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Many objects used in daily life can be modeled as three-dimensional figures. In this lesson, geometric figures will be used to describe some daily life objects.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Cone
Volume of a Cone
Cylinder
Volume of a Cylinder
Sphere
Volume of a Sphere

Challenge

Comparing Aquarium Volumes

Dominika wants to buy a new aquarium for her fish. She is interested in two types of aquariums that have equal linear measurements.

Aquarium

A

is a right cylinder with a diameter of

10

feet and a height of

5

feet. Additionally, a right cone sits inside of it. Aquarium

B

is a hemisphere with a diameter of

10

feet. Help Dominika answer the following questions.

a What is the volume of Aquarium

A ?

If necessary, round the answer to the nearest cubic foot.

b For each aquarium, what is the area of the water’s surface when filled to a height of

h

feet? Write the answer in terms of

h

and

π .

c Use the answers found above to find the volume of the other aquarium.

Example

Comparing Aquarium Volumes II

Recall how the formula for the volume of a sphere is proven. The same thought process used in the proof can be applied to solve the challenge.

Two types of aquariums attract the attention of Dominica: Aquarium

A

and Aquarium

B .

Aquarium

A

is a right cylinder with a diameter of

10

feet and a height of

5

feet. Additionally, its bottom base is a right cone. Aquarium

B,

on the other hand, is a hemisphere with a diameter of

10

feet.

a What is the volume of Aquarium

A ?

Round the answer to the nearest cubic foot.

b For each aquarium, what is the area of the water’s surface when filled to a height of

h

feet? Write the answer in terms of

h

and

π .

c Use the answers found above and find the volume of the other aquarium.

Hint

a Subtract the volume of the cone from the volume of the cylinder.

b Examine how the cross-sections that are parallel to the bases change.

c What does the Cavalieri's Principle states?

Solution

a By subtracting the volume of the cone from the volume of the cylinder, the volume of the Aquarium

A

can be found.

\underline{Aquarium A} V_{A} = \underline{Cylinder} V_{1} - \underline{Cone ∣} V_{2}

Recall the formulas for the volumes of a cylinder and a cone.

	Volume of the Cylinder	Volume of the Cone
Formula	$V_{1} = π r^{2} h$	$V_{2} = \frac{1}{3} π r^{2} h$

Both the cone and the cylinder forming Aquarium $A$ have a $10$ foot diameter, and therefore both have a radius of $5$ feet. With that in mind, substitute $r = 5$ and $h = 5 .$

	Volume of the Cylinder	Volume of the Cone
Formula	$V_{1} = π r^{2} h$	$V_{2} = \frac{1}{3} π r^{2} h$
Substitute Values	$V_{1} = π (5)^{2} 5$	$V_{2} = \frac{1}{3} π (5)^{2} 5$
Calculate	$V_{1} = 125 π$	$V_{2} = \frac{1 2 5 π}{3}$

The difference between

V_{1}

and

V_{2}

will give the number of cubic feet of water that Aquarium

A

can hold.

V_{A} = V_{1} - V_{2}

SubstituteII

$V_{1} = 125 π$ , $V_{2} = \frac{1 2 5 π}{3}$

V_{A} = 125 π - \frac{1 2 5 π}{3}

Evaluate right-hand side

NumberToFrac

$a = \frac{3 \cdot a}{3}$

V_{A} = \frac{3 7 5 π}{3} - \frac{1 2 5 π}{3}

SubFrac

Subtract fractions

V_{A} = \frac{2 5 0 π}{3}

UseCalc

Use a calculator

V_{A} = 261.799387 \dots

RoundInt

Round to nearest integer

V_{A} \approx 262

Aquarium

A

can hold about

262

cubic feet of water.

b Begin by examining the cross-sections of each aquarium. Then, write an expression for the area of the water’s surface at height

h .

Aquarium $A$

Since the cone inside the aquarium is a right cone, its vertex is directly above the center of its base. Furthermore, its height and radius have the same length. Therefore, an isosceles right triangle inside the cone can be formed, as indicated in the diagram.

With this in mind, consider the vertical and horizontal cross-sections of the aquarium.

Vertical and horizontal cross-sections of Aquarium A

As seen on the diagram, the vertical cross-sections of the water — the shaded triangles — are isosceles right triangles, and the horizontal cross-sections form two concentric circles. Using the horizontal cross-sections, the area of the water’s surface at height

h

will be found. To do so, subtract the area of the inner circle

A_{inner}

from the area of the outer circle

A_{outer} .

A_{outer} - A_{inner}

SubstituteValues

Substitute values

π (5)^{2} - π (5 - h)^{2}

Evaluate

CalcPow

Calculate power

π (25) - π (5 - h)^{2}

CommutativePropMult

Commutative Property of Multiplication

25 π - π (5 - h)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

25 π - π (25 - 10 h + h^{2})

Distr

Distribute $- π$

25 π - 25 π + 10 π h - π h^{2}

SubTerm

Subtract term

10 π h - π h^{2}

When Aquarium

A

is filled to a height of

h

feet, the area of the water’s surface is

10 π h - π h^{2}

square feet.

Aquarium $B$

Examine the vertical cross-sections of the hemisphere.

The diagram shows a right triangle

A B C .

Its hypotenuse and length of one of its legs is known in terms of

h .

Therefore, using the Pythagorean Theorem,

B C

can also be expressed in terms of

h .

A B^{2} + B C^{2} = A C^{2}

SubstituteII

$A B = 5 - h$ , $A C = 5$

(5 - h)^{2} + B C^{2} = 5^{2}

Solve for

B C

SubEqn

$LHS - (5 - h)^{2} = RHS - (5 - h)^{2}$

B C^{2} = 5^{2} - (5 - h)^{2}

CalcPow

Calculate power

B C^{2} = 25 - (5 - h)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

B C^{2} = 25 - (25 - 10 h + h^{2})

Distr

Distribute $- 1$

B C^{2} = 25 - 25 + 10 h - h^{2}

SubTerm

Subtract term

B C^{2} = 10 h - h^{2}

SqrtEqn

$LHS = RHS$

B C^{2} = 10 h - h^{2}

SqrtPowToNumber

$a^{2} = a$

B C = 10 h - h^{2}

Now, considering the horizontal cross-sections of Aquarium

B,

B C

will be the radius of the inner circle.

In this instance, the area of the water’s surface at height

h

is the area of a circle with a radius of

10 h - h^{2} .

π (10 h - h^{2})^{2}

PowSqrt

$(a)^{2} = a$

π (10 h - h^{2})

Distr

Distribute $π$

10 π h - π h^{2}

For Aquarium

B,

the area of the water’s surface at height

h

is the same as the other aquarium,

10 π h - π h^{2}

square feet.

c Recall what the Cavalieri Principle states.

Cavalieri Principle
Two solids with the same height and the same cross-sectional area at every altitude have the same volume.

Both aquariums have a height of $5$ feet, and the area of the water’s surface when filled to a height of $h$ feet is the same for each aquarium.

	Aquarium $A$	Aquarium $B$
Height $(ft)$	$5$
Cross-Sectional Area $(ft^{2})$	$10 π h - π h^{2}$

According to the Cavalieri Principle, the volumes of water must be equal when both aquariums are filled. Therefore, no further calculations are necessary — Aquarium

B

will hold about

262

cubic feet of water.

When solving geometry-related problems, algebraic expressions are needed. In this example, algebraic expressions were found to represent areas of varying cross-sections. The next question also exemplifies a situation where an algebraic relationship will need to be obtained using the variables stemming from geometric constraints.

Example

Finding the Length of a Toilet Paper Roll

Tiffaniqua wants to calculate the length of the a toilet paper roll. Hey! It is on a great sale, Okay. She draws a diagram and denotes the thickness of the paper, the inner radius, and the outer radius by $t,$ $r,$ and $R,$ respectively.

a Write an expression for the length

L

of the paper roll in terms of

t,

r,

and

R .

b Find the length of the paper roll if

t = 0.03,

r = 2.5,

and

R = 5.5,

all measured in centimeters.

Hint

a Use the area of a base of the toilet roll.

b Substitute the given values into the equation found in the previous step.

Solution

a Consider the following horizontal cross-section of the toilet roll. The cross-section consists of concentric circles.

The area of the shaded region

A

can be calculated in two ways. It can be expressed as the area of the circle of radius

R

minus the area of the circle of radius

r .

A = π R^{2} - π r^{2}

Alternatively, it can be expressed as the area of the front face of the long thin rectangular prism, which is created when the paper is unrolled.

The area of the front face is the product of

t

and

L .

That product is also equal to

A .

Therefore,

t \cdot L

can be substituted for

A

in the derived formula for the area of the shaded region.

A = π R^{2} - π r^{2}

Substitute

$A = t \cdot L$

t \cdot L = π R^{2} - π r^{2}

DivEqn

$LHS / t = RHS / t$

L = \frac{π R ^{2} - π r ^{2}}{t}

Note that there might be several alternative ways to find the equation. Only one possible way was shown here.

b By substituting

t = 0.03,

r = 2.5,

and

R = 5.5

into the previously derived equation, the length of the paper towel will be calculated.

L = \frac{π R ^{2} - π r ^{2}}{t}

SubstituteValues

Substitute values

L = \frac{π ( 5 . 5 ) ^{2} - π ( 2 . 5 ) ^{2}}{0 . 0 3}

Evaluate right-hand side

CalcPow

Calculate power

L = \frac{π ( 3 0 . 2 5 ) - π ( 6 . 2 5 )}{0 . 0 3}

Multiply

L = \frac{3 0 . 2 5 π - 6 . 2 5 π}{0 . 0 3}

SubTerm

Subtract term

L = \frac{2 4 π}{0 . 0 3}

UseCalc

Use a calculator

L = 2513.274122 \dots

RoundInt

Round to nearest integer

L \approx 2513

The length of the paper is about

2513

centimeters.

Example

Cylindrical Soda Can

A cylindrical soda can is made of aluminum. It is $6$ inches high and its bases have a radius of approximately $1.2$ inches.

Give a go at answering the following set of questions. If necessary, round the answer to two decimal places.

a Find the surface area of the soda can.

b The density of aluminum is approximately

2.7

grams per cubic centimeter. If the mass of the soda can is approximately

21

grams, how many cubic centimeters of aluminum does it contain?

c Suppose that the thickness of the soda can is uniform throughout its body. Estimate the soda can's thickness.

Hint

a Use the formula for the surface area of a cylinder.

b The density of a substance is equal to its mass divided by its volume.

c How might the surface area and volume of the soda can be related?

Solution

a Since the soda can is a cylinder, the formula for the surface area of a cylinder will be used.

S = 2 π r h + 2 π r^{2}

Here,

r

and

h

are the radius and the height of the cylinder, respectively. The radius is

1.2

inches and the height is

6

inches. Substitute these values in the formula and evaluate it.

S = 2 π r h + 2 π r^{2}

SubstituteII

$r = 1.2$ , $h = 6$

S = 2 π (1.2) (6) + 2 π (1.2)^{2}

Evaluate right-hand side

CalcPow

Calculate power

S = 2 π (1.2) (6) + 2 π (1.44)

Multiply

S = 14.4 π + 2.88 π

AddTerms

Add terms

S = 17.28 π

UseCalc

Use a calculator

S = 54.286721 \dots

RoundDec

Round to $2$ decimal place(s)

S \approx 54.29

Therefore, the surface area of the soda can is about

54.29

square inches.

b Recall that the density of a substance is equal to its mass divided by its volume. In other words, the volume of a substance is its mass divided by its density.

d = \frac{m}{V} \Leftrightarrow V = \frac{m}{d}

Since the mass

m

of the aluminum can and the density

d

of aluminum are given, the volume can be calculated. Substitute

m = 21

and

d = 2.7

into the equation.

V = \frac{m}{d}

SubstituteII

$m = 21$ , $d = 2.7$

V = \frac{2 1}{2 . 7}

Evaluate right-hand side

CalcQuot

Calculate quotient

V = 7.777777 \dots

RoundDec

Round to $2$ decimal place(s)

V \approx 7.78

The soda can is made of about

7.78

cubic centimeters of aluminum.

c To determine the thickness of the soda can, its surface area and volume are good to use.

\underline{Surface Area} 54.29 in^{2} \underline{Volume} 7.78 cm^{3}

Let

t

denote the thickness. Then, the volume of the aluminum part of the soda can will be equal to the surface area of the soda can multiplied by its thickness. Therefore, to find

t,

the the aluminum part's volume should be divided by the soda can's surface area.

V = S \cdot t \Leftrightarrow t = \frac{V}{S}

However, since the amount of aluminum is found in cubic centimeters and the surface area of the soda can in square inches, a conversion factor must be used. Use

\frac{6 . 5 cm ^{2}}{in ^{2}}

to convert square inches to square centimeters.

54.29 in^{2} \cdot \frac{6 . 5 cm ^{2}}{in ^{2}}

Simplify

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

\frac{5 4 . 2 9 in ^{2} \cdot 6 . 5 cm ^{2}}{in ^{2}}

CrossCommonFac

Cross out common factors

\frac{5 4 . 2 9 in ^{2} \cdot 6 . 5 cm ^{2}}{in ^{2}}

SimpQuot

Simplify quotient

54.29 \cdot 6.5 cm^{2}

Multiply

352.885 cm^{2}

Now, by substituting

V = 7.78

and

S = 352.885,

t

can be found.

t = \frac{V}{S}

SubstituteII

$V = 7.78$ , $S = 352.885$

t = \frac{7 . 7 8}{3 5 2 . 8 8 5}

Evaluate right-hand side

CalcQuot

Calculate quotient

t = 0.022046 \dots

RoundDec

Round to $2$ decimal place(s)

t \approx 0.02

This means that the thickness of the soda can is about

0.02

centimeters.

Example

Choosing Type of Glass

By modeling real-life objects using geometric shapes, various characteristics of the objects can be determined. These characteristics can then be compared to make inferences which could impact real decisions to be made.

Emily is attending a fair and wants to sell $1.5$ liters of homemade orange juice she is naming Oranjya Thirsty. She needs to decide the type of glass she will use to serve the juice — a cocktail glass or a Collins glass.

A carton of orange juice and two glasses

A cocktail glass is a type of glass that has an inverted cone bowl. The cone bowl's height is $5.8$ centimeters and the radius of its base is $5.4$ centimeters. A collins glass is a cylindrical glass with a height of $9$ centimeters and a radius of $3.2$ centimeters. Help Emily make a decision by answering the following questions.

a How many cocktail glasses of orange juice can she sell?

b How many Collins glasses of orange juice can she sell?

c If Emily chooses cocktail glasses, she will sell each for

6

dollars. If the Collins glasses are chosen, each will be sold for

10

dollars. Which glass type selection makes Emily more money?

Hint

a Use the formula for the volume of a cone to determine how many liters of orange juice a cocktail glass can hold.

b Use the formula for the volume of a cylinder to determine how many liters of orange juice a Collins glass can hold.

c Use the answers found in the previous parts.

Solution

a First, the volume of the bowl of a cocktail glass will be calculated. Then, it will be used to find the number of glasses. The bowl of a cocktail glass can be modeled by a cone as shown.

The height of the cone bowl and the radius of its base are

5.8

and

5.4

centimeters, respectively. Substitute these values into the volume formula of a cone.

V = \frac{1}{3} π r^{2} h

SubstituteII

$r = 5.4$ , $h = 5.8$

V = \frac{1}{3} π (5.4)^{2} 5.8

Evaluate right-hand side

CalcPow

Calculate power

V = \frac{1}{3} π (29.16) 5.8

Multiply

V = \frac{1}{3} 169.128 π

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

V = \frac{1 6 9 . 1 2 8 π}{3}

UseCalc

Use a calculator

V = 177.110427 \dots

RoundInt

Round to nearest integer

V \approx 177

This means that a cocktail glass can hold approximately

177

cubic centimeters orange juice. Conversion needs to be made. The conversion factor

\frac{1 L}{1 0 0 0 cm ^{3}}

will convert cubic centimeters into liters.

177 cm^{3} \cdot \frac{1 L}{1 0 0 0 cm ^{3}} = 0.177 L

Finally, the number of cocktail glasses can be calculated.

\frac{1 . 5}{0 . 1 7 7}

CalcQuot

Calculate quotient

8.474576 \dots

Therefore,

1.5

liters of orange juice fully fills

8

cocktail glasses.

b Similarly, start by finding the volume of the cylindrical glass.

Its height is

9

centimeters and its base radius is

3.2

centimeters. Substitute these values into the formula.

V_{2} = π r^{2} h

SubstituteII

$r = 3.2$ , $h = 9$

V_{2} = π (3.2)^{2} 9

Evaluate right-hand side

CalcPow

Calculate power

V_{2} = π (10.24) 9

Multiply

V_{2} = 92.16 π

UseCalc

Use a calculator

V_{2} = 289.529178 \dots

RoundInt

Round to nearest integer

V_{2} \approx 290

This means that the volume of a collins glass is about

290

cubic centimeters. After converting its unit of measurement into liters, it is

0.29

liters. Now, the volume of a carton of orange juice can be divided by the volume of a collins glass.

\frac{1 . 5}{0 . 2 9} = 5.172413 \dots

Therefore,

1.5

liters of orange juice fully fills

5

collins glasses.

c Emily can sell

8

cocktail glasses or

5

collins glasses. Knowing the prices, the revenue from each sale can be calculated.

Type of Glass
Cocktail Glass	Collins Glass
$8 \cdot 6$	$5 \cdot 10$
$$ 48$	$$ 50$

As a result, Emily should choose the Collins glass, as the revenue from this selection is greater.

Example

Estimating How Many Grains of Sand a Hand Can Hold

With the help of geometric modeling, any number of objects can be approximated regardless of whether they are super large or tiny minuscule grains of sand.

Take, for example, Ramsha's situation. She is looking through photos from her trip to the beach to post on her social media page. A photo that shows her holding sand sparks her curiosity. She wonders how many individual grains of sand is she holding. Ramsha thinks she can model a grain of sand using a sphere. She then assumes that each grain has a diameter of

0.06

centimeters.

Ramsha figures she can hold

300

grams of sand in her hands. If the density of sand is approximately

1.52

grams per cubic centimeter, help Ramsha approximate the number of grains of sand in her hands. Write the answer in scientific notation.

Answer

About $1.7 \times 1 0^{6}$

Hint

The formula for the volume of a sphere is $V = \frac{4}{3} π r^{3},$ where $r$ is the radius of the sphere.

Solution

To find the number

N

of grains of sand in Ramsha's hands, the mass

M

of the sand in her hands should be divided by the mass

m

of a grain.

N = \frac{M}{m}

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 = \frac{m}{V} \Leftrightarrow m = d \cdot V

Since the density of a grain of sand is given, the volume of a grain of sand will be calculated first.

Finding the Volume of a Grain

The radius of a grain is

\frac{0 . 0 6}{2} = 0.03

centimeters. Use the formula for the volume of a sphere to find the volume of a grain.

V = \frac{4}{3} π r^{3}

Substitute

$r = 0.03$

V = \frac{4}{3} π (0.03)^{3}

Evaluate right-hand side

CalcPow

Calculate power

V = \frac{4}{3} π (0.000027)

CommutativePropMult

Commutative Property of Multiplication

V = \frac{4}{3} 0.000027 π

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V = \frac{0 . 0 0 0 1 0 8 π}{3}

UseCalc

Use a calculator

V = 0.000113 \dots

RoundDec

Round to $5$ decimal place(s)

V \approx 0.00011

Write in scientific notation

V \approx 1.1 \times 1 0^{- 4}

The volume of a grain is about

1.1 \times 1 0^{- 4}

cubic centimeters.

Finding the Mass of a Grain

The density of a grain is

1.52 g / cm^{3}

and its volume is

1.1 \times 1 0^{- 4}

cubic centimeters. By multiplying these values, the mass of a grain can be found.

\frac{1 . 5 2 g}{cm ^{3}} \cdot 1.1 \times 1 0^{- 4} cm^{3}

Evaluate

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

\frac{1 . 5 2 g \cdot 1 . 1 \times 1 0 ^{- 4} cm ^{3}}{cm ^{3}}

CancelCommonFac

Cancel out common factors

\frac{1 . 5 2 g \cdot 1 . 1 \times 1 0 ^{- 4} cm ^{3}}{cm ^{3}}

SimpQuot

Simplify quotient

1.52 g \cdot 1.1 \times 1 0^{- 4}

CommutativePropMult

Commutative Property of Multiplication

1.52 \cdot 1.1 \times 1 0^{- 4} g

Multiply

1.672 \times 1 0^{- 4} g

RoundDec

Round to $1$ decimal place(s)

1.7 \times 1 0^{- 4} g

Finding the Number of Grains

Finally, substitute the values into the formula mentioned at the beginning to calculate the number of grains of sand.

N = \frac{M}{m}

SubstituteII

$M = 300 g$ , $m = 1.7 \times 1 0^{- 4} g$

N = \frac{3 0 0 g}{1 . 7 \times 1 0 ^{- 4} g}

Evaluate right-hand side

CrossCommonFac

Cross out common factors

N = \frac{3 0 0 g}{1 . 7 \times 1 0 ^{- 4} g}

SimpQuot

Simplify quotient

N = \frac{3 0 0}{1 . 7 \times 1 0 ^{- 4}}

WriteProdFrac

Write as a product of fractions

N = \frac{3 0 0}{1 . 7} (\frac{1}{1 0 ^{- 4}})

FracToNegExponent

$\frac{1}{a ^{m}} = a^{- m}$

N = \frac{3 0 0}{1 . 7} (1 0^{4})

CalcQuot

Calculate quotient

N = (176.470588 \dots) (1 0^{4})

Multiply

N = 1764705.882352 \dots

RoundSigDig

Round to $2$ significant digit(s)

N \approx 1700000

Write in scientific notation

N \approx 1.7 \times 1 0^{6}

The number of grains of sand is approximately

1.7 \times 1 0^{6},

or about 1.7 million.

The astronomer Carl Sagan once said, the total number of stars in the universe is greater than all the grains of sand on all the beaches of the planet Earth.

Closure

Modeling the Human Eye

Research projects usually require an interdisciplinary approach. That is, people from different disciplines work together to develop and test hypothesis, run experiments, and test theories.

Biologists, for example, can work with mathematicians to model a part of an organism. By doing so, researchers can predict how these parts function, grow, and change. For example, the human eye was able to be modeled as a sphere. Move the slider to rotate the eye.

These interdisciplinary studies may sometimes result in new discoveries, such as the scutoid, a new geometric solid introduced in

2018 .

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Catch-Up and Review

Hint

Solution

Aquarium A

Aquarium B

Hint

Solution

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Finding the Volume of a Grain

Finding the Mass of a Grain

Finding the Number of Grains

Aquarium $A$

Aquarium $B$