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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.

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.

{"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":"km","answer":{"text":["692,000","692000"]}}

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.
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.

$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 ⋅310_{-6} $

UseCalc

Use a calculator

$Rr =(0.914958…)(10_{-2})$

RoundDec

Round to $2$ decimal place(s)

$Rr ≈(0.91)(10_{-2})$

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 blue solid to the orange solid.

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 $

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

Finally, knowing that the volume of the larger silo is about $6750$ cubic meters, the volume of the smaller silo $V_{s}$ can be calculated. Similar to the areas, the ratio of the volumes should be equal to the volume scale factor.
$6750V_{s} =(3 1 )_{3}$

$V_{s}≈1299$

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.{"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:0.8238736249999999em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">cm<\/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":["160"]}}

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 $21 .$
The volume of the larger model is $160$ cubic centimeters.

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

Substitute

$V_{1}=20$

$81 =V_{2}20 $

MultEqn

$LHS⋅8=RHS⋅8$

$1=V_{2}160 $

MultEqn

$LHS⋅V_{2}=RHS⋅V_{2}$

$V_{2}=160$