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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.**

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.

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?

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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 $KLMN$ and $PQRS$ be similar figures, and $A_{1}$ and $A_{2}$ be their respective areas. The length scale factor between corresponding side lengths is $ba .$ In that case, the following conditional statement holds true.

$KLMN∼PQRS⇒A_{2}A_{1} =(ba )_{2}$

The theorem will be proven for similar rectangles. This proof can be adapted to other similar figures.

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

Area of $KLMN$ | Area of $PQRS$ |
---|---|

$A_{1}=KL⋅LM$ | $A_{2}=PQ⋅QR$ |

$⎩⎪⎪⎪⎨⎪⎪⎪⎧ PQKL =ba QRLM =ba ⇔⎩⎪⎪⎨⎪⎪⎧ KL=PQ⋅ba LM=QR⋅ba $

The next step is to substitute the expressions for $KL$ and $LM$ into the formula for $A_{1},$ which represents the area of $KLMN.$ $A_{1}=KL⋅LM$

SubstituteII

$KL=PQ⋅ba $, $LM=QR⋅ba $

$A_{1}=(PQ⋅ba )(QR⋅ba )$

Simplify right-hand side

RemovePar

Remove parentheses

$A_{1}=PQ⋅ba ⋅QR⋅ba $

CommutativePropMult

Commutative Property of Multiplication

$A_{1}=ba ⋅ba ⋅PQ⋅QR$

ProdToPowTwoFac

$a⋅a=a_{2}$

$A_{1}=(ba )_{2}⋅PQ⋅QR$

AssociativePropMult

Associative Property of Multiplication

$A_{1}=(ba )_{2}(PQ⋅QR)$

$A_{1}=(ba )_{2}(PQ⋅QR)$

Substitute

$PQ⋅QR=A_{2}$

$A_{1}=(ba )_{2}A_{2}$

DivEqn

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

$A_{2}A_{1} =(ba )_{2}$

$Scale Factor ba ⇒ Area Scale Factor A_{2}A_{1} =(ba )_{2} $

Dominika is making a flyer for a concert and wants to compare the areas of the two pieces of paper. The widths of $A4$ and $A2$ papers are $21$ and $42$ centimeters, respectively.

If these two pieces of paper are similar and the area of $A4$ paper is about $624$ square centimeters, find the area of the $A2$ paper. Round the answer to the nearest integer.

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If the scale factor of two similar figures is $ba ,$ then the ratio of their areas is $b_{2}a_{2} .$

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:4221 =21 $

Using this information, the ratio of the areas, or area scale factor, can be calculated by the theorem about the areas of similar figures.
$Scale Factor 21 ⇒ Area Scale Factor 2_{2}1_{2} =41 $

Now, let $A_{2}$ be the area of the piece of $A2$ paper. Then, a proportion can be written using the area scale factor and the area of the $A4$ paper, which is $624$ square centimeters.
$A_{2}624 =41 $

Solve for $A_{2}$

MultEqn

$LHS⋅A_{2}=RHS⋅A_{2}$

$A_{2}624 ⋅A_{2}=41 ⋅A_{2}$

FracMultDenomToNumber

$A_{2}a ⋅A_{2}=a$

$624=41 ⋅A_{2}$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$624=4A_{2} $

MultEqn

$LHS⋅4=RHS⋅4$

$2496=A_{2}$

RearrangeEqn

Rearrange equation

$A_{2}=2496$

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 $B$ is $2.5$ inches, find the length of the corresponding side in the other shape.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.720703125em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">in<\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span><\/span>","answer":{"text":["1.5","3\/2"]}}

If the scale factor of two similar figures is $ba ,$ then the ratio of their areas is $b_{2}a_{2} .$

Recall that the ratio of areas of two similar figures is equal to the square of the ratio of their corresponding side lengths.
The corresponding length in Figure $A$ is $1.5$ inches.

$Scale Factor ba ⇒ Area Scale Factor (ba )_{2} $

Since the areas are given, the following proportion can be written.
$(ba )_{2}=259 $

Now, take square roots of both sides of the equation to find the value of $ba ,$ the scale factor. Keep in mind that only the principal roots will be considered because only positive numbers make sense in this situation.
The scale factor of the figures is $3:5.$ Finally, with the scale factor and knowing that the side length of Figure $B$ is $2.5$ inches, the length of the corresponding side in Figure $A$ represented by $x$ can be found.
$53 =2.5x $

Solve for $x$

MultEqn

$LHS⋅2.5=RHS⋅2.5$

$53 ⋅2.5=x$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$57.5 =x$

CalcQuot

Calculate quotient

$1.5=x$

RearrangeEqn

Rearrange equation

$x=1.5$

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

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 linear measures.

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 $ba .$ Given those characteristics, the following conditional statement holds true.

$SolidA∼SolidB⇒V_{2}V_{1} =(ba )_{3}$

The theorem will be proven for similar rectangular prisms. Take into consideration that the 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 volume of a rectangular prism is the product of its base area and its height.

Volume of Solid $A$ | Volume of Solid $B$ |
---|---|

$V_{1}=a_{1}⋅a_{2}⋅a_{3}$ | $V_{2}=b_{1}⋅b_{2}⋅b_{3}$ |

$⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧ b_{1}a_{1} =ba b_{2}a_{2} =ba b_{3}a_{3} =ba ⇔⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧ a_{1}=b_{1}⋅ba a_{2}=b_{2}⋅ba a_{3}=b_{3}⋅ba $

The next step will be to substitute the expressions for $a_{1},$ $a_{3},$ and $a_{4}$ into the formula for $V_{1},$ the volume of Solid $A.$ $V_{1}=a_{1}⋅a_{2}⋅a_{3}$

SubstituteExpressions

Substitute expressions

$V_{1}=(b_{1}⋅ba )(b_{2}⋅ba )(b_{3}⋅ba )$

Simplify right-hand side

RemovePar

Remove parentheses

$V_{1}=b_{1}⋅ba ⋅b_{2}⋅ba ⋅b_{3}⋅ba $

CommutativePropMult

Commutative Property of Multiplication

$V_{1}=ba ⋅ba ⋅ba ⋅b_{1}⋅b_{2}⋅b_{3}$

ProdToPowThreeFac

$a⋅a⋅a=a_{3}$

$V_{1}=(ba )_{3}⋅b_{1}⋅b_{2}⋅b_{3}$

AssociativePropMult

Associative Property of Multiplication

$V_{1}=(ba )_{3}(b_{1}⋅b_{2}⋅b_{3})$

$V_{1}=(ba )_{3}(b_{1}⋅b_{2}⋅b_{3})$

Substitute

$b_{1}⋅b_{2}⋅b_{3}=V_{2}$

$V_{1}=(ba )_{3}V_{2}$

DivEqn

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

$V_{2}V_{1} =(ba )_{3}$

$Scale Factor ba ⇒ Volume Scale Factor V_{2}V_{1} =(ba )_{3} $

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"L","answer":{"text":["26.7"]}}

If the scale factor of two similar figures is $ba ,$ then the ratio of their volumes is $(ba )_{3}.$

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.$Height of big suitcaseHeight of small suitcase =2718 $

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.
$ba =2718 $

ReduceFrac

$ba =b/9a/9 $

$ba =32 $

RaiseEqn

$LHS_{3}=RHS_{3}$

$(ba )_{3}=(32 )_{3}$

PowQuot

$(ba )_{m}=b_{m}a_{m} $

$b_{3}a_{3} =3_{3}2_{3} $

CalcPow

Calculate power

$b_{3}a_{3} =278 $

FracToScale

$ba =a:b$

$a_{3}:b_{3}=8:27$

$90V_{1} =278 $

Solve for $V_{1}$

MultEqn

$LHS⋅90=RHS⋅90$

$V_{1}=278 ⋅90$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V_{1}=27720 $

CalcQuot

Calculate quotient

$V_{1}=26.6666666…$

RoundDec

Round to $1$ decimal place(s)

$V_{1}≈26.7$

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 $1.41×10_{18}$ and $1.08×10_{12}$ cubic kilometers, respectively.

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

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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.

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.

$Scale Factor Rr ⇒ Ratio of the Volumes (Rr )_{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. $(Rr )_{3}=1.41×10_{18}1.08×10_{12} $

Next, take the cube roots of both sides of the equation to find the value of $Rr ,$ the length scale factor.
$(Rr )_{3}=1.41×10_{18}1.08×10_{12} $

Solve for $ba $

WriteProdFrac

Write as a product of fractions

$(Rr )_{3}=(1.411.08 )(10_{18}10_{12} )$

DivPow

$a_{n}a_{m} =a_{m−n}$

$(Rr )_{3}=(1.411.08 )(10_{-6})$

RadicalEqn

$3LHS =3RHS $

$Rr =3(1.411.08 )(10_{-6}) $

RootProd

$3a⋅b =3a ⋅3b $

$Rr =31.411.08 $