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.

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Catch-Up and Review

Proof

Hint

Solution

Proof

Hint

Solution