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

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

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

\RoundSigDig{2}

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!

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Catch-Up and Review

Hint

Solution

Answer

Hint

Solution

Finding the Volume of a Cell

Finding the Mass of a Cell

Finding the Number of Cells

Hint

Solution