Level 3 - Changes in Dimensions - Transformations, Congruence, and Similarity (Pre-Algebra)

Many people enjoy collecting miniature versions of famous characters or historical structures. These tiny copies are scale models of the originals — in other words, the model and the original figure are similar. This lesson will cover the characteristics of similar figures and explain how to calculate their measurements based on a specific scale.

Catch-Up and Review

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

Area
Perimeter
Area of a Triangle
Similar Polygons
Scale Factor
Volume
Surface Area
Pythagorean Theorem
Surface Area of a Sphere

Explore

How Do the Dimensions of Similar Figures Change After a Scale?

Tadeo is interested in making miniature models of various objects. He wants to know more about how the dimensions of similar figures change depending on the scale. Consider the applet below, which displays two similar triangles.

Tadeo changes the length scale factor (lsf) and examines how the area of the blue triangle changes.

What conclusions could Tadeo draw from the applet?
How are the scale factor and the ratio between the areas of the triangles related?

Discussion

Relations Between the Similar Polygons

Two polygons are said to be similar if their corresponding sides are proportional and their corresponding angles are congruent. Because of this, there is a relation between the perimeters of similar polygons.

Rule

Perimeters of Similar Polygons Theorem

If two polygons are similar, then the ratio of their perimeters is equal to the ratio of their corresponding side lengths.

Let $P_{1}$ and $P_{2}$ be the perimeters of $Q R S T$ and $A B C D,$ respectively. Let $\frac{a}{b}$ be the scale factor between corresponding side lengths. Then, based on the above diagram, the following relation holds true.

$A B C D \sim Q R S T \Rightarrow \frac{P _{1}}{P _{2}} = \frac{a}{b}$

Proof

Let

Q R S T

and

A B C D

be two similar polygons with perimeters

P_{1}

and

P_{2},

respectively. By the definition of similar polygons, corresponding side lengths are proportional, with the scale factor

\frac{a}{b}

as the common ratio.

⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{Q R}{A B} = \frac{a}{b} \frac{R S}{B C} = \frac{a}{b} \frac{S T}{C D} = \frac{a}{b} \frac{T Q}{D A} = \frac{a}{b} \Leftrightarrow ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ Q R = \frac{a}{b} \cdot A B R S = \frac{a}{b} \cdot B C S T = \frac{a}{b} \cdot C D T Q = \frac{a}{b} \cdot D A

If the equations are added together, the left-hand side of the resulting equation will give the perimeter of

Q R S T .

The right-hand side will give the scale factor times the perimeter of

A B C D .

Q R + R S + S T + T Q = \frac{a}{b} \cdot A B + \frac{a}{b} \cdot B C + \frac{a}{b} \cdot C D + \frac{a}{b} \cdot D A

FactorOut

Factor out $\frac{a}{b}$

Q R + R S + S T + T Q = \frac{a}{b} (A B + B C + C D + D A)

SubstituteII

$Q R + R S + S T + T Q = P_{1}$ , $A B + B C + C D + D A = P_{2}$

P_{1} = \frac{a}{b} \cdot P_{2}

Finally, the Perimeters of Similar Polygons Theorem is obtained by dividing both sides by

P_{2}

$P_{1} = \frac{a}{b} \cdot P_{2} \Leftrightarrow \frac{P _{1}}{P _{2}} = \frac{a}{b}$

Example

Making a Miniature Model of a Basketball Court

Tadeo likes playing basketball. He decides to make a miniature model of a basketball court with a length of $32$ centimeters.

External credits: freepik

A standard basketball court has a length of $28$ meters and a width of $15$ meters.

a Calculate the scale factor in this model. Write the answer in the simplest fraction form.

b Calculate the perimeter of the miniature basketball court. Round the answer to the nearest integer.

Hint

a The ratio of the length of the model basketball court to the actual length is equal to the length scale factor.

b If two polygons are similar, then the ratio of their perimeters is equal to their length scale factor.

Solution

a The miniature model of the basketball court will be a scale model of a real court, so the model and its original can be thought of as similar polygons. Therefore, their side lengths will be proportional and equal to the scale factor.

Scale factor = \frac{Length of model}{Actual length}

Before the ratio can be calculated, both lengths need to have the same units of measure. The length of the real basketball court is

28

meters and the length of the scale model is

32

centimeters. Convert meters to centimeters to have same units of measure.

28 m \cdot \frac{1 0 0 cm}{1 m} = 2800 cm

Next, calculate the ratio of the length of the model to the corresponding actual length in centimeters.

Scale factor = \frac{Length of model}{Actual length}

SubstituteValues

Substitute values

Scale factor = \frac{3 2}{2 8 0 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 6}{b / 1 6}$

Scale factor = \frac{2}{1 7 5}

The scale factor of the model will be

\frac{2}{1 7 5}

2 : 175 .

b To calculate the perimeter of the model, first find the perimeter of the real basketball court.

Recall that the perimeter of a rectangle is two times the sum of its length and width.

P = 2 (15 + 28) = 86 m

The perimeter of the real basketball court is

86

meters. Since the miniature model of the court and the real court are similar rectangles, the ratio of their perimeters is equal to the ratio of their corresponding side lengths, or the scale factor.

\frac{P _{1}}{P _{2}} = Scale Factor

The scale factor is

\frac{2}{1 7 5} .

Substitute the scale factor and the perimeter of the real basketball court into the equation and solve for

P_{1} .

\frac{P _{1}}{P _{2}} = Scale factor

SubstituteII

$P_{2} = 86$ , $Scale factor = \frac{2}{1 7 5}$

\frac{P _{1}}{8 6} = \frac{2}{1 7 5}

MultEqn

$LHS \cdot 86 = RHS \cdot 86$

P_{1} = \frac{2}{1 7 5} \cdot 86

MoveRightFacToNum

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

P_{1} = \frac{1 7 2}{1 7 5}

CalcQuot

Calculate quotient

P_{1} = 0.982857 \dots

RoundDec

Round to $2$ decimal place(s)

P_{1} \approx 0.98

The perimeter of the model is about

0.98

meters. Finally, convert it to centimeters.

0.98 m \cdot \frac{1 0 0 cm}{1 m} = 98 cm

The model basketball court has a perimeter of about

98

centimeters.

Discussion

Areas of Similar Figures

Like with perimeters, there is a relation between the areas of similar polygons.

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}

Example

Making a Miniature Model of a Spectator Stand

Now Tadeo wants to build a miniature spectator stand for his miniature basketball stadium. He plans to use the following stand as a model.

a spectator stand with two lights; when viewed from the side, the spectator stand can be modeled as a right triangle with a leg of 3 m and a hypotenuse of 5 m.

External credits: macrovector

When viewed from the side, the stand looks like a right triangle. Tadeo knows that its hypotenuse is

5

meters long and its height is

3

meters.

a Tadeo makes his scale model stand with a height of

12

centimeters. Calculate the scale factor in this model. Write the answer in the simplest fraction form.

b Calculate the triangular side area of the miniature spectator stand.

Hint

a The ratio of the hypotenuse of the model spectator stand to the actual hypotenuse is equal to the length scale factor.

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

Solution

a The real spectator stand looks like a right triangle from a side perspective.

The scale model of the stand and the real stand have to be similar figures, so they have the same ratio between their corresponding side lengths. This ratio is called the scale factor.

Scale factor = \frac{Length of model}{Actual length}

Tadeo wants to create the model with a

12 -

centimeter height. He knows that the height of the real stand is

3

meters. Before finding the scale factor, both lengths must be in the same units. To achieve this, convert the height of the real stands from meters to centimeters.

3 m \cdot \frac{1 0 0 cm}{1 m} = 300 cm

Now the ratio of the height of the model to the corresponding actual height can be calculated.

Scale factor = \frac{Length of model}{Actual length}

SubstituteValues

Substitute values

Scale factor = \frac{1 2}{3 0 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 2}{b / 1 2}$

Scale factor = \frac{1}{2 5}

The scale factor of the model will be

\frac{1}{2 5},

1 : 25 .

This means that the dimensions of the model stand will be

25

times smaller than the dimensions of the real stand.

b Before calculating the side area of the model spectator stand, calculate the side area of the actual stand. This requires finding the length of the other leg.

Since it is a right triangle, the Pythagorean theorem can be used to find

x .

x^{2} + y^{2} = z^{2}

SubstituteII

$y = 3$ , $z = 5$

x^{2} + (3)^{2} = (5)^{2}

▼

Solve for

x

CalcPow

Calculate power

x^{2} + 9 = 25

SubEqn

$LHS - 9 = RHS - 9$

x^{2} = 16

SqrtEqn

$LHS = RHS$

x^{2} = 16

SqrtPowToNumber

$a^{2} = a$

x = 16

CalcPow

Calculate power

x = 4

The base length of the triangular figure is

4

meters. Now, calculate the area of the triangle as half of the product of the base and height.

A = \frac{1}{2} b h

SubstituteII

$b = 4$ , $h = 3$

A = \frac{1}{2} \cdot 4 \cdot 3

Multiply

A = \frac{1}{2} \cdot 12

MoveRightFacToNumOne

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

A = \frac{1 2}{2}

CalcQuot

Calculate quotient

A = 6

The area of the triangular side of the real stand is

6

square meters. Recall that if two figures are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding side lengths.

\frac{A _{1}}{A _{2}} = (\frac{ℓ _{1}}{ℓ _{2}})^{2} = k^{2}

The ratio of the corresponding side lengths is equal to the length scale factor, so in this case,

k = \frac{1}{2 5} .

The square of this scale factor can be used to find the area of the model's triangular side.

\frac{A _{1}}{A _{2}} = k^{2}

SubstituteII

$A_{2} = 6$ , $k = \frac{1}{2 5}$

\frac{A _{1}}{6} = (\frac{1}{2 5})^{2}

▼

Solve for

A_{1}

PowQuot

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

\frac{A _{1}}{6} = \frac{1 ^{2}}{2 5 ^{2}}

CalcPow

Calculate power

\frac{A _{1}}{6} = \frac{1}{6 2 5}

MultEqn

$LHS \cdot 6 = RHS \cdot 6$

A_{1} = \frac{1}{6 2 5} \cdot 6

MoveRightFacToNumOne

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

A_{1} = \frac{6}{6 2 5}

CalcQuot

Calculate quotient

A_{1} = 0.0096

The area of the model's triangular side is

0.0096

square meters. Finally, convert it to square centimeters.

0.0096 m^{2} \cdot \frac{1 0 0 0 0 cm ^{2}}{1 m ^{2}} = 96 cm^{2}

The area of the triangular side of Tadeo's model is

96

square centimeters.

Pop Quiz

Finding the Length Scale, Perimeter, or Area of Similar Polygons

Consider two similar polygons. Use the given information to find the scale factor, perimeter, or area of either of the polygons. Keep in mind that the given length scale factor corresponds to the ratio of Polygon $2$ to Polygon $1 .$ Round the answer to two decimal places if necessary.

Applet that randomly generates the two polygons with the given scale factor or the length of one figure.

Discussion

Surface Areas of Similar Solids

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

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

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

Proof

The statement will be proven for similar rectangular prisms, but this proof can be adapted to prove other similar solids as well. 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 surface area of a rectangular prism is the sum of the lateral area and the combined areas of the two identical bases. The lateral area of a rectangular prism consists of its four rectangular side areas. Notice that the areas of opposite faces are congruent.

Surface Area of Solid $A$	Surface Area of Solid $B$
$S A_{1} = 2 (a_{1} \cdot a_{2} + a_{1} \cdot a_{3} + a_{2} \cdot a_{3})$	$S A_{2} = 2 (b_{1} \cdot b_{2} + b_{1} \cdot b_{3} + b_{2} \cdot b_{3})$

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

S A_{1},

the surface area of Solid

A .

S A_{1} = 2 (a_{1} \cdot a_{2} + a_{1} \cdot a_{3} + a_{2} \cdot a_{3})

SubstituteExpressions

Substitute expressions

S A_{1} = 2 ((b_{1} \cdot \frac{a}{b} \cdot b_{2} \cdot \frac{a}{b}) + (b_{1} \cdot \frac{a}{b} \cdot b_{3} \cdot \frac{a}{b}) + (b_{2} \cdot \frac{a}{b} \cdot b_{3} \cdot \frac{a}{b}))

▼

Simplify right-hand side

CommutativePropMult

Commutative Property of Multiplication

S A_{1} = 2 ((\frac{a}{b} \cdot \frac{a}{b} \cdot b_{1} \cdot b_{2}) + (\frac{a}{b} \cdot \frac{a}{b} \cdot b_{1} \cdot b_{3}) + (\frac{a}{b} \cdot \frac{a}{b} \cdot b_{2} \cdot b_{3}))

RemovePar

Remove parentheses

S A_{1} = 2 (\frac{a}{b} \cdot \frac{a}{b} \cdot b_{1} \cdot b_{2} + \frac{a}{b} \cdot \frac{a}{b} \cdot b_{1} \cdot b_{3} + \frac{a}{b} \cdot \frac{a}{b} \cdot b_{2} \cdot b_{3})

ProdToPowTwoFac

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

S A_{1} = 2 ((\frac{a}{b})^{2} \cdot b_{1} \cdot b_{2} + (\frac{a}{b})^{2} \cdot b_{1} \cdot b_{3} + (\frac{a}{b})^{2} \cdot b_{2} \cdot b_{3})

FactorOut

Factor out $(\frac{a}{b})^{2}$

S A_{1} = (\frac{a}{b})^{2} \cdot 2 (b_{1} \cdot b_{2} + b_{1} \cdot b_{3} + b_{2} \cdot b_{3})

Notice that the expression on the right-hand side is

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

times the surface area of Solid

B .

S A_{1} = (\frac{a}{b})^{2} \cdot 2 (b_{1} \cdot b_{2} + b_{1} \cdot b_{3} + b_{2} \cdot b_{3})

Substitute

$2 (b_{1} \cdot b_{2} + b_{1} \cdot b_{3} + b_{2} \cdot b_{3}) = S A_{2}$

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

DivEqn

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

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

As shown, the ratio of the surface areas of the similar prisms is equal to the square of the ratio of their corresponding linear measures.

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

Example

Designing a Miniature Basketball

What does a basketball court need if not a basketball? Tadeo turns his attention to designing a miniature basketball for his miniature stadium.

a basketball with a radius of 4.7 in and the length scale of the basketbals is 3:19

External credits: freepik

He knows that the radius of a real basketball is $4.7$ inches.

a If Tadeo determines the length scale factor of the miniature ball to the real ball to be

3 : 19,

what will the surface area of the miniature ball be? Round the answer to the nearest integer.

b Calculate the radius of the miniature ball. Round the answer to two decimal places.

Hint

a Use the formula for the surface area of a sphere. If two figures are similar, then the ratio of their surface areas is equal to the square of the ratio of their corresponding side lengths.

b The ratio of the radius of the model ball to the radius of the actual ball is equal to the length scale factor.

Solution

a Since the model basketball is a scale model of a real one, the real basketball and the model are similar spheres. As such, the ratio of their surface areas is equal to the square of the ratio of their corresponding side lengths, or the scale factor

k .

\frac{S A _{1}}{S A _{2}} = k^{2}

While a sphere does not have any sides, its radius can be considered as a side measurement for the scale factor. Tadeo first wants to find the surface area of the real ball. Remember the formula for the surface area of a sphere.

Surface Area of a Sphere 4 π r^{2}

Since the radius of the real basketball is given to be

4.7

inches, substitute

r = 4.7

into the formula and simplify to find the surface area of the basketball.

4 π r^{2}

Substitute

$r = 4.7$

4 π (4.7)^{2}

CalcPow

Calculate power

4 π (22.09)

Multiply

88.36 π

The surface area of the real basketball is

88.36 π .

Next, use the square of the given scale factor to find the surface area of the miniature ball. Since the scale factor is given as the ratio of the miniature ball to the real ball,

S A_{1}

should corresponds to the surface area of the miniature ball.

\frac{S A _{1}}{S A _{2}} = k^{2}

SubstituteII

$S_{2} = 88.36 π$ , $k = \frac{3}{1 9}$

\frac{S A _{1}}{8 8 . 3 6 π} = (\frac{3}{1 9})^{2}

▼

Solve for

S A_{1}

PowQuot

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

\frac{S A _{1}}{8 8 . 3 6 π} = \frac{3 ^{2}}{1 9 ^{2}}

CalcPow

Calculate power

\frac{S A _{1}}{8 8 . 3 6 π} = \frac{9}{3 6 1}

MultEqn

$LHS \cdot 88.36 π = RHS \cdot 88.36 π$

S A_{1} = \frac{9}{3 6 1} \cdot 88.36 π

MoveRightFacToNum

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

S A_{1} = \frac{9 \cdot 8 8 . 3 6 π}{3 6 1}

UseCalc

Use a calculator

S A_{1} = 6.920554 \dots

RoundInt

Round to nearest integer

S A_{1} \approx 7

The surface area of the miniature ball will be about

7

square inches.

b Once again, use the scale factor to find the radius of the miniature ball. Since the miniature model of the ball and the real ball are similar figures, the ratio between their radii is equal to the scale factor.

\frac{r _{1}}{r _{2}} = Scale Factor

The radius of the real basketball is

4.7

inches and the length scale factor is

3 : 19 .

Substitute these values into the equation and solve it for

r_{1} .

\frac{r _{1}}{r _{2}} = Scale Factor

SubstituteII

$r_{2} = 4.7$ , $Scale Factor = \frac{3}{1 9}$

\frac{r _{1}}{4 . 7} = \frac{3}{1 9}

▼

Solve for

r_{1}

MultEqn

$LHS \cdot 4.7 = RHS \cdot 4.7$

r_{1} = \frac{3}{1 9} \cdot 4.7

MoveRightFacToNum

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

r_{1} = \frac{3 \cdot 4 . 7}{1 9}

Multiply

r_{1} = \frac{1 4 . 1}{1 9}

CalcQuot

Calculate quotient

r_{1} = 0.742105 \dots

RoundDec

Round to $2$ decimal place(s)

r_{1} \approx 0.74

The radius of the miniature ball will be about

0.74

inches.

Discussion

Volumes of Similar Solids

As with side lengths and perimeters, there is a relation between the volumes of the similar figures.

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

Designing a Miniature Model of a Stadium

Finally, Tadeo plans to model the exterior of his miniature stadium after his favorite basketball team's stadium.

The actual stadium has a volume of

6000

cubic meters. Calculate the volume of the miniature stadium if he uses the length scale of

7 : 122 .

Round the answer to two decimal places.

Hint

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

Solution

Since the miniature stadium is a scale model, the actual stadium and the miniature stadium are similar solids. Recall that the ratio of the volumes of similar solids is equal to the cube of the length scale factor

k .

Note that the scale factor is equal to the ratio of the corresponding side lengths.

\frac{V _{1}}{V _{2}} = (\frac{ℓ _{1}}{ℓ _{2}})^{3} = k^{3}

The length scale factor is

k = 7 : 122

and the volume of the real stadium is

6000

cubic meters. Then, the volume of the miniature stadium can be found by substituting

V_{2} = 6000

and

k = 7 : 122

into the formula and solving for

V_{1} .

\frac{V _{1}}{V _{2}} = k^{3}

SubstituteII

$V_{2} = 6000$ , $k = \frac{7}{1 2 2}$

\frac{V _{1}}{6 0 0 0} = (\frac{7}{1 2 2})^{3}

▼

Solve for

V_{1}

PowQuot

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

\frac{V _{1}}{6 0 0 0} = \frac{3 4 3}{1 8 1 5 8 4 8}

MultEqn

$LHS \cdot 6000 = RHS \cdot 6000$

V_{1} = \frac{3 4 3}{1 8 1 5 8 4 8} \cdot 6000

MoveRightFacToNum

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

V_{1} = \frac{2 0 5 8 0 0 0}{1 8 1 5 8 4 8}

CalcQuot

Calculate quotient

V_{1} = 1.133354 \dots

RoundDec

Round to $2$ decimal place(s)

V_{1} \approx 1.13

The volume of the miniature stadium will be about

1.13

cubic meters.

Closure

Miniature Basketball Players

Tadeo wants to complete his miniature basketball stadium with the tiny basketball players. As he places his collection of action figures of his favorite team, he thinks about the real heights and weights of the players. For one particular player, the toy figure is $3.1$ centimeters tall, while the real life counterpart player is about $2.06$ meters tall.

Tadeo supposes that the if bodies of the action figure and the real player can be modeled with two similar solids, then the weights of similar figures is related by the cube of the scale factor

k .

\frac{w _{1}}{w _{2}} = k^{3}

If the action figure weighs

0.02

pounds, is Tadeo correct?

Hint

What is the length scale factor between the action figure and the human basketball player? Find the basketball player's weight using this equation. Does it make sense for a basketball player to weigh this much?

Solution

Start by determining the length scale factor, the ratio of the given lengths. Both lengths need to have the same units of measure, so remember to convert meters to centimeters using the conversion factor

\frac{1 0 0 cm}{1 m} .

2.06 m \cdot \frac{1 0 0 cm}{1 m} = 206 cm

Now the two centimeter lengths can be used to write the length scale factor

k .

k = \frac{3 . 1}{2 0 6}

Tadeo claims that the weights of the action figure and the human player are proportional to the cube of the scale factor

k

because they are similar solids.

\frac{w _{1}}{w _{2}} = k^{3}

Calculate the weight of the player using this information. The action figure of player weighs

0.02

pounds, so substitute the weight of the action figure and the scale factor into the equation.

\frac{w _{1}}{w _{2}} = k^{3}

SubstituteII

$V_{1} = 0.02$ , $k = \frac{3 . 1}{2 0 6}$

\frac{0 . 0 2}{w _{2}} = (\frac{3 . 1}{2 0 6})^{3}

▼

Solve for

w_{2}

CalcPow

Calculate power

\frac{0 . 0 2}{w _{2}} = \frac{2 9 . 7 9 1}{8 7 4 1 8 1 6}

CrossMult

Cross multiply

0.02 \cdot 8741816 = 29.791 \cdot w_{2}

DivEqn

$LHS / 29.791 = RHS / 29.791$

\frac{0 . 0 2 \cdot 8 7 4 1 8 1 6}{2 9 . 7 9 1} = w_{2}

UseCalc

Use a calculator

5868.763049 \dots = w_{2}

RoundInt

Round to nearest integer

5869 \approx w_{2}

RearrangeEqn

Rearrange equation

w_{2} \approx 5869

According to these calculations, the human basketball player must weigh about

5869

pounds! 😱 Tadeo knows that even the heaviest NBA player of all time weighed about

375

pounds.

Calculated Weight of Similar Figures 5869 Maximum Actual Weight 375

The calculated value does not make sense because the density of an action figure and the density of a person are different. This means that the weights of the action figure and the player are not directly related to the volume scale factor.

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Catch-Up and Review

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