Unlocking Area and Volume with Scale Factors: A Guide to Proportional Measures

In this lesson, similar figures will be compared in terms of their surface areas and volumes.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Similar Figures
Surface Area
Volume
Dilation
Length Scale Factor

Challenge

Using Scale Factor of Similar Pyramids

On a trip to Egypt, Emily is visiting the Great Egyptian Museum. She sees two models of Pyramids, Khafra and Menkaure, on display. She decides to buy the model of Menkaure, but now as she exits the gift shop, she has become even more curious about the volume of Khafra.

On the left is a square pyramid named Khafra with a height of 2h, and on the right stands another square pyramid named Menkaure with a height of h.

Suppose that the models of the pyramids are similar. If the scale factor of the corresponding side lengths is $1 : 2$ and the volume of the smaller pyramid is $20$ cubic centimeters, what is the volume of the larger model?

Discussion

Area Scale Factor

If two figures are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding side lengths.

Let $K L M N$ and $P Q R S$ be similar figures, and $A_{1}$ and $A_{2}$ be their respective areas. The length scale factor between corresponding side lengths is $\frac{a}{b} .$ Here, the following conditional statement holds true.

$K L M N \sim P Q R S \Rightarrow \frac{A _{1}}{A _{2}} = (\frac{a}{b})^{2}$

Proof

The statement will be proven for similar rectangles, but this proof can be adapted for other similar figures.

The area of a rectangle is the product of its length and its width.

Area of $K L M N$	Area of $P Q R S$
$A_{1} = K L \cdot L M$	$A_{2} = P Q \cdot Q R$

By the definition of similar polygons, the corresponding side lengths are proportional and equal to the scale factor

\frac{a}{b} .

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ \frac{K L}{P Q} = \frac{a}{b} \frac{L M}{Q R} = \frac{a}{b} \Leftrightarrow ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ K L = P Q \cdot \frac{a}{b} L M = Q R \cdot \frac{a}{b}

The next step is to substitute the expressions for

K L

and

L M

into the formula for

A_{1},

which represents the area of

K L M N .

A_{1} = K L \cdot L M

SubstituteII

$K L = P Q \cdot \frac{a}{b}$ , $L M = Q R \cdot \frac{a}{b}$

A_{1} = (P Q \cdot \frac{a}{b}) (Q R \cdot \frac{a}{b})

▼

Simplify right-hand side

RemovePar

Remove parentheses

A_{1} = P Q \cdot \frac{a}{b} \cdot Q R \cdot \frac{a}{b}

CommutativePropMult

Commutative Property of Multiplication

A_{1} = \frac{a}{b} \cdot \frac{a}{b} \cdot P Q \cdot Q R

ProdToPowTwoFac

$a \cdot a = a^{2}$

A_{1} = (\frac{a}{b})^{2} \cdot P Q \cdot Q R

AssociativePropMult

Associative Property of Multiplication

A_{1} = (\frac{a}{b})^{2} (P Q \cdot Q R)

Notice that the expression on the right-hand side is

(\frac{a}{b})^{2}

times the area of

P Q R S,

A_{2} .

A_{1} = (\frac{a}{b})^{2} (P Q \cdot Q R)

Substitute

$P Q \cdot Q R = A_{2}$

A_{1} = (\frac{a}{b})^{2} A_{2}

DivEqn

$LHS / A_{2} = RHS / A_{2}$

\frac{A _{1}}{A _{2}} = (\frac{a}{b})^{2}

This proof has shown that the ratio of the areas of the similar rectangles is equal to the square of the ratio of their corresponding side lengths. This ratio is also called the area scale factor.

\underline{Scale Factor} \frac{a}{b} \Rightarrow \underline{Area Scale Factor} \frac{A _{1}}{A _{2}} = (\frac{a}{b})^{2}

The length scale factor of two similar figures can be used to find the area of one of the two figures when the area of one of the figures is known.

Example

Using Area Scale Factor to Determine an Unknown Area

Dominika is making a flyer for a concert and wants to compare the areas of the two pieces of paper. The widths of $A 4$ and $A 2$ papers are $21$ and $42$ centimeters, respectively.

If these two pieces of paper are similar and the area of $A 4$ paper is about $624$ square centimeters, find the area of the $A 2$ paper. Round the answer to the nearest integer.

Hint

If the scale factor of two similar figures is $\frac{a}{b},$ then the ratio of their areas is $\frac{a ^{2}}{b ^{2}} .$

Solution

The two pieces of paper are similar and two corresponding sides measure $21$ centimeters and $42$ centimeters.

Therefore, the scale factor is the ratio of these corresponding sides.

Scale Factor : \frac{2 1}{4 2} = \frac{1}{2}

Using this information, the ratio of the areas, or area scale factor, can be calculated by the theorem about the areas of similar figures.

\underline{Scale Factor} \frac{1}{2} \Rightarrow \underline{Area Scale Factor} \frac{1 ^{2}}{2 ^{2}} = \frac{1}{4}

Now, let

A_{2}

be the area of the piece of

A 2

paper. Then, a proportion can be written using the area scale factor and the area of the

A 4

paper, which is

624

square centimeters.

\frac{6 2 4}{A _{2}} = \frac{1}{4}

▼

Solve for

A_{2}

MultEqn

$LHS \cdot A_{2} = RHS \cdot A_{2}$

\frac{6 2 4}{A _{2}} \cdot A_{2} = \frac{1}{4} \cdot A_{2}

FracMultDenomToNumber

$\frac{a}{A _{2}} \cdot A_{2} = a$

624 = \frac{1}{4} \cdot A_{2}

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

624 = \frac{A _{2}}{4}

MultEqn

$LHS \cdot 4 = RHS \cdot 4$

2496 = A_{2}

RearrangeEqn

Rearrange equation

A_{2} = 2496

The area of the

A 2

paper is

2496

square centimeters. Dominika figures that is just the right amount of area to promote the rock band Twenty-One Scale Breakers.

Example

Using Area Scale Factor to Determine an Unknown Side

It was just learned that if the length scale factor of two similar figures and one of the areas of the figures are known, then the unknown area can be found. Next, consider if both areas but only the side length of one figure are known. It will be possible to solve for the other similar figure's corresponding side length.

The diagram shows two similar figures. Figure $A$ has an area of $9$ square inches, and Figure $B$ has an area of $25$ square inches.

If a side length of Figure

B

2.5

inches, find the length of the corresponding side in the other shape.

Hint

If the scale factor of two similar figures is $\frac{a}{b},$ then the ratio of their areas is $\frac{a ^{2}}{b ^{2}} .$

Solution

Recall that the ratio of areas of two similar figures is equal to the square of the ratio of their corresponding side lengths.

\underline{Scale Factor} \frac{a}{b} \Rightarrow \underline{Area Scale Factor} (\frac{a}{b})^{2}

Since the areas are given, the following proportion can be written.

(\frac{a}{b})^{2} = \frac{9}{2 5}

Now, take square roots of both sides of the equation to find the value of

\frac{a}{b},

the scale factor. Keep in mind that only the principal roots will be considered because only positive numbers make sense in this situation.

(\frac{a}{b})^{2} = \frac{9}{2 5}

▼

Solve for

\frac{a}{b}

SqrtEqn

$LHS = RHS$

\frac{a}{b} = \frac{9}{2 5}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

\frac{a}{b} = \frac{3}{5}

FracToScale

$\frac{a}{b} = a : b$

a : b = 3 : 5

The scale factor of the figures is

3 : 5 .

Finally, with the scale factor and knowing that the side length of Figure

B

2.5

inches, the length of the corresponding side in Figure

A

represented by

x

can be found.

\frac{3}{5} = \frac{x}{2 . 5}

▼

Solve for

x

MultEqn

$LHS \cdot 2.5 = RHS \cdot 2.5$

\frac{3}{5} \cdot 2.5 = x

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

\frac{7 . 5}{5} = x

CalcQuot

Calculate quotient

1.5 = x

RearrangeEqn

Rearrange equation

x = 1.5

The corresponding length in Figure

A

1.5

inches.

Pop Quiz

Practice Finding Linear Scale Factor Given Areas

Determine the linear scale factor of the shape on the right to the shape on the left.

Discussion

Volume Scale Factor

For similar three-dimensional figures, the volume scale factor and the length scale factor are also related.

If two figures are similar, then the ratio of their volumes is equal to the cube of the ratio of their corresponding side lengths.

Let Solid $A$ and Solid $B$ be similar solids and $V_{1}$ and $V_{2}$ be their respective volumes. The length scale factor between corresponding linear measures is $\frac{a}{b} .$ Given these characteristics, the following conditional statement holds true.

$Solid A \sim Solid B \Rightarrow \frac{V _{1}}{V _{2}} = (\frac{a}{b})^{3}$

Proof

The statement will be proven for similar rectangular prisms, but this proof can be adapted to prove other similar solids. As shown in the diagram, let $a_{1},$ $a_{2},$ and $a_{3}$ be the dimensions of Solid $A$ and $b_{1},$ $b_{2},$ and $b_{3}$ be the dimensions of Solid $B .$

The volume of a rectangular prism is the product of its base area and its height.

Volume of Solid $A$	Volume of Solid $B$
$V_{1} = a_{1} \cdot a_{2} \cdot a_{3}$	$V_{2} = b_{1} \cdot b_{2} \cdot b_{3}$

By the definition of similar solids, the side lengths are proportional and equal to the scale factor

\frac{a}{b} .

⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{a _{1}}{b _{1}} = \frac{a}{b} \frac{a _{2}}{b _{2}} = \frac{a}{b} \frac{a _{3}}{b _{3}} = \frac{a}{b} \Leftrightarrow ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a_{1} = b_{1} \cdot \frac{a}{b} a_{2} = b_{2} \cdot \frac{a}{b} a_{3} = b_{3} \cdot \frac{a}{b}

The next step is to substitute the expressions for

a_{1},

a_{2},

and

a_{3}

into the formula for

V_{1},

the volume of Solid

A .

V_{1} = a_{1} \cdot a_{2} \cdot a_{3}

SubstituteExpressions

Substitute expressions

V_{1} = (b_{1} \cdot \frac{a}{b}) (b_{2} \cdot \frac{a}{b}) (b_{3} \cdot \frac{a}{b})

▼

Simplify right-hand side

RemovePar

Remove parentheses

V_{1} = b_{1} \cdot \frac{a}{b} \cdot b_{2} \cdot \frac{a}{b} \cdot b_{3} \cdot \frac{a}{b}

CommutativePropMult

Commutative Property of Multiplication

V_{1} = \frac{a}{b} \cdot \frac{a}{b} \cdot \frac{a}{b} \cdot b_{1} \cdot b_{2} \cdot b_{3}

ProdToPowThreeFac

$a \cdot a \cdot a = a^{3}$

V_{1} = (\frac{a}{b})^{3} \cdot b_{1} \cdot b_{2} \cdot b_{3}

AssociativePropMult

Associative Property of Multiplication

V_{1} = (\frac{a}{b})^{3} (b_{1} \cdot b_{2} \cdot b_{3})

Notice that the expression on the right-hand side is

(\frac{a}{b})^{3}

times the volume of Solid

B .

V_{1} = (\frac{a}{b})^{3} (b_{1} \cdot b_{2} \cdot b_{3})

Substitute

$b_{1} \cdot b_{2} \cdot b_{3} = V_{2}$

V_{1} = (\frac{a}{b})^{3} V_{2}

DivEqn

$LHS / V_{2} = RHS / V_{2}$

\frac{V _{1}}{V _{2}} = (\frac{a}{b})^{3}

As shown, the ratio of the volumes of the similar prisms is equal to the cube of the ratio of their corresponding linear measures. This ratio is also called the volume scale factor.

\underline{Scale Factor} \frac{a}{b} \Rightarrow \underline{Volume Scale Factor} \frac{V _{1}}{V _{2}} = (\frac{a}{b})^{3}

Example

Using Volume Scale Factor to Determine an Unknown Volume

The scale factor of two similar figures can be used to find the volume of one of the figures when the volume of the other figure is known.

The suitcase company Case-O-La produces snazzy suitcases of various sizes. When a large-sized suitcase is bought, the company offers its cabin-sized version at a discounted price to the same customer.

The large-sized suitcase has a height of

27

inches and a volume of

90

liters. If the cabin-sized suitcase has a height of

18

inches, determine its volume. Round the answer to one decimal place.

Hint

If the scale factor of two similar figures is $\frac{a}{b},$ then the ratio of their volumes is $(\frac{a}{b})^{3} .$

Solution

The suitcases can be considered as two similar rectangular prisms with heights $27$ and $18$ inches.

Similar solids have the same shape and all of their corresponding sides are proportional. The ratio of the corresponding linear dimensions of the similar solids is the scale factor.

\frac{Height of small suitcase}{Height of big suitcase} = \frac{1 8}{2 7}

If the scale factor of two similar solids is

a : b,

then the ratio of their corresponding volumes is

a^{3} : b^{3} .

Now, raise the scale factor to the third power to obtain the ratio of the volumes.

\frac{a}{b} = \frac{1 8}{2 7}

ReduceFrac

$\frac{a}{b} = \frac{a / 9}{b / 9}$

\frac{a}{b} = \frac{2}{3}

RaiseEqn

$LHS^{3} = RHS^{3}$

(\frac{a}{b})^{3} = (\frac{2}{3})^{3}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{a ^{3}}{b ^{3}} = \frac{2 ^{3}}{3 ^{3}}

CalcPow

Calculate power

\frac{a ^{3}}{b ^{3}} = \frac{8}{2 7}

FracToScale

$\frac{a}{b} = a : b$

a^{3} : b^{3} = 8 : 27

The ratio of the volumes is

8 : 27 .

Now, let

V_{1}

be the volume of the small suitcase. Since the volume of the big suitcase is

90

liters, the ratio of

V_{1}

90

8 : 27 .

\frac{V _{1}}{9 0} = \frac{8}{2 7}

▼

Solve for

V_{1}

MultEqn

$LHS \cdot 90 = RHS \cdot 90$

V_{1} = \frac{8}{2 7} \cdot 90

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V_{1} = \frac{7 2 0}{2 7}

CalcQuot

Calculate quotient

V_{1} = 26.6666666 \dots

RoundDec

Round to $1$ decimal place(s)

V_{1} \approx 26.7

The volume of the small suitcase is about

26.7

liters.

Example

Using Volume Scale Factor to Determine an Unknown Side

After reading a physics magazine, Mark feels confident in estimating the radius of the Sun. To do so, he will use the volumes of the Sun and Earth, which are $1.41 \times 1 0^{18}$ and $1.08 \times 1 0^{12}$ cubic kilometers, respectively.

If the radius of the Earth is about $6300$ kilometers, help Mark find the radius of the Sun.

Hint

Sun and Earth can be regarded as two similar spheres. Therefore, the volume scale factor can be used to find the radius of the Sun.

Solution

The Sun and Earth are two similar spheres. Consequently, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear measures, which in this case is the ratio of their radii. Let

r

and

R

be the radii of the Earth and Sun, respectively.

\underline{Scale Factor} \frac{r}{R} \Rightarrow \underline{Ratio of the Volumes} (\frac{r}{R})^{3}

Given that the volumes are known, the volume scale factor can be used to find the scale factor of the Earth to the Sun.

(\frac{r}{R})^{3} = \frac{1 . 0 8 \times 1 0 ^{12}}{1 . 4 1 \times 1 0 ^{18}}

Next, take the cube roots of both sides of the equation to find the value of

\frac{r}{R},

the length scale factor.

(\frac{r}{R})^{3} = \frac{1 . 0 8 \times 1 0 ^{12}}{1 . 4 1 \times 1 0 ^{18}}

▼

Solve for

\frac{a}{b}

WriteProdFrac

Write as a product of fractions

(\frac{r}{R})^{3} = (\frac{1 . 0 8}{1 . 4 1}) (\frac{1 0 ^{12}}{1 0 ^{18}})

DivPow

$\frac{a ^{m}}{a ^{n}} = a^{m - n}$

(\frac{r}{R})^{3} = (\frac{1 . 0 8}{1 . 4 1}) (1 0^{- 6})

RadicalEqn

$3 LHS = 3 RHS$

\frac{r}{R} = 3 (\frac{1 . 0 8}{1 . 4 1}) (1 0^{- 6})

RootProd

$3 a \cdot b = 3 a \cdot 3 b$

\frac{r}{R} = 3 \frac{1 . 0 8}{1 . 4 1}

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Catch-Up and Review

Proof

Hint

Solution

Hint

Solution

Proof

Hint

Solution

Hint

Solution