Unlock Density, Area Calculations, and Volume Calculations With Real-World Problems

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.

Catch-Up and Review

Here is a bundle of recommended readings before getting started with this lesson.

Trigonometric Ratios
Hexagon
Circle
Prism
Cube
Cylinder
Sphere
Pyramid

Challenge

Using Trigonometry to Determine Volume and Density

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.

A pyramid-shaped piece separates from the cube

Ali read on a scale that the part weighs

94.5

grams. What is the density of rhenium? Ali wonders.

Example

Investigating Population Density

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 $220000 .$

A diagramm representing population density with dots distributed within a 4-mile radius.

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.

Answer

Population Density: About $4376$ people per square mile
Explanation: See solution.

Hint

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

Solution

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 = \frac{Number of people}{Area of region}

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 = π r^{2}

The radius' value of

4

can be substituted for

r

in the formula to determine the area of the region under study.

A = π r^{2}

Substitute

$r = 4$

A = π (4)^{2}

▼

Evaluate right-hand side

CalcPow

Calculate power

A = π (16)

UseCalc

Use a calculator

A = 50.265482 \dots

RoundDec

Round to $2$ decimal place(s)

A \approx 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 = \frac{Number of people}{Area of a region}

SubstituteValues

Substitute values

Population Density = \frac{2 2 0 0 0 0}{5 0 . 2 7}

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

Population Density = 4376.367614 \dots

RoundInt

Round to nearest integer

Population Density \approx 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.

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.

Water Density 1 g / cm^{3} > Cooking Oil Density 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 \cdot 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 = \frac{m}{V} \Leftrightarrow m = d \cdot V

With this in mind, the mass of the water can be found by substituting

\frac{1 g}{cm ^{3}}

for

d

and

5000 cm^{3}

for

V .

m = d \cdot V

SubstituteII

$d = \frac{1 g}{cm ^{3}}$ , $V = 5000 cm^{3}$

m = \frac{1 g}{cm ^{3}} \cdot 5000 cm^{3}

▼

Evaluate right-hand side

MoveRightFacToNum

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

m = \frac{1 g \cdot 5 0 0 0 cm ^{3}}{cm ^{3}}

SimpQuot

Simplify quotient

m = 1 g \cdot 5000

Multiply

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 \cdot V

SubstituteII

$d = \frac{0 . 9 2 g}{cm ^{3}}$ , $V = 5000 cm^{3}$

m = \frac{0 . 9 2 g}{cm ^{3}} \cdot 5000 cm^{3}

▼

Evaluate right-hand side

MoveRightFacToNum

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

m = \frac{0 . 9 2 g \cdot 5 0 0 0 cm ^{3}}{cm ^{3}}

CancelCommonFac

Cancel out common factors

m = \frac{0 . 9 2 g \cdot 5 0 0 0 cm ^{3}}{cm ^{3}}

SimpQuot

Simplify quotient

m = 0.92 g \cdot 5000

Multiply

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.

A hexagon ABCDEF with sides of length s and center O is divided into six 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 \cdot 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.

Hexagon divided into six equilateral triangles

Now, the height

O G

can be found using the sine ratio of

∠ G A O,

which measures

6 0^{\circ} .

sin m ∠ G A O = \frac{O G}{O A}

Substitute the values and solve for

O G .

sin m ∠ G A O = \frac{O G}{O A}

SubstituteValues

Substitute values

sin 6 0^{\circ} = \frac{O G}{s}

▼

Solve for

O G

\frac{3}{2} = \frac{O G}{s}

MultEqn

$LHS \cdot s = RHS \cdot s$

s \cdot \frac{3}{2} = O G

MoveLeftFacToNum

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

\frac{s 3}{2} = O G

RearrangeEqn

Rearrange equation

O G = \frac{s 3}{2}

Now the area of an equilateral triangle with a side length of

s

can be calculated.

A = \frac{1}{2} \cdot A B \cdot O G

SubstituteII

$A B = s$ , $O G = \frac{s 3}{2}$

A = \frac{1}{2} (s) (\frac{s 3}{2})

▼

Evaluate right-hand side

MoveRightFacToNumOne

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

A = \frac{s}{2} \cdot \frac{s 3}{2}

MultFrac

Multiply fractions

A = \frac{s ^{2} 3}{4}

Since

6

of these triangles form the regular hexagon, the area

A_{H}

of the hexagon is

6

times

\frac{s ^{2} 3}{4} .

A_{H} = 6 \cdot \frac{s ^{2} 3}{4} \Leftrightarrow A_{H} = \frac{3 s ^{2} 3}{2}

By substituting

0.3

for

s

into the above equation, a cell's base area can be calculated.

A_{H} = \frac{3 3 s ^{2}}{2}

Substitute

$s = 0.3$

A_{H} = \frac{3 3 ( 0 . 3 ) ^{2}}{2}

▼

Evaluate right-hand side

CalcPow

Calculate power

A_{H} = \frac{3 3 ( 0 . 0 9 )}{2}

UseCalc

Use a calculator

A_{H} = 0.233826 \dots

RoundDec

Round to $2$ decimal place(s)

A_{H} \approx 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.

A hexagonal pyramid oriented horizontally, resembling the structure of a honeycomb.

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 = B h

SubstituteII

$B = 0.23$ , $h = 0.5$

V = (0.23) (0.5)

Multiply

V = 0.115

RoundDec

Round to $2$ decimal place(s)

V \approx 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 = \frac{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 = \frac{m}{V}

SubstituteValues

Substitute values

1.43 = \frac{m}{( 0 . 1 2 ) ( 2 0 0 0 )}

▼

Solve for

m

Multiply

1.43 = \frac{m}{2 4 0}

MultEqn

$LHS \cdot 240 = RHS \cdot 240$

343.2 = m

RearrangeEqn

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

3 7^{\circ},

the hypotenuse

h

can be found.

cos 3 7^{\circ} = \frac{Adjacent side}{Hypotenuse}

SubstituteValues

Substitute values

cos 3 7^{\circ} = \frac{4}{h}

▼

Solve for

h

MultEqn

$LHS \cdot h = RHS \cdot h$

h \cdot cos 3 7^{\circ} = 4

DivEqn

$LHS / cos 3 7^{\circ} = RHS / cos 3 7^{\circ}$

h = \frac{4}{cos 3 7 ^{\circ}}

UseCalc

Use a calculator

h = 5.008542 \dots

RoundInt

Round to nearest integer

h \approx 5

Now, the sine ratio of

3 7^{\circ}

can be used to calculate the length

b

of the opposite side.

sin 3 7^{\circ} = \frac{Opposite side}{Hypotenuse}

SubstituteValues

Substitute values

sin 3 7^{\circ} = \frac{b}{5}

▼

Solve for

b

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

5 \cdot sin 3 7^{\circ} = b

RearrangeEqn

Rearrange equation

b = 5 \cdot sin 3 7^{\circ}

UseCalc

Use a calculator

b = 3.009075 \dots

RoundInt

Round to nearest integer

b \approx 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 = B h

SubstituteValues

Substitute values

V = (5^{2}) (7)

▼

Evaluate right-hand side

CalcPow

Calculate power

V = (25) (7)

Multiply

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 52.5 + Sawdust 27.5 = Total 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 = \frac{M}{V}

SubstituteII

$M = 80$ , $C = 175$

d = \frac{8 0}{1 7 5}

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

d = 0.457142 \dots

RoundDec

Round to $2$ decimal place(s)

d \approx 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.

A cylindrical water tank lying horizontally. The water level inside the tank is at 10 inches, and the distance between the two bases of the tank 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

K O L

and

M P N ?

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

Hint

a Determine if

△ M P N

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

\overline{M P},

\overline{P N},

and

M N ?

Use the measure of

∠ M P N .

c Determine the length of

\overline{M N} .

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

\overline{P M}

and

\overline{P N}

are radii of the circle, they have the same length. From here, it can be concluded that the triangle

△ M P N

is an isosceles triangle with base angles

3 0^{\circ} .

By the Interior Angles Theorem,

∠ M P N

can be concluded to measure

12 0^{\circ} .

An isosceles triangle MPN inside the circular base of the water tank

Now, draw a segment from

P

to the ground. This segment will be perpendicular to the ground and

\overline{M N} .

Since the height of an isosceles triangle is also the median of the triangle's base, this segment is also the perpendicular bisector of

\overline{M N} .

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,

∠ M P A

and

∠ N P A

are congruent angles each measuring

6 0^{\circ} .

Using the sine of

∠ A M P,

which measures

3 0^{\circ},

P M

can be written in terms of

P A .

sin 3 0^{\circ} = \frac{P A}{P M}

▼

Solve for

P M

Substitute

$sin 3 0^{\circ} = \frac{1}{2}$

\frac{1}{2} = \frac{P A}{P M}

MultEqn

$LHS \cdot P M = RHS \cdot P M$

P M \cdot \frac{1}{2} = P A

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

P M = 2 P A

Notice that

\overline{P B}

is also a radius. This implies that

P B

and

P M

are equal.

P B = P M P M = 2 P A \Rightarrow P B = 2 P A

Moreover, by the Segment Addition Postulate, the sum of

P A

and

A B

gives

P B .

Knowing

A B = 10

and

P B = 2 P A,

P A

can be calculated.

P B = P A + A B

SubstituteII

$P B = 2 P A$ , $A B = 10$

2 P A = P A + 10

SubEqn

$LHS - P A = RHS - P A$

P A = 10

Finally, by substituting

P A

with

10,

the radius of the circular bases

P M

can be calculated.

P M P M = 2 P A ⇓ = 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}

\frac{1 2 0 ^{\circ}}{3 6 0 ^{\circ}},

\frac{1}{3},

of the whole base area

A_{b}

A_{s} = \frac{1}{3} A_{b}

Substitute

$A_{b} = π (20)^{2}$

A_{s} = \frac{1}{3} π (20)^{2}

CalcPow

Calculate power

A_{s} = \frac{1}{3} π (400)

MoveRightFacToNumOne

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

A_{s} = \frac{4 0 0 π}{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} \cdot h

SubstituteII

$A_{s} = \frac{4 0 0 π}{3}$ , $h = 50$

V_{s} = \frac{4 0 0 π}{3} (50)

▼

Evaluate right-hand side

MoveRightFacToNum

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

V_{s} = \frac{2 0 0 0 0 π}{3}

UseCalc

Use a calculator

V_{s} = 20943.951023 \dots

RoundInt

Round to nearest integer

V_{s} \approx 20944

c In Part A, the height of

△ M P N

was found to be

10

inches.

The tangent ratio of

∠ A M P

can be used to find

M A .

Note that

∠ A M P

measures

3 0^{\circ} .

tan 3 0^{\circ} = \frac{P A}{M A}

SubstituteII

$tan 3 0^{\circ} = \frac{1}{3}$ , $P A = 10$

\frac{1}{3} = \frac{1 0}{M A}

▼

Solve for

M A

MultEqn

$LHS \cdot M A = RHS \cdot M A$

M A \cdot \frac{1}{3} = 10

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

M A = 103

Since

\overline{M A}

and

\overline{A N}

are congruent segments,

A N

is also

103

inches. Therefore, by the Segment Addition Postulate, the length of

\overline{M N}

203

inches. Now that

P A

and

M N

are known, the area

A_{t}

△ M P N

can be calculated.

A_{t} = \frac{1}{2} M N \cdot P A

SubstituteII

$M N = 203$ , $P A = 10$

A_{t} = \frac{1}{2} (203) (10)

Multiply

A_{t} = 1003

The height of the triangular prism is

50

inches.

The volume of this triangular prism is

A_{t}

times

50 .

V_{t} = A_{t} \cdot h

SubstituteII

$A_{t} = 1003$ , $h = 50$

V_{t} = (1003) (50)

UseCalc

Use a calculator

V_{t} = 8660.254037 \dots

RoundInt

Round to nearest integer

V_{t} \approx 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.

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} \cdot h

SubstituteII

$V_{w} = 12284$ , $A_{b} = π (20)^{2}$

12284 = π (20)^{2} \cdot h

▼

Solve for

h

CalcPow

Calculate power

12284 = π (400) \cdot h

CommutativePropMult

Commutative Property of Multiplication

12284 = 400 π \cdot h

DivEqn

$LHS / 400 π = RHS / 400 π$

\frac{1 2 2 8 4}{4 0 0 π} = h

RearrangeEqn

Rearrange equation

h = \frac{1 2 2 8 4}{4 0 0 π}

UseCalc

Use a calculator

h = 9.775296 \dots

RoundInt

Round to nearest integer

h \approx 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 $2 \times 1 0^{- 3}$ 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 $1.4 \times 1 0^{13}$ cells

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 of cells

N,

Diego's weight

M

should be divided by the mass of a cell

m .

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 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 .

\frac{2 \times 1 0 ^{- 3}}{2} = 1 0^{- 3} cm

Now, use the formula for the volume of a sphere.

V = \frac{4}{3} π r^{3}

Substitute

$r = 1 0^{- 3}$

V = \frac{4}{3} π (1 0^{- 3})^{3}

▼

Evaluate right-hand side

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

V = \frac{4}{3} π (1 0^{- 9})

MoveRightFacToNum

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

V = \frac{4 π}{3} (1 0^{- 9})

UseCalc

Use a calculator

V = (4.188790 \dots) (1 0^{- 9})

RoundDec

Round to $1$ decimal place(s)

V \approx (4.2) (1 0^{- 9})

Rewrite

Rewrite $(4.2) (1 0^{- 9})$ as $4.2 \times 1 0^{- 9}$

V \approx 4.2 \times 1 0^{- 9}

The volume of a cell is about

4.2 \times 1 0^{- 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 \times 1 0^{- 9} cm^{3} .

By multiplying these values, the mass of a cell can be found.

1 \frac{g}{cm ^{3}} \cdot 4.2 \times 1 0^{- 9} cm^{3} = 4.2 \times 1 0^{- 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 \frac{g}{kg} .

60 kg \cdot 1000 \frac{g}{kg} = 60000 g

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

N = \frac{M}{m}

SubstituteII

$M = 60000 g$ , $m = 4.2 \times 1 0^{- 9} g$

N = \frac{6 0 0 0 0 g}{4 . 2 \times 1 0 ^{- 9} g}

▼

Evaluate right-hand side

CrossCommonFac

Cross out common factors

N = \frac{6 0 0 0 0 g}{4 . 2 \times 1 0 ^{- 9} g}

SimpQuot

Simplify quotient

N = \frac{6 0 0 0 0}{4 . 2 \times 1 0 ^{- 9}}

WriteProdFrac

Write as a product of fractions

N = \frac{6 0 0 0 0}{4 . 2} (\frac{1}{1 0 ^{- 9}})

FracToNegExponent

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

N = \frac{6 0 0 0 0}{4 . 2} (1 0^{9})

CalcQuot

Calculate quotient

N = (14285.714285714 \dots) (1 0^{9})

Multiply

N = 14285714285714

RoundSigDig

Round to $2$ significant digit(s)

N \approx 14000000000000

Write in scientific notation

N \approx 1.4 \times 1 0^{13}

The number of cells is approximately

1.4 \times 1 0^{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 = \frac{1}{2} b h ⇓ B = \frac{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 = \frac{1}{3} B h

SubstituteII

$B = 4.5$ , $h = 3$

V = \frac{1}{3} (4.5) (3)

Multiply

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 = \frac{m}{V}

SubstituteII

$m = 94.5$ , $V = 4.5$

V = \frac{9 4 . 5}{4 . 5}

CalcQuot

Calculate quotient

V = 21

The density of rhenium is

21

grams per cubic centimeter.

What a festival experience for Ali!

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Catch-Up and Review

Answer

Hint

Solution

Answer

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Finding the Volume of a Cell

Finding the Mass of a Cell

Finding the Number of Cells

Hint

Solution