surface area and volume problem solving questions

.

216 cm.

7.1 cm long and a lateral edge 18.2 cm long. Determine its volume and surface area.

102 cm, 64 cm. Height of a pillar is 1.5 m.

do we need to paint ?

1.1 dm. How many liters of water will fill up a vase, if thickness of its bottom is 1.5 cm ?

of water. To what height reaches the water ?

of road it will flatten if it turns 35 times ?

= 8.8 / ?

6 cm, 8 cm. The side edges are all of the same length 12.5 cm. Find the surface area of the pyramid.

and the same height of 15 cm. Which of these two solids has a larger surface area ?

2.3 dm if the height of cone is 46 mm.

36 cm and a height of 46 cm. How many euros we will pay for the color, if we need 500 cm of a paint color to paint 1 m and 1 liter of the color costs 8 € ?

12 cm and 8 cm. Jane remodeled Michael's pyramid into a cone with a base diameter of 10 cm. What was the height of Jane's cone ?

50 cm and the upper edges of a rectangular base 20 cm and 30 cm. How many liters of water can the kettle hold ?

20 cm, the second one has the shape of a truncated cone with the bottom base diameter of 25 cm and the upper base diameter of 15 cm. Which vase can hold more water if the height of both two vases is 0.5 meters ?

we need 0.1 liters of varnish. How many liters of varnish do we have to buy, if the bowls are 25 cm high, the bottom of the bowls has a diameter of 20 cm and the upper base has a diameter of 30 cm ?

of gas will fit into it ?

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Surface Area and Volume in the Real World

New york state common core math grade 6, module 5, lesson 19.

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9.6 Solve Geometry Applications: Volume and Surface Area

Learning objectives.

• Find volume and surface area of rectangular solids
• Find volume and surface area of spheres
• Find volume and surface area of cylinders
• Find volume of cones

Be Prepared 9.6

Before you get started, take this readiness quiz.

• Evaluate x 3 x 3 when x = 5 . x = 5 . If you missed this problem, review Example 2.15 .
• Evaluate 2 x 2 x when x = 5 . x = 5 . If you missed this problem, review Example 2.16 .
• Find the area of a circle with radius 7 2 . 7 2 . If you missed this problem, review Example 5.39 .

In this section, we will finish our study of geometry applications. We find the volume and surface area of some three-dimensional figures. Since we will be solving applications, we will once again show our Problem-Solving Strategy for Geometry Applications.

• Step 1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
• Step 2. Identify what you are looking for.
• Step 3. Name what you are looking for. Choose a variable to represent that quantity.
• Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Step 5. Solve the equation using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

Find Volume and Surface Area of Rectangular Solids

A cheerleading coach is having the squad paint wooden crates with the school colors to stand on at the games. (See Figure 9.28 ). The amount of paint needed to cover the outside of each box is the surface area , a square measure of the total area of all the sides. The amount of space inside the crate is the volume, a cubic measure.

Each crate is in the shape of a rectangular solid . Its dimensions are the length, width, and height. The rectangular solid shown in Figure 9.29 has length 4 4 units, width 2 2 units, and height 3 3 units. Can you tell how many cubic units there are altogether? Let’s look layer by layer.

Altogether there are 24 24 cubic units. Notice that 24 24 is the length × width × height . length × width × height .

The volume, V , V , of any rectangular solid is the product of the length, width, and height.

We could also write the formula for volume of a rectangular solid in terms of the area of the base. The area of the base, B , B , is equal to length × width . length × width .

We can substitute B B for L · W L · W in the volume formula to get another form of the volume formula.

We now have another version of the volume formula for rectangular solids. Let’s see how this works with the 4 × 2 × 3 4 × 2 × 3 rectangular solid we started with. See Figure 9.29 .

To find the surface area of a rectangular solid, think about finding the area of each of its faces. How many faces does the rectangular solid above have? You can see three of them.

Notice for each of the three faces you see, there is an identical opposite face that does not show.

The surface area S S of the rectangular solid shown in Figure 9.30 is 52 52 square units.

In general, to find the surface area of a rectangular solid, remember that each face is a rectangle, so its area is the product of its length and its width (see Figure 9.31 ). Find the area of each face that you see and then multiply each area by two to account for the face on the opposite side.

Volume and Surface Area of a Rectangular Solid

For a rectangular solid with length L , L , width W , W , and height H : H :

Manipulative Mathematics

Example 9.47.

For a rectangular solid with length 14 14 cm, height 17 17 cm, and width 9 9 cm, find the ⓐ volume and ⓑ surface area.

Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

Try It 9.93

Find the ⓐ volume and ⓑ surface area of rectangular solid with the: length 8 8 feet, width 9 9 feet, and height 11 11 feet.

Try It 9.94

Find the ⓐ volume and ⓑ surface area of rectangular solid with the: length 15 15 feet, width 12 12 feet, and height 8 8 feet.

Example 9.48

A rectangular crate has a length of 30 30 inches, width of 25 25 inches, and height of 20 20 inches. Find its ⓐ volume and ⓑ surface area.

Try It 9.95

A rectangular box has length 9 9 feet, width 4 4 feet, and height 6 6 feet. Find its ⓐ volume and ⓑ surface area.

Try It 9.96

A rectangular suitcase has length 22 22 inches, width 14 14 inches, and height 9 9 inches. Find its ⓐ volume and ⓑ surface area.

Volume and Surface Area of a Cube

A cube is a rectangular solid whose length, width, and height are equal. See Volume and Surface Area of a Cube, below. Substituting, s for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get:

So for a cube, the formulas for volume and surface area are V = s 3 V = s 3 and S = 6 s 2 . S = 6 s 2 .

For any cube with sides of length s , s ,

Example 9.49

A cube is 2.5 2.5 inches on each side. Find its ⓐ volume and ⓑ surface area.

Try It 9.97

For a cube with side 4.5 meters, find the ⓐ volume and ⓑ surface area of the cube.

Try It 9.98

For a cube with side 7.3 yards, find the ⓐ volume and ⓑ surface area of the cube.

Example 9.50

A notepad cube measures 2 2 inches on each side. Find its ⓐ volume and ⓑ surface area.

Try It 9.99

A packing box is a cube measuring 4 4 feet on each side. Find its ⓐ volume and ⓑ surface area.

Try It 9.100

A wall is made up of cube-shaped bricks. Each cube is 16 16 inches on each side. Find the ⓐ volume and ⓑ surface area of each cube.

Find the Volume and Surface Area of Spheres

A sphere is the shape of a basketball, like a three-dimensional circle. Just like a circle, the size of a sphere is determined by its radius, which is the distance from the center of the sphere to any point on its surface. The formulas for the volume and surface area of a sphere are given below.

Showing where these formulas come from, like we did for a rectangular solid, is beyond the scope of this course. We will approximate π π with 3.14 . 3.14 .

Volume and Surface Area of a Sphere

For a sphere with radius r : r :

Example 9.51

A sphere has a radius 6 6 inches. Find its ⓐ volume and ⓑ surface area.

Try It 9.101

Find the ⓐ volume and ⓑ surface area of a sphere with radius 3 centimeters.

Try It 9.102

Find the ⓐ volume and ⓑ surface area of each sphere with a radius of 1 1 foot

Example 9.52

A globe of Earth is in the shape of a sphere with radius 14 14 centimeters. Find its ⓐ volume and ⓑ surface area. Round the answer to the nearest hundredth.

Try It 9.103

A beach ball is in the shape of a sphere with radius of 9 9 inches. Find its ⓐ volume and ⓑ surface area.

Try It 9.104

A Roman statue depicts Atlas holding a globe with radius of 1.5 1.5 feet. Find the ⓐ volume and ⓑ surface area of the globe.

Find the Volume and Surface Area of a Cylinder

If you have ever seen a can of soda, you know what a cylinder looks like. A cylinder is a solid figure with two parallel circles of the same size at the top and bottom. The top and bottom of a cylinder are called the bases. The height h h of a cylinder is the distance between the two bases. For all the cylinders we will work with here, the sides and the height, h h , will be perpendicular to the bases.

Rectangular solids and cylinders are somewhat similar because they both have two bases and a height. The formula for the volume of a rectangular solid, V = B h V = B h , can also be used to find the volume of a cylinder.

For the rectangular solid, the area of the base, B B , is the area of the rectangular base, length × width. For a cylinder, the area of the base, B , B , is the area of its circular base, π r 2 . π r 2 . Figure 9.33 compares how the formula V = B h V = B h is used for rectangular solids and cylinders.

To understand the formula for the surface area of a cylinder, think of a can of vegetables. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. See Figure 9.34 .

The distance around the edge of the can is the circumference of the cylinder’s base it is also the length L L of the rectangular label. The height of the cylinder is the width W W of the rectangular label. So the area of the label can be represented as

To find the total surface area of the cylinder, we add the areas of the two circles to the area of the rectangle.

The surface area of a cylinder with radius r r and height h , h , is

Volume and Surface Area of a Cylinder

For a cylinder with radius r r and height h : h :

Example 9.53

A cylinder has height 5 5 centimeters and radius 3 3 centimeters. Find the ⓐ volume and ⓑ surface area.

Try It 9.105

Find the ⓐ volume and ⓑ surface area of the cylinder with radius 4 cm and height 7cm.

Try It 9.106

Find the ⓐ volume and ⓑ surface area of the cylinder with given radius 2 ft and height 8 ft.

Example 9.54

Find the ⓐ volume and ⓑ surface area of a can of soda. The radius of the base is 4 4 centimeters and the height is 13 13 centimeters. Assume the can is shaped exactly like a cylinder.

Try It 9.107

Find the ⓐ volume and ⓑ surface area of a can of paint with radius 8 centimeters and height 19 centimeters. Assume the can is shaped exactly like a cylinder.

Try It 9.108

Find the ⓐ volume and ⓑ surface area of a cylindrical drum with radius 2.7 feet and height 4 feet. Assume the drum is shaped exactly like a cylinder.

Find the Volume of Cones

The first image that many of us have when we hear the word ‘cone’ is an ice cream cone. There are many other applications of cones (but most are not as tasty as ice cream cones). In this section, we will see how to find the volume of a cone.

In geometry, a cone is a solid figure with one circular base and a vertex. The height of a cone is the distance between its base and the vertex.The cones that we will look at in this section will always have the height perpendicular to the base. See Figure 9.35 .

Earlier in this section, we saw that the volume of a cylinder is V = π r 2 h . V = π r 2 h . We can think of a cone as part of a cylinder. Figure 9.36 shows a cone placed inside a cylinder with the same height and same base. If we compare the volume of the cone and the cylinder, we can see that the volume of the cone is less than that of the cylinder.

In fact, the volume of a cone is exactly one-third of the volume of a cylinder with the same base and height. The volume of a cone is

Since the base of a cone is a circle, we can substitute the formula of area of a circle, π r 2 π r 2 , for B B to get the formula for volume of a cone.

In this book, we will only find the volume of a cone, and not its surface area.

• Volume of a Cone

For a cone with radius r r and height h h .

Example 9.55

Find the volume of a cone with height 6 6 inches and radius of its base 2 2 inches.

Try It 9.109

Find the volume of a cone with height 7 7 inches and radius 3 3 inches

Try It 9.110

Find the volume of a cone with height 9 9 centimeters and radius 5 5 centimeters

Example 9.56

Marty’s favorite gastro pub serves french fries in a paper wrap shaped like a cone. What is the volume of a conic wrap that is 8 8 inches tall and 5 5 inches in diameter? Round the answer to the nearest hundredth.

Try It 9.111

How many cubic inches of candy will fit in a cone-shaped piñata that is 18 18 inches long and 12 12 inches across its base? Round the answer to the nearest hundredth.

Try It 9.112

What is the volume of a cone-shaped party hat that is 10 10 inches tall and 7 7 inches across at the base? Round the answer to the nearest hundredth.

Summary of Geometry Formulas

The following charts summarize all of the formulas covered in this chapter.

ACCESS ADDITIONAL ONLINE RESOURCES

Section 9.6 exercises, practice makes perfect.

In the following exercises, find ⓐ the volume and ⓑ the surface area of the rectangular solid with the given dimensions.

length 2 2 meters, width 1.5 1.5 meters, height 3 3 meters

length 5 5 feet, width 8 8 feet, height 2.5 2.5 feet

length 3.5 3.5 yards, width 2.1 2.1 yards, height 2.4 2.4 yards

length 8.8 8.8 centimeters, width 6.5 6.5 centimeters, height 4.2 4.2 centimeters

In the following exercises, solve.

Moving van A rectangular moving van has length 16 16 feet, width 8 8 feet, and height 8 8 feet. Find its ⓐ volume and ⓑ surface area.

Gift box A rectangular gift box has length 26 26 inches, width 16 16 inches, and height 4 4 inches. Find its ⓐ volume and ⓑ surface area.

Carton A rectangular carton has length 21.3 21.3 cm, width 24.2 24.2 cm, and height 6.5 6.5 cm. Find its ⓐ volume and ⓑ surface area.

Shipping container A rectangular shipping container has length 22.8 22.8 feet, width 8.5 8.5 feet, and height 8.2 8.2 feet. Find its ⓐ volume and ⓑ surface area.

In the following exercises, find ⓐ the volume and ⓑ the surface area of the cube with the given side length.

5 5 centimeters

10.4 10.4 feet

12.5 12.5 meters

Science center Each side of the cube at the Discovery Science Center in Santa Ana is 64 64 feet long. Find its ⓐ volume and ⓑ surface area.

Museum A cube-shaped museum has sides 45 45 meters long. Find its ⓐ volume and ⓑ surface area.

Base of statue The base of a statue is a cube with sides 2.8 2.8 meters long. Find its ⓐ volume and ⓑ surface area.

Tissue box A box of tissues is a cube with sides 4.5 inches long. Find its ⓐ volume and ⓑ surface area.

In the following exercises, find ⓐ the volume and ⓑ the surface area of the sphere with the given radius. Round answers to the nearest hundredth.

3 3 centimeters

7.5 7.5 feet

2.1 2.1 yards

In the following exercises, solve. Round answers to the nearest hundredth.

Exercise ball An exercise ball has a radius of 15 15 inches. Find its ⓐ volume and ⓑ surface area.

Balloon ride The Great Park Balloon is a big orange sphere with a radius of 36 36 feet . Find its ⓐ volume and ⓑ surface area.

Golf ball A golf ball has a radius of 4.5 4.5 centimeters. Find its ⓐ volume and ⓑ surface area.

Baseball A baseball has a radius of 2.9 2.9 inches. Find its ⓐ volume and ⓑ surface area.

In the following exercises, find ⓐ the volume and ⓑ the surface area of the cylinder with the given radius and height. Round answers to the nearest hundredth.

radius 3 3 feet, height 9 9 feet

radius 5 5 centimeters, height 15 15 centimeters

radius 1.5 1.5 meters, height 4.2 4.2 meters

radius 1.3 1.3 yards, height 2.8 2.8 yards

Coffee can A can of coffee has a radius of 5 5 cm and a height of 13 13 cm. Find its ⓐ volume and ⓑ surface area.

Snack pack A snack pack of cookies is shaped like a cylinder with radius 4 4 cm and height 3 3 cm. Find its ⓐ volume and ⓑ surface area.

Barber shop pole A cylindrical barber shop pole has a diameter of 6 6 inches and height of 24 24 inches. Find its ⓐ volume and ⓑ surface area.

Architecture A cylindrical column has a diameter of 8 8 feet and a height of 28 28 feet. Find its ⓐ volume and ⓑ surface area.

In the following exercises, find the volume of the cone with the given dimensions. Round answers to the nearest hundredth.

height 9 9 feet and radius 2 2 feet

height 8 8 inches and radius 6 6 inches

height 12.4 12.4 centimeters and radius 5 5 cm

height 15.2 15.2 meters and radius 4 4 meters

Teepee What is the volume of a cone-shaped teepee tent that is 10 10 feet tall and 10 10 feet across at the base?

Popcorn cup What is the volume of a cone-shaped popcorn cup that is 8 8 inches tall and 6 6 inches across at the base?

Silo What is the volume of a cone-shaped silo that is 50 50 feet tall and 70 70 feet across at the base?

Sand pile What is the volume of a cone-shaped pile of sand that is 12 12 meters tall and 30 30 meters across at the base?

Everyday Math

Street light post The post of a street light is shaped like a truncated cone, as shown in the picture below. It is a large cone minus a smaller top cone. The large cone is 30 30 feet tall with base radius 1 1 foot. The smaller cone is 10 10 feet tall with base radius of 0.5 0.5 feet. To the nearest tenth,

ⓐ find the volume of the large cone.

ⓑ find the volume of the small cone.

ⓒ find the volume of the post by subtracting the volume of the small cone from the volume of the large cone.

Ice cream cones A regular ice cream cone is 4 inches tall and has a diameter of 2.5 2.5 inches. A waffle cone is 7 7 inches tall and has a diameter of 3.25 3.25 inches. To the nearest hundredth,

ⓐ find the volume of the regular ice cream cone.

ⓑ find the volume of the waffle cone.

ⓒ how much more ice cream fits in the waffle cone compared to the regular cone?

Writing Exercises

The formulas for the volume of a cylinder and a cone are similar. Explain how you can remember which formula goes with which shape.

Which has a larger volume, a cube of sides of 8 8 feet or a sphere with a diameter of 8 8 feet? Explain your reasoning.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/prealgebra/pages/1-introduction

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• Book title: Prealgebra
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• Location: Houston, Texas
• Book URL: https://openstax.org/books/prealgebra/pages/1-introduction
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Volume and Surface Area

SOLVING SURFACE AREA AND VOLUME PROBLEMS

Problem 1 : 

Erin is making a jewelry box of wood in the  shape of a rectangular prism. The jewelry box  will have the dimensions shown below. The cost of painting the exterior of the box is $0.50 per square in. How much does Erin have to spend to paint the jewelry box ?

Solution : 

To know that total cost of painting, first we have to know the Surface area of the jewelry box. 

Find surface area of the box. 

Identify a base, and find its area and perimeter.

Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases.

Find base area. 

B  =  l x w

B  =  12 x 15

B  =  180 square in.

Find perimeter of the base. 

P  =  2(12) + 2(15)

P  =  24 + 30

P  =  54 in.

Step 2 : 

Identify the height, and find the surface area.

The height h of the prism is 6 inches. Use the formula to find the surface area.

S  =  Ph + 2B

S  =  54(6) + 2(180)

S  =  684 square inches

Total cost  =  Area x Cost per square in.

Total cost  =  684 x $0.50

Total cost  =  $342

Hence, Erin has to spend $342 to paint the jewelry box. 

Problem 2 : 

A metal box that is in the shape of rectangular prism has the following dimensions. The  length is 9 inches, width is 2 inches, and height is 1 1/ 2 inches. Find the total cost of silver coating for the entire box. 

To know that total cost of silver coating, first we have to know the Surface area of the metal box. 

B  =  9 x 2

B  =  18 square in.

P  =  2(9) + 2(2)

P  =  18 + 4

P  =  22 in.

The height h of the prism is 1 1/2 inches. Use the formula to find the surface area.

S  =  22(1 1/2) + 2(18)

S  =  22(3/2) + 36

S  =  33 + 36

S  =  69 square inches

Total cost  =  69 x $1.50

Total cost  =  $103.50

Hence, the total cost of silver coating for the entire box is $103.50. 

Problem 3 : 

Cherise is setting up her tent. Her tent is in the shape of a  trapezoidal prism shown below. How many cubic feet of space are in  her tent ?

To find the number of cubic feet of space in  the  tent, we have to find the volume of  Cherise's  tent. 

Volume of  Cherise's  tent (Trapezoidal prism) is 

  =  base area x height

V  =  b x h

Find base area.

Area of trapezoid with bases of lengths  b ₁ and b ₂ and height h. 

Base area (b)  =  (1/2) x ( b ₁ + b ₂)h

Base area (b)  =  (1/2) x (6  + 4 )4

Base area  =  20 sq.ft

Find volume of the  prism.

V  =  20 x 9

V  =  180 cubic.ft

Hence, the  number of cubic feet of space in  Cherise's  tent is 180. 

Problem 4 : 

Allie has two aquariums connected by  a small square prism. Find the volume  of the double aquarium.

Find the volume of each  of the larger aquariums.

Volume  =  Base area x Height

Volume  =  (4 x 3) x 3

Volume  =  12 x 3

Volume  =  36 cubic ft.  

Find the volume of the connecting prism.

Volume  =  (2 x 1) x 1

Volume  =  2 x 1

Volume  =  2 cubic ft.

Add the volumes of the three parts of the aquarium.

V  =  36 + 36 + 2

V  =  74 cubic ft.

The volume of the aquarium is 74 cubic ft.

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Volume Problem Solving

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To solve problems on this page, you should be familiar with the following: Volume - Cuboid Volume - Sphere Volume - Cylinder Volume - Pyramid

This wiki includes several problems motivated to enhance problem-solving skills. Before getting started, recall the following formulas:

• Volume of sphere with radius \(r:\) \( \frac43 \pi r^3 \)
• Volume of cube with side length \(L:\) \( L^3 \)
• Volume of cone with radius \(r\) and height \(h:\) \( \frac13\pi r^2h \)
• Volume of cylinder with radius \(r\) and height \(h:\) \( \pi r^2h\)
• Volume of a cuboid with length \(l\), breadth \(b\), and height \(h:\) \(lbh\)

Volume Problem Solving - Basic

Volume - problem solving - intermediate, volume problem solving - advanced.

This section revolves around the basic understanding of volume and using the formulas for finding the volume. A couple of examples are followed by several problems to try.

Find the volume of a cube of side length \(10\text{ cm}\). \[\begin{align} (\text {Volume of a cube}) & = {(\text {Side length}})^{3}\\ & = {10}^{3}\\ & = 1000 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

Find the volume of a cuboid of length \(10\text{ cm}\), breadth \(8\text{ cm}\). and height \(6\text{ cm}\). \[\begin{align} (\text {Area of a cuboid}) & = l × b × h\\ & = 10 × 8 × 6\\ & = 480 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

I made a large ice cream cone of a composite shape of a cone and a hemisphere. If the height of the cone is 10 and the diameter of both the cone and the hemisphere is 6, what is the volume of this ice cream cone? The volume of the composite figure is the sum of the volume of the cone and the volume of the hemisphere. Recall the formulas for the following two volumes: \( V_{\text{cone}} = \frac13 \pi r^2 h\) and \( V_{\text{sphere}} =\frac43 \pi r^3 \). Since the volume of a hemisphere is half the volume of a a sphere of the same radius, the total volume for this problem is \[\frac13 \pi r^2 h + \frac12 \cdot \frac43 \pi r^3. \] With height \(h =10\), and diameter \(d = 6\) or radius \(r = \frac d2 = 3 \), the total volume is \(48\pi. \ _\square \)

Find the volume of a cone having slant height \(17\text{ cm}\) and radius of the base \(15\text{ cm}\). Let \(h\) denote the height of the cone, then \[\begin{align} (\text{slant height}) &=\sqrt {h^2 + r^2}\\ 17&= \sqrt {h^2 + 15^2}\\ 289&= h^2 + 225\\ h^2&=64\\ h& = 8. \end{align}\] Since the formula for the volume of a cone is \(\dfrac {1}{3} ×\pi ×r^2×h\), the volume of the cone is \[ \frac {1}{3}×3.14× 225 × 8= 1884 ~\big(\text{cm}^{2}\big). \ _\square\]

Find the volume of the following figure which depicts a cone and an hemisphere, up to \(2\) decimal places. In this figure, the shape of the base of the cone is circular and the whole flat part of the hemisphere exactly coincides with the base of the cone (in other words, the base of the cone and the flat part of the hemisphere are the same). Use \(\pi=\frac{22}{7}.\) \[\begin{align} (\text{Volume of cone}) & = \dfrac {1}{3} \pi r^2 h\\ & = \dfrac {1 × 22 × 36 × 8}{3 × 7}\\ & = \dfrac {6336}{21} = 301.71 \\\\ (\text{Volume of hemisphere}) & = \dfrac {2}{3} \pi r^3\\ & = \dfrac {2 × 22 × 216}{3 × 7}\\ & = \dfrac {9504}{21} = 452.57 \\\\ (\text{Total volume of figure}) & = (301.71 + 452.57) \\ & = 754.28.\ _\square \end{align} \]

Try the following problems.

Find the volume (in \(\text{cm}^3\)) of a cube of side length \(5\text{ cm} \).

A spherical balloon is inflated until its volume becomes 27 times its original volume. Which of the following is true?

Bob has a pipe with a diameter of \(\frac { 6 }{ \sqrt { \pi } }\text{ cm} \) and a length of \(3\text{ m}\). How much water could be in this pipe at any one time, in \(\text{cm}^3?\)

What is the volume of the octahedron inside this \(8 \text{ in}^3\) cube?

A sector with radius \(10\text{ cm}\) and central angle \(45^\circ\) is to be made into a right circular cone. Find the volume of the cone.

\[\] Details and Assumptions:

• The arc length of the sector is equal to the circumference of the base of the cone.

Three identical tanks are shown above. The spheres in a given tank are the same size and packed wall-to-wall. If the tanks are filled to the top with water, then which tank would contain the most water?

A chocolate shop sells its products in 3 different shapes: a cylindrical bar, a spherical ball, and a cone. These 3 shapes are of the same height and radius, as shown in the picture. Which of these choices would give you the most chocolate?

\[\text{ I. A full cylindrical bar } \hspace{.4cm} \text{ or } \hspace{.45cm} \text{ II. A ball plus a cone }\]

How many cubes measuring 2 units on one side must be added to a cube measuring 8 units on one side to form a cube measuring 12 units on one side?

This section involves a deeper understanding of volume and the formulas to find the volume. Here are a couple of worked out examples followed by several "Try It Yourself" problems:

\(12\) spheres of the same size are made from melting a solid cylinder of \(16\text{ cm}\) diameter and \(2\text{ cm}\) height. Find the diameter of each sphere. Use \(\pi=\frac{22}{7}.\) The volume of the cylinder is \[\pi× r^2 × h = \frac {22×8^2×2}{7}= \frac {2816}{7}.\] Let the radius of each sphere be \(r\text{ cm}.\) Then the volume of each sphere in \(\text{cm}^3\) is \[\dfrac {4×22×r^3}{3×7} = \dfrac{88×r^3}{21}.\] Since the number of spheres is \(\frac {\text{Volume of cylinder}}{\text {Volume of 1 sphere}},\) \[\begin{align} 12 &= \dfrac{2816×21}{7×88×r^3}\\ &= \dfrac {96}{r^3}\\ r^3 &= \dfrac {96}{12}\\ &= 8\\ \Rightarrow r &= 2. \end{align}\] Therefore, the diameter of each sphere is \[2\times r = 2\times 2 = 4 ~(\text{cm}). \ _\square\]

Find the volume of a hemispherical shell whose outer radius is \(7\text{ cm}\) and inner radius is \(3\text{ cm}\), up to \(2\) decimal places. We have \[\begin{align} (\text {Volume of inner hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × R^3\\ & = \dfrac {1 × 4 × 22 × 27}{2 × 3 × 7}\\ & = \dfrac {396}{7}\\ & = 56.57 ~\big(\text{cm}^{3}\big) \\\\ (\text {Volume of outer hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × r^3\\ & = \dfrac {1 × 4 × 22 × 343}{2 × 3 × 7}\\ & = \dfrac {2156}{7}\\ & = 718.66 ~\big(\text{cm}^{3}\big) \\\\ (\text{Volume of hemispherical shell}) & = (\text{V. of outer hemisphere}) - (\text{V. of inner hemisphere})\\ & = 718.66 - 56.57 \\ & = 662.09 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

A student did an experiment using a cone, a sphere, and a cylinder each having the same radius and height. He started with the cylinder full of liquid and then poured it into the cone until the cone was full. Then, he began pouring the remaining liquid from the cylinder into the sphere. What was the result which he observed?

There are two identical right circular cones each of height \(2\text{ cm}.\) They are placed vertically, with their apex pointing downwards, and one cone is vertically above the other. At the start, the upper cone is full of water and the lower cone is empty.

Water drips down through a hole in the apex of the upper cone into the lower cone. When the height of water in the upper cone is \(1\text{ cm},\) what is the height of water in the lower cone (in \(\text{cm}\))?

On each face of a cuboid, the sum of its perimeter and its area is written. The numbers recorded this way are 16, 24, and 31, each written on a pair of opposite sides of the cuboid. The volume of the cuboid lies between \(\text{__________}.\)

A cube rests inside a sphere such that each vertex touches the sphere. The radius of the sphere is \(6 \text{ cm}.\) Determine the volume of the cube.

If the volume of the cube can be expressed in the form of \(a\sqrt{3} \text{ cm}^{3}\), find the value of \(a\).

A sphere has volume \(x \text{ m}^3 \) and surface area \(x \text{ m}^2 \). Keeping its diameter as body diagonal, a cube is made which has volume \(a \text{ m}^3 \) and surface area \(b \text{ m}^2 \). What is the ratio \(a:b?\)

Consider a glass in the shape of an inverted truncated right cone (i.e. frustrum). The radius of the base is 4, the radius of the top is 9, and the height is 7. There is enough water in the glass such that when it is tilted the water reaches from the tip of the base to the edge of the top. The proportion of the water in the cup as a ratio of the cup's volume can be expressed as the fraction \( \frac{m}{n} \), for relatively prime integers \(m\) and \(n\). Compute \(m+n\).

The square-based pyramid A is inscribed within a cube while the tetrahedral pyramid B has its sides equal to the square's diagonal (red) as shown.

Which pyramid has more volume?

Please remember this section contains highly advanced problems of volume. Here it goes:

Cube \(ABCDEFGH\), labeled as shown above, has edge length \(1\) and is cut by a plane passing through vertex \(D\) and the midpoints \(M\) and \(N\) of \(\overline{AB}\) and \(\overline{CG}\) respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).

If the American NFL regulation football

has a tip-to-tip length of \(11\) inches and a largest round circumference of \(22\) in the middle, then the volume of the American football is \(\text{____________}.\)

Note: The American NFL regulation football is not an ellipsoid. The long cross-section consists of two circular arcs meeting at the tips. Don't use the volume formula for an ellipsoid.

Answer is in cubic inches.

Consider a solid formed by the intersection of three orthogonal cylinders, each of diameter \( D = 10 \).

What is the volume of this solid?

Consider a tetrahedron with side lengths \(2, 3, 3, 4, 5, 5\). The largest possible volume of this tetrahedron has the form \( \frac {a \sqrt{b}}{c}\), where \(b\) is an integer that's not divisible by the square of any prime, \(a\) and \(c\) are positive, coprime integers. What is the value of \(a+b+c\)?

Let there be a solid characterized by the equation \[{ \left( \frac { x }{ a } \right) }^{ 2.5 }+{ \left( \frac { y }{ b } \right) }^{ 2.5 } + { \left( \frac { z }{ c } \right) }^{ 2.5 }<1.\]

Calculate the volume of this solid if \(a = b =2\) and \(c = 3\).

• Surface Area

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• Problems on Pyramid

Solved word problems on pyramid are shown below using step-by-step explanation with the help of the exact diagram in finding surface area and volume of a pyramid.

Worked-out problems on pyramid: 1. The base of a right pyramid is a square of side 24 cm. and its height is 16 cm.

(i) the area of its slant surface

(ii) area of its whole surface and

(iii) its volume. 

Let, the square WXYZ be the base of the right pyramid and its diagonals WY and XZ intersect at O. If  OP  be perpendicular to the plane of the square at O, then  OP  is the height of the pyramid.

Draw  OE  ┴  WX Then, E is the mid - point of  WX . 

By question,  OP  = 16 cm. and  WX  = 24 cm.  Therefore,  OE  =  EX  = 1/2 ∙  WX  = 12 cm Clearly,  PE  is the slant height of the pyramid.  Since  OP  ┴  OE , hence from ∆ POE we get,  PE² = OP² + OE² 

or,PE² = 16² + 12² 

or, PE² = 256 + 144 

or, PE² = 400

PE  = √400

Therefore,  PE  = 20.  Therefore, (i) the required area of slant surface of the right pyramid

= 1/2 × perimeter of the base × slant height. 

= 1/2 × 4 × 24× 20 square cm. 

= 960 square cm. 

(ii) The area of the whole surface of the right pyramid = area of slant surface + area of the base

= (960 + 24 × 24) square cm

= 1536 square cm.

(iii) the volume of the right pyramid

= 1/3 × area of the base × height

= 1/3 × 24 × 24 × 16 cubic cm 

= 3072 cubic cm.

2. The base of a right pyramid 8 m high, is an equilateral triangle of side 12√3 m. Find its volume and the slant surface. Solution:

Let equilateral ∆ WXY be the base and P, the vertex of the right pyramid.

In the plane of the ∆ WXY draw YZ perpendicular to WX and let OZ = 1/3 YZ . Then, O is the centroid of ∆ WXY. Let OP be perpendicular to the plane of ∆ WXY at O; then OP is the height of the pyramid. By question, WX = XY = YW = 8√3 m and OP = 8 m. Since ∆ WXY is equilateral and YZ ┴ WX Hence, Z bisects WX .

Therefore, XZ = 1/2 ∙ WX = 1/2 ∙ 12√3 = 6√3 m. Now, from right - angled ∆ XYZ we get,

YZ² = XY² - XZ²

or, YZ² = (12√3) ² - (6√3)²

or, YZ² = 6² (12 - 3)

or, YZ² = 6² ∙ 9

or, YZ² = 324

Therefore, YZ = 18

Therefore, OZ = 1/3 ∙ 18 = 6. Join PZ . Then, PZ is the slant height of the pyramid. Since OP is perpendicular to the plane of ∆ WXY at O, hence OP ┴ OZ . Therefore, from the right angled ∆ POZ we get,

PZ² = OZ² + OP²

or, PZ ² = 6² + 8²

or, PZ² = 36 + 64

or, PZ² = 100

Therefore, PZ = 10 Therefore, the required slant surface of the right pyramid

= 1/2 × perimetre of the base × slant height

= 1/2 × 3 × 12√3 × PZ

= 1/2 × 36√3 × 10

= 180√3 square meter.

and its volume = 1/3 × area of the base × height

= 1/3 × (√3)/4 (12√3)² × 8

[Since, area of equilateral triangle

= (√3)/4 × (length of a side)² and height = OP = 8]

= 288√3 cubic meter.

●  Mensuration

• Formulas for 3D Shapes
• Volume and Surface Area of the Prism
• Worksheet on Volume and Surface Area of Prism
• Volume and Whole Surface Area of Right Pyramid
• Volume and Whole Surface Area of Tetrahedron
• Volume of a Pyramid
• Volume and Surface Area of a Pyramid
• Worksheet on Volume and Surface Area of a Pyramid
• Worksheet on Volume of a Pyramid

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Surface area

Here you will learn about surface area, including what it is and how to calculate it for prisms and pyramids.

Students will first learn about surface area as part of geometry in 6 th grade.

What is surface area?

The surface area is the total area of all of the faces of a three-dimensional shape. This includes prisms and pyramids. The surface area is always recorded in square units.

Prisms are 3D shapes that have a polygonal base and rectangular faces. A rectangular prism has 6 rectangular faces, including 4 rectangular lateral faces and 2 rectangular bases.

For example,

Calculate the area of each face and then add them together for the surface area of the rectangular prism.

The surface area of the prism is the sum of the areas. Add each area twice, since each rectangle appears twice in the prism:

8+8+12+12+6+6=52 \, f t^2

You can also find the surface area by multiplying each area by 2 and then adding.

(2 \times 8)+(2 \times 12)+(2 \times 6)=52 \, f t^2

Step-by-step guide: Surface area of rectangular prism

[FREE] Surface Area Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of surface area. 15 questions with answers to identify areas of strength and support!

Another type of prism is a triangular prism.

A triangular prism is made up of 5 faces, including 2 triangular bases and 3 rectangular lateral faces.

Calculate the area of each face and then add them together for the surface area of the triangular prism.

The surface area of the prism is the sum of the areas. Add the area of the triangular base twice (or you can multiply it by 2 ), since it appears twice in the prism:

37.2+60+38.4+21+21=177.6 \mathrm{~mm}^2

Step-by-step guide: Surface area of triangular prism

Step-by-step guide: Surface area of a prism

Pyramids are another type of 3D shape. A pyramid is made up of a polygonal base and triangular lateral sides.

All lateral faces (sides) of this square pyramid are congruent.

To calculate the surface area of a pyramid , calculate the area of each face of the pyramid and then add the areas together.

\text {Area of the base }=2.5 \times 2.5=6.25 \mathrm{~cm}^2

\text {Area of a triangular face }=\cfrac{1}{2} \times 2.5 \times 4=5 \mathrm{~cm}^2

Add the area of the base and the 4 congruent triangular faces:

\text {Surface area }=6.25+5+5+5+5=6.25+(4 \times 5)=26.25 \mathrm{~cm}^2

The total surface area can also be written in one equation:

​​\begin{aligned} \text {Surface area of pyramid } & =\text {Area of base }+ \text {Areas of triangular faces } \\\\ & =2.5^2+4 \times\left(\cfrac{1}{2} \, \times 2.5 \times 4\right) \\\\ & =6.25+20 \\\\ & =26.25 \mathrm{~cm}^2 \end{aligned}

Step-by-step guide: Surface area of a pyramid

Common Core State Standards

How does this relate to 6 th grade math?

• Grade 6 – Geometry (6.G.A.4) Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.

How to calculate the surface area of a prism

In order to calculate the surface area:

Calculate the area of each face.

Add the area of each face together.

Include the units.

Surface area examples

Example 1: surface area of a rectangular prism.

Calculate the surface area of the rectangular prism.

A rectangular prism has 6 faces, with 3 pairs of identical faces.

2 Add the area of each face together.

Total surface area: 14+14+21+21+6+6=82

Since opposite rectangles are always congruent, you can also use multiplication to solve:

Total surface area: 14 \times 2+21 \times 2+6 \times 2=82

3 Include the units.

The measurements on this prism are in m , so the total surface area of the prism is 82 \mathrm{~m}^2.

Example 2: surface area of a triangular prism with an equilateral triangle – using a net

Calculate the surface area of the triangular prism. The base of the prism is an equilateral triangle with a perimeter of 16.5 \, ft.

First, use the perimeter of the base to find the length of each side. Since an equilateral triangle has all equal sides, s , the perimeter is s+s+s=16.5.

s=5.5 \, ft

You can unfold the triangular prism, and use the net to find the area of each face:

Remember that the edges in a prism are always equal, so if you were to fold up the net, the 5.5 \, ft side of the triangle would combine to form an edge with each corresponding rectangle – making their lengths equal.

The area of each triangular base:

\cfrac{1}{2} \times 4.8 \times 5.5=13.2

The area of each rectangular lateral face:

10 \times 5.5=55

If you have trouble keeping track of all the calculations, use a net:

The area of the base is always equal to the area of the opposite base, in this case the triangles.

Notice, since the triangle is equilateral, all the rectangular faces are equal as well.

Total surface area: 13.2+13.2+55+55+55=191.4

The measurements on this prism are in ft , so the total surface area of the prism is 191.4 \mathrm{~ft}^2.

Example 3: surface area of a square-based pyramid in cm

All the lateral faces of the pyramid are congruent. Calculate the surface area.

The base is a square with the area 6\times{6}=36\text{~cm}^2.

All four triangular faces are identical, so calculate the area of one triangle, and then multiply the area by 4 .

\begin{aligned} A&= \cfrac{1}{2} \, \times{b}\times{h}\\\\ &=\cfrac{1}{2} \, \times{10}\times{6}\\\\ &=30 \end{aligned}

30\times{4}=120

Add the area of the base and the area of the four triangles:

SA=36+120=156

The side lengths are measured in centimeters, so the area is measured in square centimeters.

SA=156\text{~cm}^{2}

Example 4: surface area of a rectangular prism – using a net

Calculate the lateral surface area of the rectangular prism. The base of the prism is a square and one side of the base measures 3 \, \cfrac{2}{3} inches.

You can unfold the rectangular prism, and use the net to find the area of each face:

Remember that the edges in a prism are always equal, so if you were to fold up the net, the 3 \cfrac{2}{3} \mathrm{~ft} side of the square would combine to form an edge with each corresponding rectangle – making their lengths equal.

\begin{aligned} & 9 \cfrac{4}{5} \, \times 3 \, \cfrac{2}{3} \\\\ &= \cfrac{49}{5} \, \times \cfrac{11}{3} \\\\ &= \cfrac{539}{15} \\\\ &= 35 \, \cfrac{14}{15} \end{aligned}

Remember, you are only finding the area of the lateral faces, so you do not need to calculate the area of the bases.

Notice, since the square has all equal sides, all the rectangular faces are equal as well.

Total lateral surface area:

\begin{aligned} & 35 \, \cfrac{14}{15}+35 \, \cfrac{14}{15}+35 \, \cfrac{14}{15}+35 \, \cfrac{14}{15} \\\\ & =140 \, \cfrac{56}{15} \\\\ & =143 \, \cfrac{11}{15} \end{aligned}

Since all the lateral faces are congruent, you can also use multiplication to solve:

\begin{aligned} & 4 \times 35 \, \cfrac{14}{15} \\\\ & =\cfrac{4}{1} \, \times \, \cfrac{539}{15} \\\\ & =\cfrac{2,156}{15} \\\\ & =143 \, \cfrac{11}{15} \end{aligned}

The measurements on this prism are in inches, so the total lateral surface area of the prism is 143 \, \cfrac{11}{15} \text {~inches }^2.

Example 5: surface area of a parallelogram prism with different units

Calculate the surface area of the parallelogram prism.

A parallelogram prism has 6 faces and, like a rectangular prism, it has 3 pairs of identical faces. The base is a parallelogram and all of the lateral faces are rectangular.

In this example, some of the measurements are in cm and some are in m . You must convert the units so that they are the same. Convert all the units to meters (m)\text{: } 40 {~cm}=0.4 {~m} and 50 {~cm}=0.5 {~m}.

Total surface area: 0.48+0.48+1.8+1.8+0.75+0.75=6.06

The measurements that we have used are in m so the surface area of the prism is 6.06 \mathrm{~m}^2.

Example 6: surface area of a square pyramid – word problem

Mara is making a square pyramid out of cardboard. She cut out 4 acute triangles that have a base of 5 inches and a height of 7.4 inches. How much cardboard will she need to complete the entire square pyramid?

The lateral faces are all congruent, acute triangles.

\begin{aligned} \text {Area of triangle } & =\cfrac{1}{2} \, \times 5 \times 7.4 \\\\ & =18.5 \end{aligned}

Since it is a square pyramid, the base is a square. Each side of the square shares an edge with the base of the triangle, so each side of the square is 5 .

\begin{aligned} \text { Area of square } & =5 \times 5 \\\\ & =25 \end{aligned}

There is one square base and 4 congruent lateral triangular faces.

Total surface area: 25+18.5 \times 4=25+74=99

The measurements on this prism are in inches, so the total surface area of the prism is 99 \text {~inches}^2.

Teaching tips for the surface area of a prism

• Make sure that students have had time to work with physical 3D models and nets before doing activities that involve finding the surface area of pyramids and prisms.
• Choose worksheets that offer a variety of question types – a mixture of showing the full pyramid or prism versus showing the net, a mixture of solving for the missing surface area versus a missing dimension, and one that includes some word problems.

Easy mistakes to make

• Confusing lateral area with total surface area Lateral area is the area of each of the sides, and total surface area is the area of the bases plus the area of the sides. When asked to find the lateral area, be sure to only add up the area of the sides – which are always rectangles in right prisms (the types of prisms shown on this page). Note: In oblique prisms the lateral faces are parallelograms.

Practice surface area of a prism questions

1) The pyramid is composed of four congruent equilateral triangles. Find the surface area of the pyramid.

\begin{aligned} \text {Surface area of pyramid }&= \text { Area of base and faces} \\ & \quad \text{ (4 congruent triangles) } \\\\ & =4 \times\left(\cfrac{1}{2} \, \times 3 \times 2.6\right) \\\\ & =4 \times 3.9 \\\\ & =15.6 \mathrm{~ft}^2 \end{aligned}

2) Calculate the surface area of the triangular prism:

You can unfold the triangular prism, and use the net to find the area of each face.

Remember that the edges in a prism always fold up together to form the prism – making their lengths equal.

\cfrac{1}{2} \times 5 \times 4.3=10.75

7 \times 5=35

Total surface area: 10.75+10.75+35+35+35=126.5 \mathrm{~ft}^2

3) Calculate the surface area of the rectangular prism:

You can unfold the rectangular prism, and use the net to find the area of each face.

The area of each rectangular base:

\begin{aligned} & 1 \cfrac{2}{5} \, \times 8 \\\\ &= \cfrac{7}{5} \, \times \cfrac{8}{1} \\\\ &= \cfrac{56}{5} \\\\ &= 11 \, \cfrac{1}{5} \end{aligned}

\begin{aligned} & 4 \, \cfrac{2}{3} \, \times 8 \\\\ &= \cfrac{14}{3} \, \times \cfrac{8}{1} \\\\ &= \cfrac{112}{3} \\\\ &= 37 \, \cfrac{1}{3} \end{aligned}

\begin{aligned} & 4 \, \cfrac{2}{3} \, \times 1 \cfrac{2}{5} \\\\ &= \cfrac{14}{3} \, \times \cfrac{7}{5} \\\\ &= \cfrac{98}{15} \\\\ &= 6 \cfrac{8}{15} \end{aligned}

Total surface area:

\begin{aligned} & 6 \, \cfrac{8}{15} \, +6 \, \cfrac{8}{15} \, +11 \, \cfrac{1}{5} \, +11 \, \cfrac{1}{5} \, +37 \, \cfrac{1}{3} \, +37 \, \cfrac{1}{3} \\\\ & =6 \, \cfrac{8}{15} \, +6 \, \cfrac{8}{15} \, +11 \, \cfrac{5}{15} , +11 \, \cfrac{5}{15} \, +37 \, \cfrac{3}{15} \, +37 \, \cfrac{3}{15} \\\\ & =108 \, \cfrac{32}{15} \\\\ & =110 \, \cfrac{2}{15} \mathrm{~m}^2 \end{aligned}

4) Here is a net of a square pyramid. Calculate the surface area.

\begin{aligned} \text {Area of triangle } & =\cfrac{1}{2} \, \times 6.5 \times 3.8 \\\\ & =12.35 \end{aligned}

Since it is a square pyramid, the base is a square.

\begin{aligned} \text {Area of square } & =6.5 \times 6.5 \\\\ & =42.25 \end{aligned}

Total surface area = 12.35+12.35+12.35+12.35+42.25=91.65 \mathrm{~m}^2

5) Calculate the surface area of the prism.

The congruent bases (front and back faces) are composed of a rectangle and a right triangle.

Total surface area = 87.5+87.5+330+220+154+189.2=1,068 .2 \text { units}^2 

6) Malika was painting the hexagonal prism below. It took 140.8 \text { inches}^2 to cover the entire shape. If the area of the base is \text {10.4 inches}^2 and each side of the hexagon is 2 \text { inches} , what is the height of the prism?

You can unfold the hexagonal prism, and use the net to find the area of each face:

Total area of the bases: 10.4+10.4=20.8

Subtract the area of the bases from the total amount of paint Malika used, to see how much was used on the lateral faces:

140.8-20.8=120

The total area of the faces left is 120 \text { inches}^2. 

Since the 6  faces are congruent, the total for each face can be found by dividing by 6\text{:} 

120 \div 6=20 

Labeling the missing length as x , means the area of each face can be written as 2 \times x or 2 x .

Since each face has an area of 20 \text{ inches}^2 , the missing height can be found with the equation: 2 x=20.

Since 2 \times 10=20 , the missing height is 10 inches.

Surface area FAQs

A cuboid is a prism with a rectangular base and rectangular lateral sides. It is also known as a rectangular prism.

Some shapes do have a general formula that you can use. For example, the surface area of a rectangular prism uses the formula 2 \: (l b+b h+l h) . There are other formulas, but for all prisms, the general formula is \text {area of } 2 \text { bases }+ \text {area of all lateral faces} .

Since all the faces have the same area, find the area of the square base and multiply it by 6 . Step-by-step guide : Surface area of a cube

The surface area of a cylinder is the area of a circle (the two congruent bases) plus the the curved surface area (2 \pi r h). . This will give you the surface area of the cylinder. Step-by-step guide: Surface area of a cylinder

To find the curved surface area, square the radius of the sphere and multiply it by 4 \pi . This will give you the surface area of the sphere. Step-by-step guide : Surface area of a sphere

The next lessons are

• Pythagorean theorem
• Trigonometry
• Circle math
• Surface area of a cone
• Surface area of a hemisphere

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GKT101: General Knowledge for Teachers – Math

Surface Area

The boundary of three-dimensional objects consists of several two-dimensional shapes. The total area of these shapes is called the surface area of the object. It gives some idea of how large an object is, but now how much space it takes up. This is measured by volume, which we will discuss next. Watch this lecture series and complete the interactive exercises to practice calculating the surface area of rectangular prisms.

Surface area word problems - Questions

1. A group of students made trees out of paper for a scene in a school play. The trees are shaped like hollow square pyramids.

They created this net to represent the paper they needed per tree.

How much paper will it take to make each tree, including the bottom?

_________ cm 2

2. Dmitri's mom is making him a tent to use for backyard camp outs with his friends.

She created this net to represent the material she needed.

How much material will Dmitri's mom need for the tent, including the floor?

_________ meters 2

What is the surface area of the door?

What is the surface area of the roof, including the bottom?

Give an exact answer (do not round).

_________ m 2

• Math Article
• Surface Areas Volumes

Surface Areas and Volume

Surface area and volume are calculated for any three-dimensional geometrical shape. The surface area of any given object is the area or region occupied by the surface of the object. Whereas volume is the amount of space available in an object.

In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc. Each shape has its surface area as well as volume. But in the case of two-dimensional figures like square, circle, rectangle, triangle, etc., we can measure only the area covered by these figures and there is no volume available. Now, let us see the formulas of surface areas and volumes for different 3d-shapes.

What is Surface Area?

The space occupied by a two-dimensional flat surface is called the area. It is measured in square units. The area occupied by a three-dimensional object by its outer surface is called the surface area. It is also measured in square units.

Generally, Area can be of two types:

(i) Total Surface Area

(ii) Curved Surface Area/Lateral Surface Area

Total Surface Area

Total surface area refers to the area including the base(s) and the curved part. It is the total area covered by the surface of the object. If the shape has a curved surface and base, then the total area will be the sum of the two areas.

Curved Surface Area/Lateral Surface Area

Curved surface area refers to the area of only the curved part of the shape excluding its base(s). It is also referred to as lateral surface area for shapes such as a cylinder.

What is Volume?

The amount of space, measured in cubic units, that an object or substance occupies is called volume. Two-dimensional doesn’t have volume but has area only. For example, the  Volume of the Circle cannot be found, though the Volume of the sphere can be. It is so because a sphere is a three-dimensional shape.

Learn more: Mathematics Grade 10

Surface Area and Volume Formulas

Below given is the table for calculating Surface area and Volume for the basic geometrical figures:

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• Surface Area and Volume Class 9
• Surface Areas and Volumes Class 10 Notes

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Solved Examples

What is the surface area of a cuboid with length, width and height equal to 4.4 cm, 2.3 cm and 5 cm, respectively?

Given, the dimensions of cuboid are:

length, l = 4.4 cm

width, w = 2.3 cm

height, h = 5 cm

Surface area of cuboid = 2(wl+hl+hw)

= 2·(2.3 x 4.4 + 5 x 4.4 + 5 x 2.3)

= 87.24 square cm.

What is the volume of a cylinder whose base radii are 2.1 cm and height is 30 cm?

Radius of bases, r = 2.1 cm

Height of cylinder = 30 cm

Volume of cylinder = πr 2 h = π·(2.1) 2 ·30 ≈ 416.

Practice Questions on Surface Areas and Volumes

• Find the volume of a cube whose side length is 5 cm.
• Find the CSA of the hemisphere, if the radius is 7 cm.
• If the radius of the sphere is 4 cm, find its surface area.

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Frequently Asked Questions on Surface Area and Volume

What are the formulas for surface area and volume of cuboid, what is the total surface area of the cylinder, how to calculate the volume of a cone-shaped object, what is the total surface area of the hemisphere.

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