5. Volume and Surface Area of Composite Solids
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Chapter 10
5.
Volume and Surface Area of Composite Solids
Composite solids are three-dimensional shapes made up of simpler geometric figures. Understanding how to find their volume and surface area is essential for solving real-world problems in areas like architecture, engineering, and design. The process involves breaking down the composite solid into individual shapes, calculating their volume and surface area, and then combining the results. This method ensures accuracy and simplifies complex calculations. Mastery of composite solids aids in the practical application of geometric concepts and is critical for advanced studies in spatial reasoning and problem-solving. By developing these skills, individuals are better equipped to tackle challenges in various fields, such as construction and manufacturing, where accurate measurements are essential.
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| | 10 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Volume and Surface Area of Composite Solids
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Sometimes, a solid may not have a recognizable shape, which makes determining its volume or surface area difficult. However, certain solids can be broken down into familiar shapes like prisms, pyramids, cones, spheres, and more, allowing the application of familiar formulas. This lesson delves into the computations of volumes and surface areas for this type of solids.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Explore
Constructing New Solids
Consider a hemisphere, a cone, and a cylinder, all of which have the same radius. Each solid can be dragged and rotated. Create new solids by combining the given ones.
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Discussion
Composite Solid
A solid that is made up of more than one solid is called a composite solid. The individual solids can be combined either by adding or subtracting them from one another. For instance, a hemisphere can be combined with a cone to make something that resembles a snow cone, or it could be used to dig a bowl shape out of a cylinder.
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Example
Finding the Volume of a Traffic Cone
Ramsha has recently learned how to find the volume of composite solids. She is curious about finding the volumes of composite solids that she encounters in her daily life. Consider the diagram of a traffic cone she passed during her walk to school.
The height of the cone part is 30 inches and its radius is 5 inches. The prism below the cone is a square prism with side lengths of 14 inches and a height of 1 inch. Help Ramsha find the volume of the traffic cone. Use a calculator for calculations and round the result to the nearest whole number.
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Hint
The volume occupied by the traffic cone is the sum of the volume of the prism base and the volume of the cone part.
Solution
The traffic cone is basically composed of two solids — a cone and a square prism. This means that the volume of the traffic cone is the sum of the volume of the prism V_p and the volume of the cone V_c.
V = V_p + V_c
This solution will begin by calculating the volume of the prism. After that, it will find the volume of the cone.
Finding the Volume of the Prism
The base of the prism is a square with side lengths of 14 inches, so its area is the square of 14. B =14^2 ⇔ B=196 Since the volume of a prism is its base area times its height, the volume of the square prism can be found as follows.
The volume of the prism part of the traffic cone is 196 cubic inches.
Finding the Volume of the Cone
Use the formula for the volume of a cone to find the volume of the part that has a conic shape. V_c = 1/3 π r^2 h Substitute 5 for r and 30 for h into the formula and solve for V_c.
V_c = 1/3 π r^2 h
SubstituteII
r= 5, h= 30
V_c = 1/3 π ( 5)^2 ( 30)
▼
Simplify right-hand side
CalcPow
Calculate power
V_c = 1/3 π (25) (30)
MoveRightFacToNumOne
1/b* a = a/b
V_c = π (25) (30)/3
UseCalc
Use a calculator
V_c = 785.398163 ...
RoundInt
Round to nearest integer
V_c ≈ 785
The volume of the cone is about 785 cubic inches. Now the volume of the traffic cone can be found.
Volume of the Traffic Cone
The volume of the prism part — 196 cubic inches — can be added to the volume of the cone part — 785 cubic inches — to determine the volume of the traffic cone. & 196 + & 785 [-0.25em] &981 The volume of the traffic cone is about 981 cubic inches.
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Example
A Double-Walled Glass Cup
A double-walled glass cup is a special cup with two layers of glass that help keep the drink at the right temperature, whether hot or cold. Ramsha has one of these cups. Her cup is cylindrical with a radius of 4 centimeters and a height of 12 centimeters. The second wall of the cup creates a cone.
Ramsha fills the cup with water.
a
Ramsha wants to find the volume of the water filling her cup and the volume of the air between the cup walls. Help her in calculate these volumes. Round the answers to the nearest whole number.
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b
What percent of the volume of the entire cup is the volume of the air between the walls of the double-walled glass cup? Round the answer to one decimal place.
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Hint
a
The volume of a cone is one third the product of π, the square of the radius, and the height. The volume of a cylinder is the product of π, the square of the radius, and the height.
b
Use the exact values for the volumes from Part A.
Solution
a
The volumes of each solid will be calculated one at a time.
Volume of Water
Ramsha will fill the cone with water, so the volume of the cone is needed. The cone has the same height and radius as the cylinder, measuring 12 centimeters and 4 centimeters, respectively.
The volume of a cone is one third the product of π, the square of the radius, and the height. V_(cone) = 1/3π r^2h To find the volume of this cone, substitute 12 and 4 into the formula for h and r, respectively, and evaluate.
V_(cone) = 1/3π r^2h
SubstituteII
h= 12, r= 4
V_(cone) = 1/3π ( 4^2)( 12)
▼
Evaluate right-hand side
CalcPow
Calculate power
V_(cone) = 1/3π (16)(12)
Multiply
Multiply
V_(cone) = 1/3π (192)
CommutativePropMult
Commutative Property of Multiplication
V_(cone) = 1/3(192)π
MoveRightFacToNumOne
1/b* a = a/b
V_(cone) = 192/3π
CalcQuot
Calculate quotient
V_(cone) = 64π
The volume of the water filling the cup is 64π cubic centimeters. Using a calculator, find the result of 64π and then round it to the nearest whole number.
Ramsha filled the cup with approximately 201 cubic centimeters of water.
Volume of Air
Next, Ramsha needs to find the volume of the air between the two walls of the cup. The first step is to find the volume of the shell of the cup. The cup has a cylindrical shape with a height of 12 centimeters and a radius of 4 centimeters.
The volume of a cylinder is the product of π, the square of the radius, and the height. V_(cylinder) =π r^2h Substitute 12 and 4 into the formula for h and r, respectively, again and evaluate.
The volume of the cylindrical shell of the cup is 192π cubic centimeters. The volume of the air between the walls of the double-walled glass cup is the difference between the volume of the cylinder and the volume of the cone. Remember, the volume of the cone was previously determined to be 64π cubic centimeters. 192π- 64π = 128 π The volume of the region between the cone and the cylinder is 128π cubic centimeters. This implies that the volume of the air between the walls of the double-walled glass cup is also 128π cubic centimeters. Use a calculator to find the nearest integer value of the volume.
The volume of the air between the walls of the double-walled glass cup is 402 cubic centimeters.
b
In Part A it was found that the volume of the cylindrical-shaped cup is 192π cubic centimeters and the volume of the portion of the cylinder not occupied by the cone is 128π cubic centimeters. Calculating the ratio of the second value to the first value will provide the desired percentage.
128π/192π
▼
Evaluate
CancelCommonFac
Cancel out common factors
128π/192π
SimpQuot
Simplify quotient
128/192
ReduceFrac
a/b=.a /64./.b /64.
2/3
▼
Convert to percent
FracToDiv
a/b=a÷ b
0.666666...
WritePercent
Convert to percent
66.666666... %
RoundDec
Round to 1 decimal place(s)
66.7 %
The volume of the air between the walls of the double-walled glass cup makes up about 66.7 % of the volume of the entire cup.
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Example
Surface Area of the Double-Walled Glass Cup
Ramsha also wants to calculate the surface area of her double-walled glass cup.
Calculate the surface area of the cup with her. Round the answer to two decimal places.
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Hint
The surface area of the cup consists of the lateral area of the cylinder, one of the bases of the cylinder, and the lateral area of the cone.
Solution
The double-walled glass cup is made up of a cylinder with a cone inside. To calculate its surface area, the lateral areas of the cylinder and the cone, along with one base area of the cylinder, need to be calculated.
Lateral and Base Areas of the Cylinder
Notice that only one end of the cylinder is closed, so only the sum of the lateral area and the base area of the cylinder will be calculated.
The lateral area of a cylinder is twice the product of π, the radius, and the height. LA_(cylinder)=2π r h To find the lateral area of this cylinder, substitute 12 and 4 for h and r, respectively.
LA_(cylinder)=2π r h
SubstituteII
h= 12, r= 4
LA_(cylinder)=2π ( 4) ( 12)
▼
Evaluate right-hand side
LA_(cylinder) = 96π
The lateral area of the cylinder is about 96 π square centimeters. The base area of the cylinder is calculated by finding the area of a circle with a radius of 4 centimeters.
BA_(cylinder)=π r^2
Substitute
r= 4
BA_(cylinder)=π ( 4)^2
CalcPow
Calculate power
BA_(cylinder)=π (16)
CommutativePropMult
Commutative Property of Multiplication
BA_(cylinder)=16π
Next, calculate the total by adding the lateral area of the cylinder to one of its base areas. Surface Area of the Cylindrical Part of the Cup 96π+16π=112π
Lateral Area of the Cone
The lateral area of a cone is the product of π, the radius, and the slant height of the cone. LA_(cone) = π r l The slant height l is the hypotenuse of the right triangle formed by the radius, the height, and the segment connecting the center of the base of the cylinder with a point on the circumference of the opposite base.
The missing value can be found by using the Pythagorean Theorem.
a^2+b^2=c^2
SubstituteValues
Substitute values
4^2+ 12^2= l^2
▼
Solve for l
CalcPow
Calculate power
16+144=l ^2
AddTerms
Add terms
160=l ^2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(160)=l
RearrangeEqn
Rearrange equation
l=sqrt(160)
The slant height of the cone is sqrt(160) centimeters. Now the formula for the lateral area of a cone can be used. Substitute sqrt(160) for l and 4 for r and simplify.
LA_(cone)=π r l
SubstituteII
l= sqrt(160), r= 4
LA_(cone)=π ( 4) ( sqrt(160))
CommutativePropMult
Commutative Property of Multiplication
LA_(cone)= 4 sqrt(160) π
Total Surface Area
Finally, the combined areas of the cylinder and the cone will provide the total surface area of the cup. Use a calculator to make the calculations and round the result to two decimal places.
112π+ 4 sqrt(160) π
≈ 510.81
The surface area of the double-walled glass cup is about 510.81 square centimeters.
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Example
Volume of a Pencil
Ramsha bought a pencil with a radius of 3 millimeters. The total length of the pencil, excluding the eraser, is 160 millimeters. The tip of the pencil is a 10-millimeter high cone.
Assuming the eraser is half of a sphere, what is the volume of the pencil? Round the answer to one decimal place.
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Hint
The pencil is made of a cone, a cylinder, and a hemisphere, all with the same radius. The volume of the pencil equals the sum of the volumes of each solid.
Solution
The given pencil can be seen to consist of a cone, a cylinder, and half of a sphere — all with the same radius.
As such, the volume of the pencil equals the sum of the volumes of each of these solids.
| Volume of a Cone | Volume of a Cylinder | Volume of a Hemisphere |
|---|---|---|
| V_1 = 1/3π r^2h | V_2 = π r^2h | V_3 = 2/3π r^3 |
Use a calculator to make the calculations easier.
Volume of the Pencil's Tip
The tip of the pencil is a cone with a radius of 3 millimeters and a height of 10 millimeters. Substituting these values into the first formula will give the volume of the tip.
V_1 = 1/3π r^2h
SubstituteII
r= 3, h= 10
V_1 = 1/3π ( 3)^2( 10)
▼
Simplify right-hand side
CalcPow
Calculate power
V_1 = 1/3π (9)(10)
Multiply
Multiply
V_1 = 1/3π (90)
CommutativePropMult
Commutative Property of Multiplication
V_1 = 1/3(90)π
MoveRightFacToNumOne
1/b* a = a/b
V_1 = 90/3π
CalcQuot
Calculate quotient
V_1 = 30π
The tip of the pencil has a volume of 30π cubic millimeters.
Volume of the Pencil's Body
The body of the pencil is a cylinder with a radius of 3 millimeters. To find the height of the cylinder, subtract the height of the tip of the pencil from the original length of the pencil. 160 mm - 10 mm = 150 mm Next, substitute r=3 and h=150 into the formula for the volume a cylinder.
The body of the pencil has a volume of 1350π cubic millimeters.
Volume of the Pencil's Eraser
The eraser is a hemisphere with a radius of 3 millimeters. To find its volume, substitute r=3 into the hemisphere volume formula.
V_3 = 2/3π r^3
Substitute
r= 3
V_3 = 2/3π ( 3)^3
▼
Simplify right-hand side
CalcPow
Calculate power
V_3 = 2/3π (27)
CommutativePropMult
Commutative Property of Multiplication
V_3 = 2/3(27)π
MoveRightFacToNum
a/c* b = a* b/c
V_3 = 2(27)/3π
Multiply
Multiply
V_3 = 54/3π
CalcQuot
Calculate quotient
V_3 = 18π
The eraser of the pencil has a volume of 18π cubic millimeters.
Pencil's Volume
Finally, the total volume of the pencil is equal to the sum of the volumes of its parts.
V_P = V_1 + V_2 + V_3
SubstituteValues
Substitute values
V_P = 30π + 1350π + 18π
▼
Evaluate right-hand side
FactorOut
Factor out π
V_P = π(30 + 1350 + 18)
AddTerms
Add terms
V_P = π(1398)
CommutativePropMult
Commutative Property of Multiplication
V_P = 1398π
UseCalc
Use a calculator
V_P = 4391.946529...
RoundDec
Round to 1 decimal place(s)
V_P ≈ 4391.9
In conclusion, the volume of Ramsha's pencil is approximately 4391.9 cubic millimeters.
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Example
Finding the Lateral Area of a Roof
Ramsha's house is a rough composite solid consisting of a square pyramid with a height of 8 feet and a base side length of 30 feet on top of a square prism.
Ramsha's father decides to cover the roof of their house with waterproof insulation material. Help Ramsha and her father calculate how many square feet of insulation material are needed.
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Hint
Ramsha only needs to know the lateral area of the pyramid. Use the Pythagorean Theorem to find the slant height.
Solution
The amount of insulation material to cover the roof corresponds to the lateral area of the pyramid. Recall the formula for the lateral area of a pyramid.
LA = 1/2pl
In this formula, p is the perimeter of the base and l is the slant height. The slant height of this pyramid can be found by using the Pythagorean Theorem.
The height h of the pyramid is the distance between the vertex and the base, so h=8 for this pyramid. The value of b is half the base side length, so b= 302=15 feet.
Since a negative value does not make sense in this context, only the principal root is considered. This means that the slant height is 17 feet. The next step is to find the perimeter of the base. Since the base is a square, its perimeter p is 4 times the base side length. p = 4 * 30 =120 Finally, the lateral area of the pyramid can be found by substituting p=120 and l = 17 into the formula.
LA = 1/2pl
SubstituteII
p= 120, l= 17
LA = 1/2( 120)( 17)
▼
Evaluate right-hand side
Multiply
Multiply
LA = 1/2 * 2040
MoveRightFacToNumOne
1/b* a = a/b
LA = 2040/2
CalcQuot
Calculate quotient
LA = 1020
Ramsha's father will need 1020 square feet of insulation material to cover the entire roof.
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Example
Finding the Volume of a Deck Prism
While her father is busy installing the insulation material, Ramsha decides to explore the attic. She discovers her grandfather's old deck prism, a captivating object designed to illuminate cabins below the deck of a ship before electric lighting. The deck prism is a composite solid made up of a base prism and a pyramid, both with regular hexagonal bases.
Feeling a connection with her grandfather, Ramsha takes a closer look at the deck prism. Find the volume of the deck prism. Round the answer to the two decimal places.
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Hint
The formula for the area of a regular hexagon with side lengths a is B= 3a^2sqrt(3)2.
Solution
The deck prism is composed of two solids:
- a regular hexagonal prism with an edge length of 4 centimeters and a height of 2 centimeters, and
- a regular hexagonal pyramid with an edge length of 3 centimeters and a height of 4 centimeters.
This means that the volume of the deck prism is the sum of the volumes of the two solids. The volume of each solid will be found one at a time.
Volume of the Base Prism
The base of the prism is a regular hexagon with a side length of 4 centimeters. Recall the formula for the area of a regular hexagon with side lengths a. B =3a^2sqrt(3)/2 Substitute 4 for a into the formula and evaluate its value.
The area of the hexagonal base is 24sqrt(3) square centimeters. Now the volume of the prism can be found by multiplying the base area by the height of the hexagonal prism. V_1 = 24sqrt(3) * 2 ⇒ V_1 = 48sqrt(3)
Volume of the Pyramid
The base of the pyramid is a regular hexagon with side lengths of 3 centimeters. Use the formula for the area of the hexagonal base again, this time substituting 3 for a.
Now the volume can be found. Recall that the volume of a pyramid is one third of the product of its base area and height. The height of the pyramid is 4 centimeters.
V_2 =1/3Bh
SubstituteII
B= 27sqrt(3)/2, h= 4
V_2 = 1/3 * 27sqrt(3)/2 * 4
▼
Evaluate right-hand side
MoveRightFacToNum
a/c* b = a* b/c
V_2 = 1/3 * 108sqrt(3)/2
MultFrac
Multiply fractions
V_2 = 108sqrt(3)/6
SimpQuot
Simplify quotient
V_2 = 18 sqrt(3)
Volume of the Deck Prism
The sum of the volumes of the solids will give the total volume of the deck prism.
The deck prism has a volume of about 114.32 cubic centimeters. A sense of wonder washes over Ramsha as she holds the relic of maritime history in her hands and thinks about the stories her grandfather told her about his life at sea.
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Closure
Composite Solids in Real Life
This lesson explored a few real-life examples of composite solids. The calculation of volumes and surface areas for these combined shapes were examined. However, more composite solids can be found everywhere in daily life.
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Volume and Surface Area of Composite Solids
Exercise 2.1
Harmony Harvest Farms is designing a silo to store the wheat harvested from the fields.
Determine the square footage of steel needed to create the silo. Round the result to the nearest whole number.
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We want to to calculate the required square footage of steel for the creation of the silo. This means determining the surface area of the silo, which is a composite solid made up of a cone and a cylinder.
To determine the surface area of the silo, we need to calculate the lateral areas of the cone and the cylinder and the area of one base of the cylinder. The sum of these areas will give us the total surface area of the silo. Notice that the base of the cone and the other base of the cylinder are not included in the calculations. Let's start with the lateral area of the cone.
Lateral Area of the Cone
The lateral area of a cone can be calculated using the following formula. LA_(cone)=π rl In this formula, r is the cone's radius and l is its slant height. We will first find the slant height by using the Pythagorean Theorem. The height of the cone is 15 feet and the radius is 20 feet. Let's use this information to find the slant height.
The slant height of the cone is 25 feet. Now we can calculate the lateral area of the cone.
The lateral area of the cone is 500π square feet.
Surface Area of the Cylinder
Now we will calculate the surface area of the cylinder. Note that only one end of the the cylinder is included in the surface area of the composite solid, so we need to calculate the sum of the lateral area and one base area of the cylinder. The lateral area of a cylinder is twice the product of π, the radius, and the height. LA_(cylinder)=2π r h Let's substitute 30 and 20 into the formula for h and r, respectively.
The lateral area of the cylinder is 1200 π square feet. The base area of the cylinder is calculated by finding the area of a circle with a radius of 20 feet.
Next, calculate the total by adding the lateral area of the cylinder to one of its base areas. Surface Area of the Cylinder Part of the Silo 1200π+400π=1600π The related surface area of the cylinder part of the silo is 1600π square feet.
Total Surface Area of the Silo
Finally, let's add the surface areas of the cylinder and the cone together to find the total surface area of the silo. We can use a calculator to make the calculations easier and round the result to the nearest whole number.
The surface area of the silo is about 6597 square feet, so Harmony Harvest Farms needs about 6597 square feet of steel to build the silo.
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Izabella and her family are planning to build a pool in their yard. They picture it as a combination of a rectangular prism and half of a cylinder. Izabella made a sketch of the pool to show her friends.
Find the amount of water needed to fill the pool, expressed in cubic meters. Round the result to the nearest whole number.
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We are given a sketch of the pool that Izabella and her family plan to build, and we want to determine the amount of water needed to fill it.
We need to calculate the volume of the pool, since the volume of a solid is the measure of the amount of space inside it. The pool is is a composite solid, so we will calculate the volume of the prism part and the volume of a cylinder part separately, then add them together.
Volume of the Prism
The prism section of the pool has side lengths of 2 meters, 8 meters, and 10 meters.
We can calculate the volume of the prism part by multiplying the base area by the height. V_p=Bh The base of the prism is a rectangle with a length of 10 meters and a width of 8 meters. Let's multiply these values together to find the area of the base. B= 10( 8) ⇒ B=80m^2 The base area of the prism is 80 square meters. Now we can substitute 80 for B and 2 for h into the prism volume formula and simplify. Let's do it!
The volume of the prism part is 160 cubic meters.
Volume of the Half Cylinder
The half cylinder part of the pool has a radius of 2 meters and a height of 2 meters. Let's use the volume formula for a cylinder. Since the part of the pool is only half a cylinder, we will divide the formula by 2.
The volume of the half cylinder part of the pool is 4π cubic meters.
Total Volume of the Pool
As a final step, let's add the volume of the prism part and the volume of the half cylinder part together to get the total volume of the pool. Use a calculator to make the calculations easier and and remember to round the answer to the nearest whole number!
The pool can hold around 173 cubic meters of water since its volume is approximately 173 cubic meters.
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Volume and Surface Area of Composite Solids
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