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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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. The 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 $PQRS$ |
---|---|

$A_{1}=KL⋅LM$ | $A_{2}=PQ⋅QR$ |

$A_{1}=KL⋅LM$

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.
$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.
The corresponding length in Figure $A$ is $1.5$ inches.

$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 $B$ |
---|---|

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

$V_{1}=a_{1}⋅a_{2}⋅a_{3}$

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 ${\textstyle 1 \, \ifnumequal{1}{1}{\text{decimal}}{\text{decimals}}}$

$V_{1}≈26.7$

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.
Finally, with the scale factor and knowing that the radius of the Earth is about $6300$ kilometers, the radius of the Sun $R$ can be found. The ratio of $6300$ to $R$ is equal to the scale factor.
The radius of Sun is about $6.92×10_{5},$ or $692000,$ kilometers.

$(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 ⋅310_{-6} $

UseCalc

Use a calculator

$Rr =(0.914958…)(10_{-2})$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$Rr ≈(0.91)(10_{-2})$

Write in scientific notation

$Rr ≈9.1×10_{-3}$

$R6300 =9.1×10_{-3}$

Solve for $R$

MultEqn

$LHS⋅R=RHS⋅R$

$6300=(9.1×10_{-3})(R)$

DivEqn

$LHS/(9.1×10_{-3})=RHS/(9.1×10_{-3})$

$9.1×10_{-3}6300 =R$

WriteProdFrac

Write as a product of fractions

$(9.16300 )(10_{-3}1 )=R$

CalcQuot

Calculate quotient

$(692.307692…)(10_{-3}1 )=R$

FracToNegExponent

$a_{m}1 =a_{-m}$

$(692.307692…)(10_{3})=R$

RearrangeEqn

Rearrange equation

$R=(692.307692…)(10_{3})$

RoundInt

Round to nearest integer

$R≈(692)(10_{3})$

Write in scientific notation

$R≈6.92×10_{5}$

The applet shows the volumes of two similar solids. Determine the scale factor of the solid on the right to the solid on the left.

The corresponding faces of two similar three-dimensional figures are also similar. Subsequently, the ratio of the areas of the corresponding faces is proportional to the square of the length scale factor of the figures.

Dylan has a golden retriever and a chihuahua. He buys two similar doghouses, whose corresponding side lengths are proportional. When he is about to finish painting the doghouses, he realizes that there is not enough paint for the front face of the small doghouse!

Dylan knows the volumes of each doghouse. They are about $63500$ and $34500$ cubic inches. He also knows that the front face of the big doghouse has an area of $650$ square inches. Help Dylan determine the area of the front face of the small doghouse. This will help him determine how much more paint to buy. Round the answer to the nearest integer.

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Start by finding the length scale factor. Then, use the ratio for the areas of similar figures.

The length scale factor will be found first. To do so, the cube root of the volume scale factor will be calculated. Recall that the ratio of volumes of two similar solids is equal to the cube of the ratio of their corresponding side lengths.
$Scale Factor ba ⇒ Volume Scale Factor (ba )_{3} $
A proportion can be written using the ratio of the volumes.
$(ba )_{3}=6350034500 $
Now, take cube roots of both sides of the equation to find the value of $ba ,$ the length scale factor.
The area scale factor can be found by squaring the scale factor for length.
$Scale Factor 0.82 ⇒ Area Scale Factor 0.82_{2} $
Finally, knowing that the larger doghouse's front face area is about $650$ square inches, the corresponding area in the other one can be calculated. Let $A_{1}$ be that area. Then, $0.82_{2}$ should be equal to the ratio of $A_{1}$ to $650.$
The smaller doghouse has a front face with an area about $437$ square inches. He can now go shopping for more paint to appease his cool chihuahua.

$(ba )_{3}=6350034500 $

Solve for $ba $

RadicalEqn

$3LHS =3RHS $

$ba =36350034500 $

CalcQuot

Calculate quotient

$ba =30.543307… $

UseCalc

Use a calculator

$ba =0.815984…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$ba ≈0.82$

$0.82_{2}=650A_{1} $

Solve for $A_{1}$

MultEqn

$LHS⋅650=RHS⋅650$

$0.82_{2}⋅650=A_{1}$

CalcPow

Calculate power

$0.6724⋅650=A_{1}$

Multiply

Multiply

$437.06=A_{1}$

RearrangeEqn

Rearrange equation

$A_{1}=437.06$

RoundInt

Round to nearest integer

$A_{1}≈437$

A three-dimensional figure is called a composite solid if it is the combination of two or more solids.

Like similar solids, if the corresponding linear measures of two composite solids are proportional, the composite solids are said to be similar. Therefore, their length scale factor can be determined and used to find certain characteristics of the shapes.

The amount of material used to construct the larger silo is three times that of the smaller one. That is, the surface area of the larger silo is three times as large as the surface area of the smaller silo. These two silos can be considered to be similar solids. Furthermore, each silo is composed of a cone and a cylinder as shown.

If the volume of the larger silo is $6750$ cubic meters, find the volume of the smaller silo. Round the answer to the nearest integer.{"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:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">m<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["1299"]}}

Use the area scale factor to determine the length scale factor.

The given silos are similar composite solids. Since they are similar, their corresponding linear measures are proportional. Therefore, each silo can be considered as a whole. To find the volume of the smaller silo, these steps will be followed.

- The ratio of their surface areas will be used to determine the length scale factor.
- The length scale factor will be used to determine the volume scale factor.
- Finally, the volume scale factor will be used to determine the volume of the smaller silo.

It is given that the surface area of the larger silo is three times as large as the surface area of the smaller silo.

To find the length scale factor, consider its relationship to the surface areas. Recall that for areas of similar figures, the ratio of their areas is equal to the square of the ratio of their corresponding side lengths. Refer to the smaller silo's side length as $a$ and the larger silo's side length as $b.$ $Length Scale Factor ba ⇒ Area Scale Factor (ba )_{2} $ Since the surface area of the larger silo is three times the surface area of the smaller silo, the ratio of the surface area, small to large, is $1:3.$ With that information, the following equation can be expressed. $(ba )_{2}=31 $ Next, this equation can be simplified to solve for the length scale factor. Begin by taking the square root of both sides of the equation.$(ba )_{2}=31 $

Solve for $ba $

SqrtEqn

$LHS =RHS $

$(ba )_{2} =31 $

SqrtPowToNumber

$a_{2} =a$

$ba =31 $

SqrtQuot

$ba =b a $

$ba =3 1 $

CalcRoot

Calculate root

$ba =3 1 $

$6750V_{s} =(3 1 )_{3}$

$V_{s}≈1299$

In this course, the relationships between the length scale factor, area scale factor and volume scale factor have been discussed. If the scale factor between two similar figures is $ba ,$ then the ratio for their areas and volumes can be expressed as the table shows.

Length Scale Factor | Area Scale Factor | Volume Scale Factor |
---|---|---|

$ba $ | $(ba )_{2}$ | $(ba )_{3}$ |

Considering these expressions, the challenge presented at the beginning can be solved with more confidence.

Emily knows that the models in the museum are similar pyramids and the scale factor between the corresponding side lengths is $1:2.$

If the volume of the smaller model is $20$ cubic centimeters, find the volume of the larger model.If the scale factor of two similar figures is $ba ,$ then the ratio of their volumes is $(ba )_{3}.$

To find the volume of the larger model, first, find the volume scale factor. Note that the volume scale factor is equal to the cube of the length scale factor. The length scale factor is given as $1:2,$ or $21 .$ $Length Scale Factor21 ⇒ Volume Scale Factor(21 )_{3}=81 $
An equation can now be written using the known volume scale factor and the ratio of the model's volumes. Recall that it is given that the volume of the smaller model is $20$ cubic centimeters. Substitute that value into the equation to solve for the volume of the larger model.
The volume of the larger model is $160$ cubic centimeters.

$81 =V_{2}V_{1} $

Substitute

$V_{1}=20$

$81 =V_{2}20 $

MultEqn

$LHS⋅8=RHS⋅8$

$1=V_{2}160 $

MultEqn

$LHS⋅V_{2}=RHS⋅V_{2}$

$V_{2}=160$

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