Modeling Using Area and Volume

r= 4

A=π( 4)^2

▼

Evaluate right-hand side

Calculate power

A=π(16)

Use a calculator

A=50.265482 ...

Round to 2 decimal place(s)

A≈ 50.27

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.

Population Density=Number of people/Area of a region

Substitute values

Population Density=220 000/50.27

▼

Evaluate right-hand side

Calculate quotient

Population Density= 4376.367614 ...

Round to nearest integer

Population Density ≈ 4376

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.

Example

Investigating Density of Liquids

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. st1

Two 5-liters bottles of water and cooking oil

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.

Answer

a See solution.

b 400 grams

Hint

a Compare the densities of the substances.

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

Solution

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.

ccc Water Density & & Cooking Oil Density [0.7em] 1 g/ cm^3 & > & 0.92 g/ 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 = 5000 cm^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= m/V ⇔ m = d * V With this in mind, the mass of the water can be found by substituting 1 gcm^3 for d and 5000 cm^3 for V.

m = d * V

d= 1 g/cm^3, V= 5000 cm^3

m = 1 g/cm^3* 5000 cm^3

▼

Evaluate right-hand side

a/c* b = a* b/c

m = 1 g * 5000 cm^3/cm^3

SimpQuot

Simplify quotient

m = 1 g * 5000

m = 5000 g

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.

m = d * V

d= 0.92 g/cm^3, V= 5000 cm^3

m = 0.92 g/cm^3* 5000 cm^3

▼

Evaluate right-hand side

a/c* b = a* b/c

m = 0.92 g * 5000 cm^3/cm^3

CancelCommonFac

Cancel out common factors

m = 0.92 g * 5000 cm^3/cm^3

SimpQuot

Simplify quotient

m = 0.92 g * 5000

m = 4600 g

One bottle of cooking oil weighs 4600 grams. Therefore, the difference between the bottles' weights is 400 grams.

5000-4600 = 400 g

Example

Solving Problems Involving Hexagonals

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.

a What is the area of a regular hexagonal base? Round the answer to two decimal places.

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

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.

Hint

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.

Solution

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^(∘). sin m∠ GAO = OG/OA

Substitute the values and solve for OG.

sin m∠ GAO = OG/OA

Substitute values

sin 60^(∘) = OG/s

▼

Solve for OG

\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}{}

sqrt(3)/2 = OG/s

LHS * s=RHS* s

s * sqrt(3)/2 = OG

MoveLeftFacToNum

a*b/c= a* b/c

ssqrt(3)/2 = OG

Rearrange equation

OG = ssqrt(3)/2

Now the area of an equilateral triangle with a side length of s can be calculated.

A = 1/2 * AB * OG

AB= s, OG= ssqrt(3)/2

A = 1/2 ( s) ( ssqrt(3)/2 )

▼

Evaluate right-hand side

MoveRightFacToNumOne

1/b* a = a/b

A = s/2 * ssqrt(3)/2

MultFrac

Multiply fractions

A = s^2 sqrt(3)/4

Since 6 of these triangles form the regular hexagon, the area A_H of the hexagon is 6 times s^2 sqrt(3)4. A_H = 6 * s^2sqrt(3)/4 ⇔ A_H = 3s^2sqrt(3)/2 By substituting 0.3 for s into the above equation, a cell's base area can be calculated.

A_H = 3 sqrt(3)s^2/2

s= 0.3

A_H = 3 sqrt(3)( 0.3)^2/2

▼

Evaluate right-hand side

Calculate power

A_H = 3 sqrt(3)(0.09)/2

Use a calculator

A_H = 0.233826 ...

Round to 2 decimal place(s)

A_H ≈ 0.23

The base area of a cell is about 0.23 square centimeters.

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.

V = Bh

B= 0.23, h= 0.5

V = ( 0.23)( 0.5)

V= 0.115

Round to 2 decimal place(s)

V ≈ 0.12

The volume of a cell is about 0.12 cubic centimeters.

c Recall that the density of a substance is defined as its mass per unit volume. d = m/V

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.

d = m/V

Substitute values

1.43 = m/( 0.12)( 2000)

▼

Solve for m

1.43= m/240

LHS * 240=RHS* 240

343.2 = m

Rearrange equation

m = 343.2

The mass of the honeycomb is 343.2 grams.

Example

Mixing Substances With Different Densities

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.

Cube-shaped mole in assembled and disassembled forms

a Calculate the volume of the hollow part.

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.

Hint

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

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

Solution

a Consider the top view of the mold.

Using the cosine ratio of 37^(∘), the hypotenuse h can be found.

cos 37^(∘) = Adjacent side/Hypotenuse

Substitute values

cos 37^(∘) = 4/h

▼

Solve for h

LHS * h=RHS* h

h * cos 37^(∘) = 4

DivEqn

.LHS /cos 37^(∘).=.RHS /cos 37^(∘).

h= 4/cos 37^(∘)

Use a calculator

h= 5.008542 ...

Round to nearest integer

h ≈ 5

Now, the sine ratio of 37^(∘) can be used to calculate the length b of the opposite side.

sin 37^(∘) = Opposite side/Hypotenuse

Substitute values

sin 37^(∘) = b/5

▼

Solve for b

LHS * 5=RHS* 5

5 * sin 37^(∘) = b

Rearrange equation

b = 5 * sin 37^(∘)

Use a calculator

b = 3.009075 ...

Round to nearest integer

b ≈ 3

Since identical triangular prisms are used, the other prisms' bases have the same side lengths.

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.

V = Bh

Substitute values

V = ( 5^2) ( 7)

▼

Evaluate right-hand side

Calculate power

V = (25) (7)

V = 175

The volume of the hollow is 175 cm^3.

b Start by finding the mass of the mixture that the carpenter prepared.

Glue & & Sawdust & & Total 52.5 & + & 27.5 & = & 80 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.

d = M/V

M= 80, C= 175

d = 80/175

▼

Evaluate right-hand side

Calculate quotient

d = 0.457142 ...

Round to 2 decimal place(s)

d ≈ 0.46

The density of the mixture is about 0.46 g/cm^3.

Example

Investigating the Volume of a Cylinder in Different Orientations

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.

a What is the radius of the circular base?

b What is the volume of the portion of the cylinder?

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

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

Hint

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?

Solution

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.

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.

sin 30^(∘) = PA/PM

▼

Solve for PM

sin 30^(∘)= 1/2

1/2 = PA/PM

LHS * PM=RHS* PM

PM * 1/2 = PA

LHS * 2=RHS* 2

PM = 2 PA

Notice that PB is also a radius. This implies that PB and PM are equal. l PB=PM PM = 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.

PB = PA+AB

PB= 2PA, AB= 10

2PA = PA + 10

SubEqn

LHS-PA=RHS-PA

PA = 10

Finally, by substituting PA with 10, the radius of the circular bases PM can be calculated. PM &= 2 PA &⇓ PM &= 2 ( 10)=20

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 120^(∘)360^(∘), or 13, of the whole base area A_b.

A_s = 1/3 A_b

A_b= π (20)^2

A_s = 1/3 π(20)^2

Calculate power

A_s = 1/3 π(400)

MoveRightFacToNumOne

1/b* a = a/b

A_s = 400π/3

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

A_s= 400π/3, h= 50

V_s = 400π/3( 50)

▼

Evaluate right-hand side

a/c* b = a* b/c

V_s = 20 000π/3

Use a calculator

V_s = 20 943.951023 ...

Round to nearest integer

V_s ≈ 20 944

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^(∘).

tan 30^(∘) = PA/MA

tan 30^(∘) = 1/sqrt(3), PA= 10

1/sqrt(3) = 10/MA

▼

Solve for MA

LHS * MA=RHS* MA

MA * 1/sqrt(3) = 10

LHS * sqrt(3)=RHS* sqrt(3)

MA = 10sqrt(3)

Since MA and AN are congruent segments, AN is also 10sqrt(3) inches. Therefore, by the Segment Addition Postulate, the length of MN is 20sqrt(3) inches. Now that PA and MN are known, the area A_t of △ MPN can be calculated.

A_t = 1/2 MN* PA

MN= 20sqrt(3), PA= 10

A_t = 1/2 ( 20sqrt(3))( 10)

A_t = 100sqrt(3)

The height of the triangular prism is 50 inches. The volume of this triangular prism is A_t times 50.

V_t = A_t * h

A_t= 100sqrt(3), h= 50

V_t = ( 100sqrt(3))( 50)

Use a calculator

V_t = 8660.254037 ...

Round to nearest integer

V_t ≈ 8660

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 difference between these volumes gives the volume of the water in the tank. ccccc V_p & - & V_t & = & V_w 20 944 & - & 8660 & = & 12 284

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

V_w= 12 284, A_b= π (20)^2

12 284= π(20)^2 * h

▼

Solve for h

Calculate power

12 284= π(400) * h

CommutativePropMult

Commutative Property of Multiplication

12 284= 400π* h

DivEqn

.LHS /400π.=.RHS /400π.

12 284/400π= h

Rearrange equation

h= 12 284/400π

Use a calculator

h= 9.775296 ...

Round to nearest integer

h ≈ 10

The water will reach the height of about 10 inches when placed vertically.

Example

Approximating Volumes

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.

Answer

About 1.4 * 10^(13) cells

Hint

The formula for the volume of a sphere is V = 43π r^3, where r is the radius of the sphere.

Solution

To find the number of cells N, Diego's weight M should be divided by the mass of a cell m. N = 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= m/V ⇔ m = d * V Since the density of a cell is given, the volume of a cell can be calculated first.

Finding the Volume of a Cell

With knowledge of the diameter of a cell, its radius can be calculated by dividing the diameter by 2. 2 * 10^(- 3)/2 = 10^(- 3) cm Now, use the formula for the volume of a sphere.

V = 4/3 π r^3

r= 10^(- 3)

V = 4/3 π ( 10^(- 3))^3

▼

Evaluate right-hand side

PowPow

(a^m)^n=a^(m* n)

V = 4/3 π (10^(- 9))

a/c* b = a* b/c

V = 4 π/3 (10^(- 9))

Use a calculator

V = (4.188790 ...)(10^(- 9))

Round to 1 decimal place(s)

V ≈ (4.2)(10^(- 9))

Rewrite

Rewrite (4.2)(10^(- 9)) as 4.2 * 10^(- 9)

V ≈ 4.2 * 10^(- 9)

The volume of a cell is about 4.2 * 10^(- 9) cubic centimeters.

Finding the Mass of a Cell

The density of a cell is 1 g/cm^3 and its volume is 4.2 * 10^(- 9) cm^3. By multiplying these values, the mass of a cell can be found. 1 g/cm^3 * 4.2 * 10^(- 9) cm^3= 4.2 * 10^(- 9) g

Finding the Number of Cells

The mass of a cell was found in grams. Therefore, Diego's weight should also be written in grams. To do so, use the conversion factor of 1000 gkg. 60 kg * 1000 g/kg= 60 000 g Finally, substitute the values into the formula mentioned at the beginning to calculate the number of cells.

N = M/m

M= 60 000 g, m= 4.2 * 10^(- 9) g

N = 60 000 g/4.2 * 10^(- 9) g

▼

Evaluate right-hand side

CrossCommonFac

Cross out common factors

N = 60 000 g/4.2 * 10^(- 9) g

SimpQuot

Simplify quotient

N = 60 000/4.2 * 10^(- 9)

WriteProdFrac

Write as a product of fractions

N = 60 000/4.2 ( 1/10^(- 9) )

FracToNegExponent

1/a^m=a^(- m)

N = 60 000/4.2 (10^9)

Calculate quotient

N = (14 285.714285714 ... )(10^9)

N = 14 285 714 285 714

RoundSigDig

Round to 2 significant digit(s)

N ≈ 14 000 000 000 000

Write in scientific notation

N ≈ 1.4 * 10^(13)

The number of cells is approximately 1.4 * 10^(13).

Closure

Determining the Density of a Cube

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.

Piramid-shaped part is separated from the cube

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

Hint

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

Solution

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 = 1/2 b h ⇓ B = 1/2 ( 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.

V = 1/3 Bh

B= 4.5, h= 3

V = 1/3( 4.5)( 3)

V= 4.5

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.

d = m/V

m= 94.5, V= 4.5

V = 94.5/4.5