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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.**

Explore

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.

Discussion

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.

The volume of a composite solid is either the sum or the difference between the volumes of the individual solids, whichever is applicable. The surface area of a composite solid is the sum of the faces that enclose the solid.Example

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.924711125em;vertical-align:0em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">in<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.924711125em;\"><span style=\"top:-3.173603125em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["981"]}}

The volume occupied by the traffic cone is the sum of the volume of the prism base and the volume of the cone part.

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 $V_{c}.$
### 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 $14.$
### 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.
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.

$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. $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. $V_{c}=31 πr_{2}h $

Substitute $5$ for $r$ and $30$ for $h$ into the formula and solve for $V_{c}.$
$V_{c}=31 πr_{2}h$

SubstituteII

$r=5$, $h=30$

$V_{c}=31 π(5)_{2}(30)$

▼

Simplify right-hand side

CalcPow

Calculate power

$V_{c}=31 π(25)(30)$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V_{c}=3π(25)(30) $

UseCalc

Use a calculator

$V_{c}=785.398163…$

RoundInt

Round to nearest integer

$V_{c}≈785$

$+ 196785981 $

The volume of the traffic cone is about $981$ cubic inches.
Example

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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Water:","formTextAfter":"cm<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["201"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Air:","formTextAfter":"cm<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["402"]}}

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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mspace\" style=\"margin-right:0.16666666666666666em;\"><\/span><span class=\"mord\">%<\/span><\/span><\/span><\/span>","answer":{"text":["66.7"]}}

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.

a The volumes of each solid will be calculated one at a time.

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}=31 πr_{2}h $

To find the volume of this cone, substitute $12$ and $4$ into the formula for $h$ and $r,$ respectively, and evaluate.
$V_{cone}=31 πr_{2}h$

SubstituteII

$h=12$, $r=4$

$V_{cone}=31 π(4_{2})(12)$

▼

Evaluate right-hand side

CalcPow

Calculate power

$V_{cone}=31 π(16)(12)$

Multiply

Multiply

$V_{cone}=31 π(192)$

CommutativePropMult

Commutative Property of Multiplication

$V_{cone}=31 (192)π$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V_{cone}=3192 π$

CalcQuot

Calculate quotient

$V_{cone}=64π$

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}=πr_{2}h $

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=128π $

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.

$192π128π $

▼

Evaluate

CancelCommonFac

Cancel out common factors

$192π 128π $

SimpQuot

Simplify quotient

$192128 $

ReduceFrac

$ba =b/64a/64 $

$32 $

▼

Convert to percent

FracToDiv

$ba =a÷b$

$0.666666…$

WritePercent

Convert to percent

$66.666666…%$

RoundDec

Round to $1$ decimal place(s)

$66.7%$

Example

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8238736249999999em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">cm<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["510.81"]}}

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.

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.

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}=2πrh $

To find the lateral area of this cylinder, substitute $12$ and $4$ for $h$ and $r,$ respectively.
$LA_{cylinder}=2πrh$

SubstituteII

$h=12$, $r=4$

$LA_{cylinder}=2π(4)(12)$

▼

Evaluate right-hand side

$LA_{cylinder}=96π$

$BA_{cylinder}=πr_{2}$

Substitute

$r=4$

$BA_{cylinder}=π(4)_{2}$

CalcPow

Calculate power

$BA_{cylinder}=π(16)$

CommutativePropMult

Commutative Property of Multiplication

$BA_{cylinder}=16π$

$Surface Area of theCylindrical Part of the Cup 96π+16π=112π $

$LA_{cone}=πrℓ $

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

$LHS =RHS $

$160 =ℓ$

RearrangeEqn

Rearrange equation

$ℓ=160 $

$LA_{cone}=πrℓ$

SubstituteII

$ℓ=160 $, $r=4$

$LA_{cone}=π(4)(160 )$

CommutativePropMult

Commutative Property of Multiplication

$LA_{cone}=4160 π$

$112π+4160 π$

$≈510.81$

Example

Ramsha bought a pencil with a radius of $3$ millimeters. The total length of the pencil, excluding the eraser, is $160$ millimeters. Moreover, 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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"mm<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["4391.9"]}}

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.

The given pencil can be seen to consist of a cone, a cylinder, and half of a sphere — all with the same radius.

### 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.
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.
### 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.
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.
In conclusion, the volume of Ramsha's pencil is approximately $4391.9$ cubic millimeters.

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 |
---|---|---|

$V_{1}=31 πr_{2}h$ | $V_{2}=πr_{2}h$ | $V_{3}=32 πr_{3}$ |

Use a calculator to make the calculations easier.

$V_{1}=31 πr_{2}h$

SubstituteII

$r=3$, $h=10$

$V_{1}=31 π(3)_{2}(10)$

▼

Simplify right-hand side

CalcPow

Calculate power

$V_{1}=31 π(9)(10)$

Multiply

Multiply

$V_{1}=31 π(90)$

CommutativePropMult

Commutative Property of Multiplication

$V_{1}=31 (90)π$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V_{1}=390 π$

CalcQuot

Calculate quotient

$V_{1}=30π$

$160mm−10mm=150mm $

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. $V_{3}=32 πr_{3}$

Substitute

$r=3$

$V_{3}=32 π(3)_{3}$

▼

Simplify right-hand side

CalcPow

Calculate power

$V_{3}=32 π(27)$

CommutativePropMult

Commutative Property of Multiplication

$V_{3}=32 (27)π$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V_{3}=32(27) π$

Multiply

Multiply

$V_{3}=354 π$

CalcQuot

Calculate quotient

$V_{3}=18π$

$V_{P}=V_{1}+V_{2}+V_{3}$

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)

$V_{P}≈4391.9$

Example

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.97402753125em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">ft<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.96426190625em;\"><span style=\"top:-3.21315390625em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["1020"]}}

Ramsha only needs to know the lateral area of the pyramid. Use the Pythagorean Theorem to find the slant height.

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=21 pℓ $

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 $b=230 =15$ 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.