitute b=5 and h=12
\displaystyle =\displaystyle 60 \text{ cm}^2Evaluate the multiplication

Number of tiles required:

\displaystyle \text{Number of tiles}\displaystyle =\displaystyle \text{Floor area} \div \text{Area of tile}Set up equation
\displaystyle =\displaystyle 2280 \div 60Substitute \text{Area of tile}=60
\displaystyle =\displaystyle 38Evaluate the division
Reflect and check

The formula for the area of a parallelogram is equal to the area of a rectangle with the same base and perpendicular height, thus A=bh. This can be observed by cutting off a triangle and rearranging to form a rectangle as shown below.

This can also be seen as a 2-Dimensional application of Cavalieri's principle. Where slices taken parallel to the base, through the rectangle and parallelogram, would produce lines of the same length at each height.

Example 5

By considering a trapezoid as a composite figure made up of two triangles, find a general formula for the area of a trapezoid in terms of its perpendicular height, h, and parallel side lengths, a and b.

Worked Solution
Create a strategy

We can split the trapezoid into two triangles using one of the diagonals. Then use the formula for the area of a triangle A=\frac{1}{2}\cdot \text{base} \times \text{height} to find a general formula for a trapezoid.

Apply the idea

Splitting the trapezoid as follows:

Triangle 1 has base length a and height h and Triangle 2 has base length b and height h.

Area of trapezoid:

\displaystyle A\displaystyle =\displaystyle A_{\triangle 1}+ A_{\triangle 2}Breaking the area into the two triangles
\displaystyle =\displaystyle \frac{1}{2}ah+\frac{1}{2}bhUsing formula for area of a triangle
\displaystyle =\displaystyle \frac{1}{2}h\left(a+b\right)Factoring \dfrac{1}{2} and h

Thus, the area of a trapezoid is one half of the product of the height and the sum of the lengths of the bases, where h is the height and a and b are the bases.

Reflect and check

There are several ways to justify the formula for the area of a trapezoid. For example, we could create a parallelogram out of two identical trapezoids as follows:

Then, area of trapezoid:

\displaystyle A\displaystyle =\displaystyle \frac{1}{2} \cdot \text{Area of parallelogram}
\displaystyle =\displaystyle \frac{1}{2}\cdot \text{base} \times \text{height}Using formula for area of a parallelogram
\displaystyle =\displaystyle \frac{1}{2}\left(a+b\right)hSubstitution

Can you think of any other ways to justify the area of a trapezoid?

Idea summary

The area of a rectangle is given by:

\displaystyle A=bh
\bm{b}
base
\bm{h}
height

The area of a triangle is given by:

\displaystyle A=\frac{1}{2}bh
\bm{b}
base
\bm{h}
height

The area of a circle is given by:

\displaystyle A=\pi r^2
\bm{r}
radius

General formulas for shapes with special properties such as parallelograms, trapezoids, kites, rhombuses, and regular polygons can be often be derived by breaking the shape down into components of simpler shapes.

Composite shapes and density

A composite figure is a figure that can be decomposed into smaller figures that have been added together or sometimes subtracted from each other.

A composite shape composed of a triangle on top of a rectangle.
A composite figure formed by adding a rectangle and triangle together.
A composite shape composed of a semicircle cut out of a rectangle.
A composite figure formed by subtracting a semicircle from a rectangle.

We can determine the area of composite figures by breaking them down into simpler shapes. After we find the area of the simpler shapes, we can add or subtract those areas to find the area of the composite shape.

Composite shape composed of a rectangle with triangle on top, parallelogram below, and small square on the lower right.

Dashed lines can be used to visualize which simple shapes make up a composite shape.

Another use for finding the area of irregular shapes is determining the population density of that region.

Population density

The number of people living in each unit of area of a location (such as a square mile)

The population density is calculated as the population in a region divided by the area of that region.

\displaystyle \text{Population density}=\frac{\text{Population}}{\text{Area}}
\bm{\text{Population}}
number of people or objects
\bm{\text{Area}}
space available for the population
This image shows a piece of a map with different amounts of people in different areas to convey the concept of population density. Ask your teacher for more information.

A high population density means there is a large number of people living in a given amount of space. Typically, cities have a high population density. A small population density means there is a small number of people living in a given amount of space. Typically rural areas have a small population density.

Examples

Example 6

Find the area of a sandbox which has been approximated by this geometric figure. All measurements are in feet.

A quadrilateral with sides of length 11, 8, 4, and 13. Between sides 11 and 8 is an included right angle. The included angle between side lengths 11 and 8 as well as the angle between side lengths 4 and 13 is a right angle.
Worked Solution
Create a strategy

We don't have a formula for the area of this type of quadrilateral. Instead, we can decompose the shape into two right-triangles as follows:

A quadrilateral with sides of length 11, 8, 4, and 13. The included angle between side lengths 11 and 8 as well as the angle between side lengths 4 and 13 is a right angle. A dashed segment from the angle between sides of length 11 and 13 and the angle between sides of length 8 and 4 serves as the hypotenuse of the right triangles. The triangle with side lengths 11 and 8 is labelled 1 and the triangle with side lengths 4 and 13 is labelled 2
Apply the idea

Area of triangle 1:

\displaystyle A_1\displaystyle =\displaystyle \dfrac{1}{2}\cdot b\cdot hFormula for area of a triangle
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2}\cdot 8\cdot11Substitute b=8 and h=11
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2} \cdot 88Evaluate the multiplication
\displaystyle {}\displaystyle =\displaystyle 44 \text{ ft}^2Evaluate the multiplication

Area of triangle 2:

\displaystyle A_2\displaystyle =\displaystyle \dfrac{1}{2}\cdot b\cdot hFormula for area of a triangle
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2}\cdot 4\cdot13Substitute b=4 and h=13
\displaystyle {}\displaystyle =\displaystyle \dfrac{1}{2} \cdot 52Evaluate the multiplication
\displaystyle {}\displaystyle =\displaystyle 26 \text{ ft}^2Evaluate the multiplication

Area of the composite shape:

\displaystyle A\displaystyle =\displaystyle A_1 + A_2Set up equation
\displaystyle {}\displaystyle =\displaystyle 44+26Substitute A_1=44 and A_2=26
\displaystyle {}\displaystyle =\displaystyle 70 \text{ ft}^2Evaluate the addition

Area of the sandbox = 70 \text{ ft}^2

Reflect and check

Consider that the sandbox was not necessarily an exact geometric figure. When modeling with 2D figures, it is reasonable to approximate the shape of an area to a geometric figure that makes our calculations easier.

Example 7

An amphitheater is designed with a semicircular viewing section and a rectangular stage. The viewing section is designed so that the furthest audience member is 10 meters from the middle of the front of the stage. The stage is 20 meters by 6 meters.

a

Find the approximate area covered by the amphitheater to the nearest square meter.

Worked Solution
Create a strategy

Using the description of the amphitheater, we can approximate the area to that of a composite figure formed by a rectangle and a semicircle.

A composite figure composed of a semicircle on top of a rectangle. The rectangle is 20 by 6 meters. The semicircle has a radius of 10 meters.
Apply the idea

Area of the semicircle:

\displaystyle A_{\text{semicircle}}\displaystyle =\displaystyle \frac{1}{2}\pi r^2Formula for area of half of a circle
\displaystyle =\displaystyle \frac{1}{2}\pi\cdot 10^2Substitute r=10
\displaystyle =\displaystyle \frac{1}{2}\pi\cdot 100Evaluate the exponent
\displaystyle =\displaystyle 50\piEvaluate the multiplication

Area of the rectangle:

\displaystyle A_{\text{rectangle}}\displaystyle =\displaystyle bhFormula for area of a rectangle
\displaystyle =\displaystyle 20\cdot6Substitute b=20 and h=6
\displaystyle =\displaystyle 120Evaluate the multiplication

Area of the amphitheater:

\displaystyle A\displaystyle =\displaystyle A_{\text{semicircle}}+A_{\text{rectangle}}Set up equation
\displaystyle =\displaystyle 50\pi +120Substitute A_{\text{semicircle}}=50 \pi and A_{\text{rectangle}}=120

If we evaluate the expression for the area of the amphitheater and round to the nearest square meter, we get that:A\approx 277\text{ m}^2

b

If there are 300 people in the audience and 20 actors on stage, find the population density of the amphitheater.

Worked Solution
Apply the idea
\displaystyle \text{Population density}\displaystyle =\displaystyle \frac{\text{Population}}{\text{Area}}Formula for population density
\displaystyle =\displaystyle \frac{300+20}{277}Substitute total number of people and area found in part (a)
\displaystyle =\displaystyle \frac{320}{277}Evaluate the addition

Rounding to two decimal places, the population density of the amphitheater is 1.16 people per square meter.

Reflect and check

Consider the following:

  • How are the units determined for a population density?

  • How the population of the stage compares to that of the audience section?

  • Why might population density at a venue be important?

The units for a population density come from the units of the population divided by the units of the area. In the example above we had a population of individual people and an area in square meters. Thus, the units were "people per square meter". Other examples could include units such at "hundreds of cats per square mile" or "mites per square inch".

The population density of the stage, 0.08 people per square meter, is significantly lower than that of the audience section, 1.91 people per square meter.

Population density calculations can play important roles in regulations such as fire safety, or restrictions during a pandemic.

Idea summary

The area of composite figures can be determined by breaking them down into simpler shapes, such as triangles, rectangles and circles and finding the sums of those areas.

The population density of an area can be calculated as follows:

\displaystyle \text{Population density}=\frac{\text{Population}}{\text{Area}}
\bm{\text{Population}}
number of people or objects
\bm{\text{Area}}
space available for the population

Outcomes

G.GMD.A.1

Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.

G.MG.A.1

Use geometric shapes, their measures, and their properties to describe objects.

G.MG.A.2

Apply concepts of density based on area and volume in modeling situations.

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