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{{ printedBook.courseTrack.name }} {{ printedBook.name }} As investigated in the lesson exploring modeling with geometric shapes, trigonometric ratios can be used to solve various real-life problems. This lesson will expand the applications of trigonometric ratios to situations that involve density based on area and volume.

Last weekend Ali visited an aviation festival near his neighborhood. In an experiment at the festival, an engineer introduced the element rhenium, which is used to make jet engine parts. The engineer showed the onlookers a small pyramid-shaped part of the element. The part was obtained by cutting one corner of a cube with edges of length $6$ centimeters as shown.

Ali read on a scale that the part weighs $94.5$ grams.

What is the density of rhenium?Ali wonders.

The concept of density can be considered in different contexts. Density is essentially a derived unit that compares a quantity per unit of area or volume.

Maya is running a population census for her home city. She estimates the number of people in a region with a $4-$mile radius to be around $220000.$

Find the population density using the unit measurement of people per square mile. Then, round the answer to the nearest integer. Be sure to explain what this number means.**Population Density:** About $4376$ people per square mile

**Explanation:** See solution.

The population density is the ratio of the number of people living in a region to region's area.

To find the density in people per square mile, the ratio of the number of people to the area of the region should be found.
$Population Density=Area of regionNumber of people $
The area of the region, which is a circle with a radius of $4$ miles, can be calculated using the formula for the area of a circle.
$A=πr_{2} $
The radius' value of $4$ can be substituted for $r$ in the formula to determine the area of the region under study.
The area of the region is about $50.27$ square miles. Now that the area is known, the population density of this region can be determined.
The population density in this region is about $4376$ people per square mile. Assuming that the total population is evenly distributed across the region, the number found, $4376,$ represents the number of people per square mile living in the region of Maya's home city.

$Population Density=Area of a regionNumber of people $

SubstituteValues

Substitute values

$Population Density=50.27220000 $

$Population Density≈4376$

In physics, density refers to the ratio of the mass of a substance to its volume. The next few examples are about the density based on volume.

Izabella is planning a to make a nice lunch, she buys a bottle of water and a bottle of cooking oil. She knows that both bottles have a volume of $5$ liters.
While carrying the bottle of water in her right hand and the bottle of cooking oil in her left, she realizes that she is walking leaning to the right.

a Explain why her walking gait has changed.

b Use the given densities to find the difference between the masses of the substances.

a See solution.

b $400$ grams

a Compare the densities of the substances.

b Use the fact that $1$ liter is $1000$ cubic centimeters.

a Although the volumes of both bottles are the same, one can weigh more than the other. This phenomena can be explained by the density of the substances. Denser substances weigh more because they have a bigger mass to volume ratio.

$Water Density1g/cm_{3} > Cooking Oil Density0.92g/cm_{3} $

Since water has a greater density than cooking oil and the volumes of the bottles are the same, the bottle of water is heavier than the bottle of oil. The heavier bottle of water is being held in her right hand. Therefore, Izabella has found herself leaning to the right. b Since $1$ liter is equivalent to $1000$ cubic centimeters, $5$ liters is equal to $5$ times $1000$ cubic centimeters. $5⋅1000=5000cm_{3} $
The density of a substance is defined as its mass divided by its volume. Equivalently, the mass of a substance is its density times its volume. $d=Vm ⇔m=d⋅V $
With this in mind, the mass of the water can be found by substituting $cm_{3}1g $ for $d$ and $5000cm_{3}$ for $V.$
It has been derived that $5$ liters of water weigh $5000$ grams. Similarly, the weight of a $5-$liter bottle of cooking oil can be found.
One bottle of cooking oil weighs $4600$ grams. Therefore, the difference between the bottles' weights is $400$ grams. $5000−4600=400g $

$m=d⋅V$

SubstituteII

$d=cm_{3}1g $, $V=5000cm_{3}$

$m=cm_{3}1g ⋅5000cm_{3}$

Evaluate right-hand side

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$m=cm_{3}1g⋅5000cm_{3} $

SimpQuot

Simplify quotient

$m=1g⋅5000$

Multiply

Multiply

$m=5000g$

$m=d⋅V$

SubstituteII

$d=cm_{3}0.92g $, $V=5000cm_{3}$

$m=cm_{3}0.92g ⋅5000cm_{3}$

Evaluate right-hand side

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$m=cm_{3}0.92g⋅5000cm_{3} $

CancelCommonFac

Cancel out common factors

$m=cm_{3} 0.92g⋅5000cm_{3} $

SimpQuot

Simplify quotient

$m=0.92g⋅5000$

Multiply

Multiply

$m=4600g$

Bees build their honeycombs in such a way that each cell is a prism with a regular hexagonal base. In these small hexagonal cells, bees are born and raised, and honey and pollen are stored.

Diego, a bee-loving biology student, discovers that the depth of a hexagonal cell is $0.5$ centimeters, and its base has a side length of $0.3$ centimeters. Help Diego answer his following research questions.

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b What is the volume of a cell? Round the answer to two decimal places.

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c Diego assumes that the honeycomb consist of $2000$ cells. Given the density of honey $1.43$ grams per cubic centimeter, find the mass of the honeycomb.

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a A regular hexagon can be decomposed into $6$ equilateral triangles that are congruent.

b The volume of a three-dimensional figure is equal to its base area times its height.

c The density of a substance is defined as its mass per unit volume.

a Consider a regular hexagon with a side length $s.$ It can be divided into six congruent equilateral triangles. Therefore, the area $A_{H}$ of the hexagon is $6$ times the area $A$ of the equilateral triangle with side length $s.$ $A_{H}=6⋅A $
Since the area of a triangle is half the product of its base and its height, the height of the equilateral triangle should be determined first. Draw the height of the triangle, which will bisect the base. Now, the height $OG$ can be found using the sine ratio of $∠GAO,$ which measures $60_{∘}.$ $sinm∠GAO=OAOG $
Substitute the values and solve for $OG.$
Now the area of an equilateral triangle with a side length of $s$ can be calculated.
Since $6$ of these triangles form the regular hexagon, the area $A_{H}$ of the hexagon is $6$ times $4s_{2}3 .$
$A_{H}=6⋅4s_{2}3 ⇔A_{H}=23s_{2}3 $
By substituting $0.3$ for $s$ into the above equation, a cell's base area can be calculated. The base area of a cell is about $0.23$ square centimeters.

$sinm∠GAO=OAOG $

SubstituteValues

Substitute values

$sin60_{∘}=sOG $

Solve for $OG$

${\textstyle \ifnumequal{60}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{60}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{60}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{60}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{60}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{60}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{60}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{360}{\sin\left(360^\circ\right)=0}{}}$

$23 =sOG $

MultEqn

$LHS⋅s=RHS⋅s$

$s⋅23 =OG$

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$2s3 =OG$

RearrangeEqn

Rearrange equation

$OG=2s3 $

b It is a given that a cell in honeycomb is a hexagonal prism with a height of $0.5$ centimeters. In Part A, its base area was found to be $0.23$ square centimeters. Recall that the volume of a three-dimensional figure is equal to its base area times its height.
The volume of a cell is about $0.12$ cubic centimeters.

$V=Bh$

SubstituteII

$B=0.23$, $h=0.5$

$V=(0.23)(0.5)$

Multiply

Multiply

$V=0.115$

RoundDec

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

$V≈0.12$

c Recall that the density of a substance is defined as its mass per unit volume. $d=Vm $
With the use of this definition, the mass of the honeycomb that consists of $2000$ cells can be calculated. The total volume of the honeycomb is the product of $one cell’s volume$ and $2000.$ The density of honey is $1.43$ grams per cubic centimeter.
The mass of the honeycomb is $343.2$ grams.

When two substances are mixed, the density of the mixture will still be the ratio of total mass to total volume. A carpenter uses four identical prisms with right triangle bases to create a cube-shaped mold as shown in the applet.
### Hint

### Solution

a Calculate the volume of the hollow part.

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b The carpenter mixes $52.5$ grams of glue and $27.5$ grams of sawdust, and pours it into the mold. What is the density of the mixture? If necessary, round the answer to two decimal places.

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a Use trigonometric ratios to find the dimensions of the mold.

b Use the volume of the hollow and the mass of the mixture.

a Consider the top view of the mold.
Using the cosine ratio of $37_{∘},$ the hypotenuse $h$ can be found.
Now, the sine ratio of $37_{∘}$ can be used to calculate the length $b$ of the opposite side.
Since identical triangular prisms are used, the other prisms' bases have the same side lengths.

$sin37_{∘}=HypotenuseOpposite side $

SubstituteValues

Substitute values

$sin37_{∘}=5b $

Solve for $b$

MultEqn

$LHS⋅5=RHS⋅5$

$5⋅sin37_{∘}=b$

RearrangeEqn

Rearrange equation

$b=5⋅sin37_{∘}$

UseCalc

Use a calculator

$b=3.009075…$

RoundInt

Round to nearest integer

$b≈3$

As the diagram indicates, the base of the hollow part is a square with a side length of $5$ centimeters. Additionally, the mold is a cube with an edge length of $7$ centimeters. Therefore, the hollow is a square prism with a base edge length of $5$ centimeters and height of $7$ centimeters.

The volume of this square prism can be calculated as the product of its base area and its height. Since the base of the prism is a square with a side length of $5$ centimeters, the base area is $5_{2}$ square centimeters. The volume of the hollow is $175cm_{3}.$ b Start by finding the mass of the mixture that the carpenter prepared.
$Glue52.5 + Sawdust27.5 = Total80 $ This amount of mixture occupies a volume of $175$ cubed centimeters. Since density is a measure of mass per volume, the density of the mixture is equal to the quotient of the mass and volume. The density of the mixture is about $0.46g/cm_{3}.$

The diagram shows a water tank that is positioned horizontally with some water inside. The water level height is $10$ inches, and the distance between two bases is $50$ inches.

Kevin wonders how high the water level will rise when the tank is positioned vertically. Help Kevin to find it by answering the following questions.

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c What is the volume of the prism whose bases are the triangles $KOL$ and $MPN?$

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d Find the height of water level when the tank is positioned vertically.

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a Determine if $△MPN$ is an isosceles triangle. Then, draw a segment from $P$ to the ground and use one of trigonometric ratios.

b How much of the circle is represented by the region enclosed by $MP,$ $PN,$ and $MN?$ Use the measure of $∠MPN.$

c Determine the length of $MN.$ Use the formula for the area of a triangle.

d Use the calculated volumes to find the volume of the water. How can the volume of the water be determined when the water tank is positioned vertically?

a Consider the circular base with center $P.$ Since $PM$ and $PN$ are radii of the circle, they have the same length. From here, it can be concluded that the triangle $△MPN$ is an isosceles triangle with base angles $30_{∘}.$ By the Interior Angles Theorem, $∠MPN$ can be concluded to measure $120_{∘}.$
Now, draw a segment from $P$ to the ground. This segment will be perpendicular to the ground and $MN.$ Since the height of an isosceles triangle is also the median of the triangle's base, this segment is also the perpendicular bisector of $MN.$
Notice that $PB$ is also a radius. This implies that $PB$ and $PM$ are equal. $PB=PMPM=2PA ⇒PB=2PA $
Moreover, by the Segment Addition Postulate, the sum of $PA$ and $AB$ gives $PB.$ Knowing $AB=10$ and $PB=2PA,$ $PA$ can be calculated.
Finally, by substituting $PA$ with $10,$ the radius of the circular bases $PM$ can be calculated. $PMPM =2PA⇓=2(10)=20 $

Additionally, the height of an isosceles triangle, which is drawn from its vertex angle to its base, is the bisector of the vertex angle. Therefore, $∠MPA$ and $∠NPA$ are congruent angles each measuring $60_{∘}.$ Using the sine of $∠AMP,$ which measures $30_{∘},$ $PM$ can be written in terms of $PA.$

$sin30_{∘}=PMPA $

Solve for $PM$

${\textstyle \ifnumequal{30}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{30}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{30}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{30}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{30}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{30}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{30}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{30}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{30}{360}{\sin\left(360^\circ\right)=0}{}}$

$21 =PMPA $

MultEqn

$LHS⋅PM=RHS⋅PM$

$PM⋅21 =PA$

MultEqn

$LHS⋅2=RHS⋅2$

$PM=2PA$

b In Part A, the radius of the base and the measures of the central angles were found. With this information, the volume of the shaded portion of the cylinder can now be calculated. Recall that the volume of a three-dimensional figure equals its base area times its height. The base area of the shaded portion $A_{s}$ is $360_{∘}120_{∘} ,$ or $31 ,$ of the whole base area $A_{b}$.

$A_{s}=31 A_{b}$

Substitute

$A_{b}=π(20)_{2}$

$A_{s}=31 π(20)_{2}$

CalcPow

Calculate power

$A_{s}=31 π(400)$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$A_{s}=3400π $

Since the height of a solid is the distance between its bases, multiplying $A_{b}$ by $50$ will give the volume of the portion of the cylinder.

$V_{s}=A_{s}⋅h$

SubstituteII

$A_{s}=3400π $, $h=50$

$V_{s}=3400π (50)$

Evaluate right-hand side

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V_{s}=320000π $

UseCalc

Use a calculator

$V_{s}=20943.951023…$

RoundInt

Round to nearest integer

$V_{s}≈20944$

c In Part A, the height of $△MPN$ was found to be $10$ inches.
The tangent ratio of $∠AMP$ can be used to find $MA.$ Note that $∠AMP$ measures $30_{∘}.$
Since $MA$ and $AN$ are congruent segments, $AN$ is also $103 $ inches. Therefore, by the Segment Addition Postulate, the length of $MN$ is $203 $ inches. Now that $PA$ and $MN$ are known, the area $A_{t}$ of $△MPN$ can be calculated.
The height of the triangular prism is $50$ inches. The volume of this triangular prism is $A_{t}$ times $50.$

d Up to this point, the radius of the cylinder, the volume of the portion, and the volume of the triangular prism have been found.
The water will reach the height of about $10$ inches when placed vertically.

The difference between these volumes gives the volume of the water in the tank. $V_{p}20944 −− V_{t}8660 == V_{w}12284 $

Now suppose the tank is positioned vertically. The water level in the tank can be shown as follows. Since there is no change in the amount of water, the volume of the water in the cylinder remains the same. Using the formula for volume of a cylinder, the height $h$ can be found.$V_{w}=A_{b}⋅h$

SubstituteII

$V_{w}=12284$, $A_{b}=π(20)_{2}$

$12284=π(20)_{2}⋅h$

Solve for $h$

CalcPow

Calculate power

$12284=π(400)⋅h$

CommutativePropMult

Commutative Property of Multiplication

$12284=400π⋅h$

DivEqn

$LHS/400π=RHS/400π$

$400π12284 =h$

RearrangeEqn

Rearrange equation

$h=400π12284 $

UseCalc

Use a calculator

$h=9.775296…$

RoundInt

Round to nearest integer

$h≈10$

After discussing hexagonal cells of beehives, Diego thinks he can approximate the number of cells in his body. His teacher recommends for Diego to treat a cell like a sphere with a diameter of $2×10_{-3}$ centimeters.

If Diego's weight is $60$ kilograms and the density of a cell is approximately the density of water, which is $1$ gram per cubic centimeter, help Diego approximate the number of cells in his body. Write the answer in scientific notation.

About $1.4×10_{13}$ cells

The formula for the volume of a sphere is $V=34 πr_{3},$ where $r$ is the radius of the sphere.

To find the number of cells $N,$ Diego's weight $M$ should be divided by the mass of a cell $m.$ $N=mM $ Recall that the density of a substance is equal to its mass divided by its volume. In other words, the mass of a substance is its density times its volume. $d=Vm ⇔m=d⋅V $ Since the density of a cell is given, the volume of a cell can be calculated first.

$V=34 πr_{3}$

Substitute

$r=10_{-3}$

$V=34 π(10_{-3})_{3}$

Evaluate right-hand side

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$V=34 π(10_{-9})$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V=34π (10_{-9})$

UseCalc

Use a calculator

$V=(4.188790…)(10_{-9})$

RoundDec

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

$V≈(4.2)(10_{-9})$

Rewrite

Rewrite $(4.2)(10_{-9})$ as $4.2×10_{-9}$

$V≈4.2×10_{-9}$

The density of a cell is $1g/cm_{3}$ and its volume is $4.2×10_{-9}cm_{3}.$ By multiplying these values, the mass of a cell can be found. $1cm_{3} g ⋅4.2×10_{-9}cm_{3} =4.2×10_{-9}g $

$N=mM $

SubstituteII

$M=60000g$, $m=4.2×10_{-9}g$

$N=4.2×10_{-9}g60000g $

Evaluate right-hand side

CrossCommonFac

Cross out common factors

$N=4.2×10_{-9}g 60000g $

SimpQuot

Simplify quotient

$N=4.2×10_{-9}60000 $

WriteProdFrac

Write as a product of fractions

$N=4.260000 (10_{-9}1 )$

FracToNegExponent

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

$N=4.260000 (10_{9})$

CalcQuot

Calculate quotient

$N=(14285.714285714…)(10_{9})$

Multiply

Multiply

$N=14285714285714$

RoundSigDig

Round to $2$ significant ${\textstyle \ifnumequal{2}{1}{\text{digit}}{\text{digits}}}$

$N≈14000000000000$

Write in scientific notation

$N≈1.4×10_{13}$

Throughout the lesson, different cases involving concepts of density based on area and volume have been discussed. Considering these situations, the challenge presented at the beginning of the lesson can be solved.

Ali knows that the small pyramid-shaped part of the element was obtained by cutting one corner of a cube with edges of length $6$ centimeters.

Given that the small piece's weight is $94.5$ grams, the density of rhenium can be calculated.

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The volume of a pyramid is one-third of the product of its base area and height.

From the diagram, it can be seen that the pyramid has three congruent edges, each is $3$ centimeters in length.

Additionally, the pyramid's base is an isosceles right triangle. Therefore, its area is half the product of its legs.
$B=21 bh⇓B=21 (3)(3)=4.5 $
Recall that the volume of a pyramid is one-third of the product of its base area and height. Using the fact that the height of the pyramid is $3$ centimeters, its volume can be determined.
Therefore, $94.5$ grams of rhenium has a volume of $4.5$ cubic centimeters. Now, using the definition of density, the volume of rhenium can be calculated.
The density of rhenium is $21$ grams per cubic centimeter.

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