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10. Modeling in 2 & 3 Dimensions
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Geometry

10.01 Area and perimeter

Lesson
Practice

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 the lengths of all the sides.

Some polygons have a formula that can be also be used to find perimeter:

A rectangle showing its labels of length and width.

\,\\\,Consider a rectangle.

If the width is w and the length is l, the perimeter is:\text{Perimeter} = l+w+l+w

\text{Perimeter} = 2 \left(l+w \right)

A square showing its label of side length.

\,\\\,Consider a square.

If each side has length l, the perimeter is:\text{Perimeter} = l+l+l+l

\text{Perimeter} = 4 l

A hexagon showing its label of side lengths.

\,\\\,Consider a regular hexagon.

If each side is l, the perimeter is:\text{Perimeter} = l+l+l+l+l+l

\text{Perimeter} = 6 l

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

A diagram of a circle with its radius labeled.

Consider a circle.

If the radius is r, the circumference is:

\text{Circumference}=2\cdot \pi \cdot r

\text{Circumference}=2 \pi r

A change in one dimension of a figure results in a predictable change in perimeter. The resulting figure may or may not be similar to the original figure.

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

First we need to find the radius which is half the diameter. So r=\dfrac{1}{2} \cdot 26=13

\displaystyle \text{Distance}\displaystyle =\displaystyle 2\pi r \cdot 240Circumference multiplied by the number of revolutions
\displaystyle =\displaystyle 2\pi \cdot 13 \cdot 240Substitute d=13
\displaystyle =\displaystyle 6240 \pi Evaluate the multiplication
\displaystyle =\displaystyle \dfrac{6240 \pi}{ 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?

Example 3

The side length of a square Q is doubled to create square R. How does this affect the perimeter of square R relative to square Q?

Worked Solution
Create a strategy

Let the side length of square Q be s. Since the side length of square R is double that of square Q, the side length of square R is 2s. Calculate the perimeters of both squares and compare them to determine the relationship between the perimeters.

Apply the idea

We know that the formula for the perimeter of a square is P=4l where l is the side length.

Perimeter of square Q with side s:

\displaystyle P_Q\displaystyle =\displaystyle 4(s)Substitute the side length s
\displaystyle =\displaystyle 4sSimplify

Perimeter of square R with side 2s:

\displaystyle P_R\displaystyle =\displaystyle 4(2s)Substitute the side length 2s
\displaystyle =\displaystyle 8sSimplify

The perimeter of square R is twice the perimeter of square Q. \dfrac{8s}{4s}=2This means doubling the side length of square Q to create square R also doubles the perimeter of the square.

Reflect and check

Doubling all sides of square Q is the same as dilating square Q by a scale factor of 2. As we saw previously, dilating all sides of a figure creates a similar figure. So, square Q is similar to square R.

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

A change in one dimension of a figure results in a predictable change in perimeter. The resulting figure may or may not be similar to the original figure.

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.

A rectangle with its length and width labeled. Opposite sides are marked as equal, and all corners are right angles.

\,\\\,\\\,The area of a rectangle is given by\text{Area} = \text{length} \cdot \text{width} \text{, or }\\ \\ A = l \cdot w

A rectangle with its side labeled. All sides are marked as equal, and all corners are right angles.

\,\\\,\\\,The area of a square is given by\text{Area} = \text{side} \cdot \text{side} \text{, or }\\ \\ A =s \cdot s=s^2

A triangle with its height and base. The height is perpendicular distance from the base to the vertex opposite.

\,\\\,\\\,The area of a triangle is given by\text{Area} = \dfrac{1}{2}\cdot \text{base} \cdot \text{height} \text{, or }\\ \\ A =\dfrac{1}{2} b h

Image of a parallelogram showing a height and parallel bases. The base is also perpendicular to the height.

\,\\\,\\\,The area of a parallelogram is given by\text{Area} = \text{base} \cdot \text{height} \text{, or }\\ \\ A =b h

A diagram of a circle with its radius labeled.

The area of a circle is given by \text{Area} = \pi \cdot \text{radius} \cdot \text{radius} \text{, or }\\ \\ A = \pi r^2

A geometric diagram of a trapezoid. The top and bottom sides, labeled as base a and base b respectively, are horizontal and of different lengths, indicating the parallel bases of the trapezoid. A dashed vertical line labeled height is drawn from the midpoint of base a to the midpoint of base b, indicating the perpendicular height of the trapezoid. A right angle is shown where the height and base b meet, emphasizing the perpendicularity.

The area of a trapezoid is given by: \text{Area} = \dfrac{1}{2}\cdot \left(\text{base }a+\text{base }b\right)\cdot \text{height, or}

A=\dfrac{1}{2}(a+b)h

A change in one dimension of a figure results in a predictable change in area. The resulting figure may or may not be similar to the original figure.

Examples

Example 4

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 \dfrac{3}{4} to find area of the given figure.

Apply the idea
\displaystyle A\displaystyle =\displaystyle \dfrac{3}{4} \cdot \pi r^2The proportion of the circle times the area of the circle
\displaystyle =\displaystyle \dfrac{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 5

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 6

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.

A trapezoid with a parallel side lengths a and b and perpendicula height h
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=\dfrac{1}{2} \text{base} \cdot \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 \dfrac{1}{2}ah+\dfrac{1}{2}bhUsing formula for area of a triangle
\displaystyle =\displaystyle \dfrac{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 \dfrac{1}{2} \text{Area of parallelogram}
\displaystyle =\displaystyle \dfrac{1}{2} \text{base} \cdot \text{height}Using formula for area of a parallelogram
\displaystyle =\displaystyle \dfrac{1}{2}\left(a+b\right)hSubstitution

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

Example 7

The area of a square is initially 25 \text{ cm}^2 before its dimensions are altered.

a

If one side of the square is increased by a factor of 4, what is the new side length of the square?

Worked Solution
Create a strategy

Since the area of a square is given by s^2, where s is the length of a side, we can find the original side length by taking the square root of the area. We can then multiply this original side length by the given factor to find the new side length.

To distinguish between the original side length and the new side length, we can use the notation s' for the new side length.

Apply the idea
\displaystyle s\displaystyle =\displaystyle \sqrt{25}Find the original side length
\displaystyle s\displaystyle =\displaystyle 5Evaluate the square root
\displaystyle s'\displaystyle =\displaystyle 4 \cdot sMultiply the original side length by 4
\displaystyle s'\displaystyle =\displaystyle 20Evaluate the multiplication

Therefore, the new side length of the square is 20\text{ cm}.

b

If one side of the square is decreased by a factor of 2, what is the new side length of the square?

Worked Solution
Create a strategy

In the previous part, we found the original side length to be s=5\text{ cm}. Since the side is now being decreased, we can divide this length by the given factor to find the new side length.

Recall s represents the original side, s' represents the side after the enlargement, so we will use s'' to represent the side after the reduction.

Apply the idea
\displaystyle s\displaystyle =\displaystyle \sqrt{25}Find the original side length
\displaystyle s\displaystyle =\displaystyle 5Evaluate the square root
\displaystyle s''\displaystyle =\displaystyle \dfrac{5}{2}Divide the side length by the factor
\displaystyle s''\displaystyle =\displaystyle 2.5Evaluate the division

Therefore, the new side length of the square is 2.5\text{ cm}.

c

Describe how these changes affect the area and perimeter of the square.

Worked Solution
Create a strategy

The area of a square is given by s^2, and the perimeter is given by 4s, where s is the side length of the square. In the first case, we multiplied the side length by 4, which is represented generally as 4s. In the second case, we divided the side length by 2, which is represented generally by \dfrac{s}{2}.

By substituting these general side lengths into the formulas, we can compare the new area and perimeter of the square in both cases to the original area and perimeter of the square.

Apply the idea

The area of a square is given by s^2. If the side length is multiplied by 4, the new area becomes (4s)^2=16s^2. This means the new area is 16 times more than the original area.

The perimeter of a square is given by 4s. Multiplying the side by 4 gives a new perimeter of 4(4s)=16s. The means the new perimeter is 4 times the original perimeter.

If the side length is halved, the new area becomes \left(\dfrac{s}{2}\right)^2=\dfrac{s^2}{4}. This means the new area is \dfrac{1}{4}th the original area.

The new perimeter becomes 4\left(\dfrac{s}{2}\right)=2s, which means the new perimeter is half the original perimeter.

In general, if the side of a square is dilated by a factor of k, the new area will be k^2 times the original area and the new perimeter will be k times the original perimeter.

Reflect and check

We can check our answers by finding the new areas and perimeters and comparing them to the original area and perimeter of the square. The original area of the square is 25\text{ cm}^2, and the original perimeter is 4\left(5\right)=20\text{ cm}.

The area of the square after the enlargement:

\displaystyle A'\displaystyle =\displaystyle (20)^2Substitute s'=20
\displaystyle A'\displaystyle =\displaystyle 400Evaluate the square
\displaystyle \dfrac{400}{25}\displaystyle =\displaystyle 16Divide new area by original area

This shows that the new area is 16 times (or 4^2 times) larger than the original area.

The perimeter of the square after the enlargement:

\displaystyle P'\displaystyle =\displaystyle 4(20)Substitute s'=20
\displaystyle P'\displaystyle =\displaystyle 80Evaluate the multiplication
\displaystyle \dfrac{80}{20}\displaystyle =\displaystyle 4Divide new perimeter by original perimeter

This shows the new perimeter is 4 times the original perimeter.

The area of the square after the reduction:

\displaystyle A''\displaystyle =\displaystyle (2.5)^2Substitute s''=2.5
\displaystyle A''\displaystyle =\displaystyle 6.25Evaluate the square
\displaystyle \dfrac{6.25}{25}\displaystyle =\displaystyle 0.25Divide new area by original area

This shows the new area is \dfrac{1}{4}th (or \left(\dfrac{1}{2}\right)^2) times the original area.

The perimeter of the square after the reduction:

\displaystyle P''\displaystyle =\displaystyle 4(2.5)Substitute s''=2.5
\displaystyle P''\displaystyle =\displaystyle 10Evaluate the multiplication
\displaystyle \dfrac{10}{20}\displaystyle =\displaystyle \dfrac{1}{2}Divide new perimeter by original perimeter

This shows the new perimeter is half the original perimeter.

Idea summary

The area of a rectangle is given by:

\displaystyle A=lw
\bm{l}
length
\bm{w}
width

The area of a triangle is given by:

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

The area of a parallelogram is given by:

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

The area of a trapezoid is given by:

\displaystyle A=\dfrac{1}{2}(a+b)h
\bm{a}
base a
\bm{b}
base b
\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.

A change in one dimension of a figure results in a predictable change in area. The resulting figure may or may not be similar to the original figure.

Outcomes

G.DF.2

The student will determine the effect of changing one or more dimensions of a three-dimensional geometric figure and describe the relationship between the original and changed figure.

G.DF.2a

Describe how changes in one or more dimensions of a figure affect other derived measures (perimeter, area, total surface area, and volume) of the figure.

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