10. Modeling Using Area and Volume
Sign In
An error ocurred, try again later!
Chapter 2
10.
Modeling Using Area and Volume
This lesson focuses on density, area calculations, and volume calculations, showing how these mathematical concepts have practical applications. For instance, it explains how to find the mass of a honeycomb based on its density and volume. It also discusses calculating the population density of a region. These mathematical tools are useful in diverse fields such as urban planning, biology, and engineering. The aim is to provide learners with the knowledge they need to apply these calculations in real-world scenarios, from designing a building to assessing the sustainability of an ecosystem
Show less Show more expand_more 
Lesson Settings & Tools
| | 9 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Modeling Using Area and Volume
Slide of 9
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.
Share
Report error
Translate
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.
What is the density of rhenium?Ali wonders.
Share
Report error
Translate
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 220 000.
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=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.
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=Number of people/Area of a region
SubstituteValues
Substitute values
Population Density=220 000/50.27
Population Density ≈ 4376
Share
Report error
Translate
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
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.
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 = 400 g
m = d * V
SubstituteII
d= 1 g/cm^3, V= 5000 cm^3
m = 1 g/cm^3* 5000 cm^3
▼
Evaluate right-hand side
MoveRightFacToNum
a/c* b = a* b/c
m = 1 g * 5000 cm^3/cm^3
SimpQuot
Simplify quotient
m = 1 g * 5000
Multiply
Multiply
m = 5000 g
m = d * V
SubstituteII
d= 0.92 g/cm^3, V= 5000 cm^3
m = 0.92 g/cm^3* 5000 cm^3
▼
Evaluate right-hand side
MoveRightFacToNum
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
Multiply
Multiply
m = 4600 g
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"cm^2","answer":{"text":["0.23"]}}
b What is the volume of a cell? Round the answer to two decimal places.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"cm^3","answer":{"text":["0.12"]}}
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"g","answer":{"text":["343.2"]}}
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.
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 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. The base area of a cell is about 0.23 square centimeters.
sin m∠ GAO = OG/OA
SubstituteValues
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
MultEqn
LHS * s=RHS* s
s * sqrt(3)/2 = OG
MoveLeftFacToNum
a*b/c= a* b/c
ssqrt(3)/2 = OG
RearrangeEqn
Rearrange equation
OG = ssqrt(3)/2
A = 1/2 * AB * OG
SubstituteII
AB= s, OG= ssqrt(3)/2
A = 1/2 ( s) ( ssqrt(3)/2 )
▼
Evaluate right-hand side
A = s^2 sqrt(3)/4
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 2 decimal place(s)
V ≈ 0.12
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.
The mass of the honeycomb is 343.2 grams.
Share
Report error
Translate
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. 
a Calculate the volume of the hollow part.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"cm^3","answer":{"text":["175"]}}
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"g\/cm^3","answer":{"text":["0.46"]}}
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.
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. 
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 175 cm^3.
sin 37^(∘) = Opposite side/Hypotenuse
SubstituteValues
Substitute values
sin 37^(∘) = b/5
▼
Solve for b
MultEqn
LHS * 5=RHS* 5
5 * sin 37^(∘) = b
RearrangeEqn
Rearrange equation
b = 5 * sin 37^(∘)
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.
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. The density of the mixture is about 0.46 g/cm^3.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"in.","answer":{"text":["20"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"in.^3","answer":{"text":["20944"]}}
c What is the volume of the prism whose bases are the triangles KOL and MPN?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"in.^3","answer":{"text":["8660"]}}
d Find the height of water level when the tank is positioned vertically.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"in.","answer":{"text":["10"]}}
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.
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.
Finally, by substituting PA with 10, the radius of the circular bases PM can be calculated. PM &= 2 PA &⇓ PM &= 2 ( 10)=20
sin 30^(∘) = PA/PM
▼
Solve for PM
PM = 2 PA
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
Substitute
A_b= π (20)^2
A_s = 1/3 π(20)^2
CalcPow
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
SubstituteII
A_s= 400π/3, h= 50
V_s = 400π/3( 50)
▼
Evaluate right-hand side
MoveRightFacToNum
a/c* b = a* b/c
V_s = 20 000π/3
UseCalc
Use a calculator
V_s = 20 943.951023 ...
RoundInt
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^(∘).
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.
The height of the triangular prism is 50 inches. 
The volume of this triangular prism is A_t times 50.
A_t = 1/2 MN* PA
SubstituteII
MN= 20sqrt(3), PA= 10
A_t = 1/2 ( 20sqrt(3))( 10)
Multiply
Multiply
A_t = 100sqrt(3)
V_t = A_t * h
SubstituteII
A_t= 100sqrt(3), h= 50
V_t = ( 100sqrt(3))( 50)
UseCalc
Use a calculator
V_t = 8660.254037 ...
RoundInt
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. 
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.
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. 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.V_w = A_b * h
SubstituteII
V_w= 12 284, A_b= π (20)^2
12 284= π(20)^2 * h
▼
Solve for h
CalcPow
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
RearrangeEqn
Rearrange equation
h= 12 284/400π
UseCalc
Use a calculator
h= 9.775296 ...
RoundInt
Round to nearest integer
h ≈ 10
Share
Report error
Translate
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
Substitute
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))
MoveRightFacToNum
a/c* b = a* b/c
V = 4 π/3 (10^(- 9))
UseCalc
Use a calculator
V = (4.188790 ...)(10^(- 9))
RoundDec
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)
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
SubstituteII
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)
CalcQuot
Calculate quotient
N = (14 285.714285714 ... )(10^9)
Multiply
Multiply
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)
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":null,"formTextAfter":"g\/cm^3","answer":{"text":["21"]}}
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.
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.
Share
Report error
Translate
Modeling Using Area and Volume
Exercises
When Magdalena and Izabella went to a trendy café, they were given milk in packages the shape of regular tetrahedrons.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"V=","formTextAfter":"cm^3","answer":{"text":["25"]}}
security
report
code
security
report
code
Let's substitute the length of the tetrahedron's side into the formula and simplify.
The package contains about 25 cm^3.
security
report
code
E
Hint & Answer
S
Solution
Legend has it that way back in the day, pirates of the Caribbean Sea used to trade in gold coins. Today, Ali decides to go for a walk on a beach in the Caribbean. Something stubs his toe. It's a gold coin!
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":" "},"formTextBefore":"$","formTextAfter":null,"answer":{"text":["1573"]}}
security
report
code
security
report
code
Let's analyze the gold coin Ali found. It is a given that the diameter of the coin is 24 millimeters. Therefore, the radius is 12 millimeters.
To calculate the volume, we must multiply the area of the base with the coin's height. Notice that the density is given in cubic centimeters. Therefore, we must first convert the lengths into centimeters.
Conversion
Since 1 centimeter equals 10 millimeters, we get the conversion factor 1 cm10 mm. Let's convert the measurements.
| Dimension | * 1 cm/10 mm | Evaluate |
|---|---|---|
| 12 mm | 12 mm * 1 cm/10 mm | 1.2 cm |
| 3 mm | 3 mm * 1 cm/10 mm | 0.3 cm |
Volume
Now we can determine the volume of the gold coin which has the form of a cylinder.
Calculating the Mass
The volume is 0.432π cubic centimeters. Density is defined as the mass divided by the volume. Density=Mass/Volume In this case, we know the density and we have also calculated the volume. This is enough information to determine the mass in grams.
When we know the weight of the gold coin, we can determine its worth today by multiplying its mass by the price per gram of gold. 26.2204...($60)≈ $1573
security
report
code
E
Hint & Answer
S
Solution
The density of an object is defined as its mass divided by its volume. Diego is holding a piece of copper with a volume of 8.47 cubic centimeters and a mass of 75.92 grams. Diego's brother is holding a piece of iron with a volume of 5 cubic centimeters and a mass of 39.37 grams. Of the two metal pieces, which has the greater density?
{"type":"choice","form":{"alts":["Copper","Iron","Same density"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
security
report
code
security
report
code
To determine density, we divide the mass by the volume. Density=Mass/Volume Let's calculate the density for each of the metals, beginning with the copper piece.
Copper Piece Density
The copper piece has a volume of 9.3 cubic centimeters and a mass of 75.92 grams.
Iron Piece Density
The iron piece has a volume of 5 cubic centimeters and a mass of 39.37 grams. Let's use that information to calculate the density.
As we can see, copper has a higher density.
security
report
code
E
Hint & Answer
S
Solution
Modeling Using Area and Volume
Recommended exercises
To get personalized recommended exercises, please login first.
Loading content
Edit Lesson
≤
>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
Loading content