problem solving 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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10.7 Volume and Surface Area

Learning objectives.

After completing this section, you should be able to:

• Calculate the surface area of right prisms and cylinders.
• Calculate the volume of right prisms and cylinders.
• Solve application problems involving surface area and volume.

Volume and surface area are two measurements that are part of our daily lives. We use volume every day, even though we do not focus on it. When you purchase groceries, volume is the key to pricing. Judging how much paint to buy or how many square feet of siding to purchase is based on surface area. The list goes on. An example is a three-dimensional rendering of a floor plan. These types of drawings make building layouts far easier to understand for the client. It allows the viewer a realistic idea of the product at completion; you can see the natural space, the volume of the rooms. This section gives you practical information you will use consistently. You may not remember every formula, but you will remember the concepts, and you will know where to go should you want to calculate volume or surface area in the future.

We will concentrate on a few particular types of three-dimensional objects: right prisms and right cylinders. The adjective “right” refers to objects such that the sides form a right angle with the base. We will look at right rectangular prisms, right triangular prisms, right hexagonal prisms, right octagonal prisms, and right cylinders. Although, the principles learned here apply to all right prisms.

Three-Dimensional Objects

In geometry, three-dimensional objects are called geometric solids . Surface area refers to the flat surfaces that surround the solid and is measured in square units. Volume refers to the space inside the solid and is measured in cubic units. Imagine that you have a square flat surface with width and length. Adding the third dimension adds depth or height, depending on your viewpoint, and now you have a box. One way to view this concept is in the Cartesian coordinate three-dimensional space. The x x -axis and the y y -axis are, as you would expect, two dimensions and suitable for plotting two-dimensional graphs and shapes. Adding the z z -axis, which shoots through the origin perpendicular to the x y x y -plane, and we have a third dimension. See Figure 10.124 .

Here is another view taking the two-dimensional square to a third dimension. See Figure 10.125 .

To study objects in three dimensions, we need to consider the formulas for surface area and volume. For example, suppose you have a box ( Figure 10.126 ) with a hinged lid that you want to use for keeping photos, and you want to cover the box with a decorative paper. You would need to find the surface area to calculate how much paper is needed. Suppose you need to know how much water will fill your swimming pool. In that case, you would need to calculate the volume of the pool. These are just a couple of examples of why these concepts should be understood, and familiarity with the formulas will allow you to make use of these ideas as related to right prisms and right cylinders.

Right Prisms

A right prism is a particular type of three-dimensional object. It has a polygon-shaped base and a congruent, regular polygon-shaped top, which are connected by the height of its lateral sides, as shown in Figure 10.127 . The lateral sides form a right angle with the base and the top. There are rectangular prisms, hexagonal prisms, octagonal prisms, triangular prisms, and so on.

Generally, to calculate surface area, we find the area of each side of the object and add the areas together. To calculate volume, we calculate the space inside the solid bounded by its sides.

The formula for the surface area of a right prism is equal to twice the area of the base plus the perimeter of the base times the height, S A = 2 B + p h , S A = 2 B + p h , where B B is equal to the area of the base and top, p p is the perimeter of the base, and h h is the height.

The formula for the volume of a rectangular prism, given in cubic units, is equal to the area of the base times the height, V = B ⋅ h , V = B ⋅ h , where B B is the area of the base and h h is the height.

Example 10.56

Calculating surface area and volume of a rectangular prism.

Find the surface area and volume of the rectangular prism that has a width of 10 cm, a length of 5 cm, and a height of 3 cm ( Figure 10.128 ).

The surface area is S A = 2 ( 10 ) ( 5 ) + 2 ( 5 ) ( 3 ) + 2 ( 10 ) ( 3 ) = 190 cm 2 . S A = 2 ( 10 ) ( 5 ) + 2 ( 5 ) ( 3 ) + 2 ( 10 ) ( 3 ) = 190 cm 2 .

The volume is V = 10 ( 5 ) ( 3 ) = 150 cm 3 . V = 10 ( 5 ) ( 3 ) = 150 cm 3 .

Your Turn 10.56

In Figure 10.129 , we have three views of a right hexagonal prism. The regular hexagon is the base and top, and the lateral faces are the rectangular regions perpendicular to the base. We call it a right prism because the angle formed by the lateral sides to the base is 90 ∘ . 90 ∘ . See Figure 10.127 .

The first image is a view of the figure straight on with no rotation in any direction. The middle figure is the base or the top. The last figure shows you the solid in three dimensions. To calculate the surface area of the right prism shown in Figure 10.129 , we first determine the area of the hexagonal base and multiply that by 2, and then add the perimeter of the base times the height. Recall the area of a regular polygon is given as A = 1 2 a p , A = 1 2 a p , where a a is the apothem and p p is the perimeter. We have that

Then, the surface area of the hexagonal prism is

To find the volume of the right hexagonal prism, we multiply the area of the base by the height using the formula V = B h . V = B h . The base is 93.6 cm 2 , 93.6 cm 2 , and the height is 20 cm cm . Thus,

Example 10.57

Calculating the surface area of a right triangular prism.

Find the surface area of the triangular prism ( Figure 10.130 ).

The area of the triangular base is A b a s e = 1 2 ( 12 ) ( 6 ) = 36 in 2 A b a s e = 1 2 ( 12 ) ( 6 ) = 36 in 2 . The perimeter of the base is p = 12 + 8.49 + 8.49 = 28.98 in . p = 12 + 8.49 + 8.49 = 28.98 in . Then, the surface area of the triangular prism is S A = 2 ( 36 ) + 28.98 ( 10 ) = 361.8 in 2 . S A = 2 ( 36 ) + 28.98 ( 10 ) = 361.8 in 2 .

Your Turn 10.57

Example 10.58, finding the surface area and volume.

Find the surface area and the volume of the right triangular prism with an equilateral triangle as the base and height ( Figure 10.131 ).

The area of the triangular base is A b a s e = 1 2 ( 6 ) ( 10.39 ) = 31.17 cm 2 A b a s e = 1 2 ( 6 ) ( 10.39 ) = 31.17 cm 2 . Then, the surface area is S A = 2 ( 31.17 ) + 36 ( 10 ) = 422.34 cm 2 . S A = 2 ( 31.17 ) + 36 ( 10 ) = 422.34 cm 2 .

The volume formula is found by multiplying the area of the base by the height. We have that V = B ⋅ h = ( 31.17 ) ( 10 ) = 311.7 cm 3 . V = B ⋅ h = ( 31.17 ) ( 10 ) = 311.7 cm 3 .

Your Turn 10.58

Example 10.59, determining surface area application.

Katherine and Romano built a greenhouse made of glass with a metal roof ( Figure 10.132 ). In order to determine the heating and cooling requirements, the surface area must be known. Calculate the total surface area of the greenhouse.

The area of the long side measures 95 ft × 6.5 ft = 1,235 ft 2 . 95 ft × 6.5 ft = 1,235 ft 2 . Multiplying by 2 gives 2,470 ft 2 . 2,470 ft 2 . The front (minus the triangular area) measures 22 ft × 6.5 ft = 286 ft 2 . 22 ft × 6.5 ft = 286 ft 2 . Multiplying by 2 gives 572 ft 2 . 572 ft 2 . The floor measures 95 ft × 22 ft = 2,090 ft 2 . 95 ft × 22 ft = 2,090 ft 2 . Each triangular region measures A = 1 2 ( 22 ) ( 5 ) = 55 ft 2 . A = 1 2 ( 22 ) ( 5 ) = 55 ft 2 . Multiplying by 2 gives 110 ft 2 . 110 ft 2 . Finally, one side of the roof measures 12.1 ft × 95 ft = 1,149.5 ft 2 . 12.1 ft × 95 ft = 1,149.5 ft 2 . Multiplying by 2 gives 2299 ft 2 . 2299 ft 2 . Add them up and we have S A = 2,470 + 572 + 2,090 + 110 + 2,299 = 7,541 ft 2 . S A = 2,470 + 572 + 2,090 + 110 + 2,299 = 7,541 ft 2 .

Your Turn 10.59

Right cylinders.

There are similarities between a prism and a cylinder. While a prism has parallel congruent polygons as the top and the base, a right cylinder is a three-dimensional object with congruent circles as the top and the base. The lateral sides of a right prism make a 90 ∘ 90 ∘ angle with the polygonal base, and the side of a cylinder, which unwraps as a rectangle, makes a 90 ∘ 90 ∘ angle with the circular base.

Right cylinders are very common in everyday life. Think about soup cans, juice cans, soft drink cans, pipes, air hoses, and the list goes on.

In Figure 10.133 , imagine that the cylinder is cut down the 12-inch side and rolled out. We can see that the cylinder side when flat forms a rectangle. The S A S A formula includes the area of the circular base, the circular top, and the area of the rectangular side. The length of the rectangular side is the circumference of the circular base. Thus, we have the formula for total surface area of a right cylinder.

The surface area of a right cylinder is given as S A = 2 π r 2 + 2 π r h . S A = 2 π r 2 + 2 π r h .

To find the volume of the cylinder, we multiply the area of the base with the height.

The volume of a right cylinder is given as V = π r 2 h . V = π r 2 h .

Example 10.60

Finding the surface area and volume of a cylinder.

Given the cylinder in Figure 10.133 , which has a radius of 5 inches and a height of 12 inches, find the surface area and the volume.

Step 1: We begin with the areas of the base and the top. The area of the circular base is

Step 2: The base and the top are congruent parallel planes. Therefore, the area for the base and the top is

Step 3: The area of the rectangular side is equal to the circumference of the base times the height:

Step 4: We add the area of the side to the areas of the base and the top to get the total surface area. We have

Step 5: The volume is equal to the area of the base times the height. Then,

Your Turn 10.60

Applications of surface area and volume.

The following are just a small handful of the types of applications in which surface area and volume are critical factors. Give this a little thought and you will realize many more practical uses for these procedures.

Example 10.61

Applying a calculation of volume.

A can of apple pie filling has a radius of 4 cm and a height of 10 cm. How many cans are needed to fill a pie pan ( Figure 10.134 ) measuring 22 cm in diameter and 3 cm deep?

The volume of the can of apple pie filling is V = π ( 4 ) 2 ( 10 ) = 502.7 cm 3 . V = π ( 4 ) 2 ( 10 ) = 502.7 cm 3 . The volume of the pan is V = π ( 11 ) 2 ( 3 ) = 1140.4 cm 3 . V = π ( 11 ) 2 ( 3 ) = 1140.4 cm 3 . To find the number of cans of apple pie filling, we divide the volume of the pan by the volume of a can of apple pie filling. Thus, 1140 ÷ 502.7 = 2.3. 1140 ÷ 502.7 = 2.3. We will need 2.3 cans of apple pie filling to fill the pan.

Your Turn 10.61

Optimization.

Problems that involve optimization are ones that look for the best solution to a situation under some given conditions. Generally, one looks to calculus to solve these problems. However, many geometric applications can be solved with the tools learned in this section. Suppose you want to make some throw pillows for your sofa, but you have a limited amount of fabric. You want to make the largest pillows you can from the fabric you have, so you would need to figure out the dimensions of the pillows that will fit these criteria. Another situation might be that you want to fence off an area in your backyard for a garden. You have a limited amount of fencing available, but you would like the garden to be as large as possible. How would you determine the shape and size of the garden? Perhaps you are looking for maximum volume or minimum surface area. Minimum cost is also a popular application of optimization. Let’s explore a few examples.

Example 10.62

Maximizing area.

Suppose you have 150 meters of fencing that you plan to use for the enclosure of a corral on a ranch. What shape would give the greatest possible area?

So, how would we start? Let’s look at this on a smaller scale. Say that you have 30 inches of string and you experiment with different shapes. The rectangle in Figure 10.135 measures 12 in long by 3 in wide. We have a perimeter of P = 2 ( 12 ) + 2 ( 3 ) = 30 in P = 2 ( 12 ) + 2 ( 3 ) = 30 in and the area calculates as A = 3 ( 12 ) = 36 in 2 . A = 3 ( 12 ) = 36 in 2 . The rectangle in Figure 10.135 , measures 8 in long and 7 in wide. The perimeter is P = 2 ( 8 ) + 2 ( 7 ) = 30 in P = 2 ( 8 ) + 2 ( 7 ) = 30 in and the area is A = 8 ( 7 ) = 56 in 2 . A = 8 ( 7 ) = 56 in 2 . In Figure 10.135 , the square measures 7.5 in on each side. The perimeter is then P = 4 ( 7.5 ) = 30 in P = 4 ( 7.5 ) = 30 in and the area is A = 7.5 ( 7.5 ) = 56.25 in 2 . A = 7.5 ( 7.5 ) = 56.25 in 2 . If you want a circular corral as in Figure 10.135 , we would consider a circumference of 30 = 2 π r , 30 = 2 π r , which gives a radius of 30 ÷ 2 π = 4.77 in 30 ÷ 2 π = 4.77 in and an area of A = π ( 4.77 ) 2 = 71.5 in 2 . A = π ( 4.77 ) 2 = 71.5 in 2 .

We see that the circular shape gives the maximum area relative to a circumference of 30 in. So, a circular corral with a circumference of 150 meters and a radius of 23.87 meters gives a maximum area of 1,790.5 m 2 . 1,790.5 m 2 .

Your Turn 10.62

Example 10.63, designing for cost.

Suppose you want to design a crate built out of wood in the shape of a rectangular prism ( Figure 10.136 ). It must have a volume of 3 cubic meters. The cost of wood is $15 per square meter. What dimensions give the most economical design while providing the necessary volume?

To choose the optimal shape for this container, you can start by experimenting with different sizes of boxes that will hold 3 cubic meters. It turns out that, similar to the maximum rectangular area example where a square gives the maximum area, a cube gives the maximum volume and the minimum surface area.

As all six sides are the same, we can use a simplified volume formula:

V = s 3 , V = s 3 ,

where s s is the length of a side. Then, to find the length of a side, we take the cube root of the volume.

3 = s 3 3 3 = s = 1.4422 m 3 = s 3 3 3 = s = 1.4422 m

The surface area is equal to the sum of the areas of the six sides. The area of one side is A = ( 1.4422 ) 2 = 2.08 m 2 . A = ( 1.4422 ) 2 = 2.08 m 2 . So, the surface area of the crate is S A = 6 ( 2.08 ) = 12.5 m 2 S A = 6 ( 2.08 ) = 12.5 m 2 . At $15 a square meter, the cost comes to 12.5 ( $ 15 ) = $ 187.50. 12.5 ( $ 15 ) = $ 187.50. Checking the volume, we have V = ( 1.4422 ) 3 = 2.99 m 3 V = ( 1.4422 ) 3 = 2.99 m 3 .

Your Turn 10.63

Check your understanding, section 10.7 exercises.

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CHAPTER 3 Measurement, Perimeter, Area, and Volume

3.3 Solve Geometry Applications: Volume and Surface Area

Learning Objectives

By the end of this section, you will be able to:

• 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 cone

In this section, we will 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.

Problem Solving Strategy for Geometry Applications

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

Find Volume and Surface Area of Rectangular Solids

A cheer leading coach is having the squad paint wooden crates with the school colors to stand on at the games. (See Figure.1 ). 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.

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.

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.4 ). 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

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

• 1,440 cu. ft

• 2,772 cu. in.
• 1,264 sq. in.

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:

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

• 91.125 cu. m
• 121.5 sq. m

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

• 389.017 cu. yd.
• 319.74 sq. yd.

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 centre of the sphere to any point on its surface. The formulas for the volume and surface area of a sphere are given below.

Volume and Surface Area of a Sphere

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

• 113.04 cu. cm
• 113.04 sq. cm

• 4.19 cu. ft
• 12.56 sq. ft

• 3052.08 cu. in.
• 1017.36 sq. in.

• 14.13 cu. ft
• 28.26 sq. ft

Find the Volume and Surface Area of a Cylinder

A cylinder has two circular bases of equal size. The height is the distance between the bases.

Seeing how a cylinder is similar to a rectangular solid may make it easier to understand the formula for the volume of a cylinder.

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.6) .

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

Volume and Surface Area of a Cylinder

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

• 351.68 cu. cm
• 276.32 sq. cm

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

• 100.48 cu. ft
• 125.6 sq. ft

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

• 3,818.24 cu. cm
• 1,356.48 sq. cm

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

• 91.5624 cu. ft
• 113.6052 sq. ft

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.6) .

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

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

• Volume of a Cone

65.94 cu. in.

235.5 cu. cm

TRY IT 10.1

678.24 cu. in.

TRY IT 10.2

128.2 cu. in.

Key Concepts

Practice Makes Perfect

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

In the following exercises, solve.

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

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

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

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

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

Everyday Math

Writing Exercises

Attributions

• This chapter has been adapted from “Solve Geometry Applications: Volume and Surface Area” in Prealgebra (OpenStax) by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, which is under a CC BY 4.0 Licence . Adapted by Izabela Mazur. See the Copyright page for more information.

Introductory Algebra Copyright © 2021 by Izabela Mazur is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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• Unit Conversions
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• Common Multiple and Divisor
• Fractions and Decimals
• Algebraic Expressions and Polynomials
• Equations and Inequalities
• Sequences and Series
• Combinatorics
• Probability and Statistics
• Volume and Surface Area of Solids
• Metric Relations in Space
• Analytical Geometry
• Limits, Derivatives, Integrals
• Analysis of Functions
• Trig table and chart

.   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 ?                 Link Partners Privacy Policy Copyright © 2015-2023 math-exercises.com - All rights reserved. Any use of website content without written permission is prohibited. SURFACE AREA AND VOLUME WORD PROBLEMS Problem 1 : Savannah has a water bottle that is a rectangular prism. The bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water. Then, she drank 1/4 of the volume of water in her water bottle. How many cubic centimeters of water were left in the water bottle? Capacity of bottle = Volume of rectangular prism = length x width x height = 18 x 5 x 7 After she drank 1/4 of the capacity, she will left over with 3/4 of the bottle Quantity of water remaining = 3/4 of 630 = 472.5 cm 3 Problem 2 : A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches. a) Find the height of the prism. b) Find the surface area of the prism. a)  Volume of prism = Base area x height 768 = 8 2 x height height = 768/64 b)  Surface area of prism = perimeter x height Perimeter of square base = 4(8) Surface area of prism = 32 x 12 = 3072 cm 2 Problem 3 : A triangular prism has the measurements shown. a) Find the volume of the prism. b) Find the surface area of the prism Volume of the triangular prism = base area x height Base area = 1/2 x base x height = 1/2 x 19.6 x 5 Volume = 49 x 16 b)  Surface area of the triangular prism = Perimeter  x height = (10 + 12 + 19.6) x 16 = 665.6 ft 2 Problem 4 : The volume of Box A is 2/5 the volume of Box B. What is the height of Box A if it has a base area of 32 square centimeters? Volume of box B = length x width x height = 16 x 8 x 10 Volume of box A = 2/5 of volume of box B Volume of box A = ((2/5) x base area x height 1280 = (2/5) x 32 x height height = 100 cm Problem 5 : The ratio of the length to the width to the height of an open rectangular tank is 10 : 5 : 8. The height of the tank is 18 feet longer than the width. a) Find the volume of the tank. b) Find the surface area of the open tank. a)  Let 10x, 5x and 8x be length, width and height of the rectangular tank respectively. height = width + 18 8x = 5x + 18 8x - 5x = 18 length (10x) = 60 ft, width (5x) = 30 ft and height (8x) = 48 ft a)  Volume = 60 x 30 x 48 = 86400 ft 3 b) Surface area of the tank = perimeter of base x height = 2(60 x 30 + 30 x 48 + 48 x 60) = 12240 ft 2 Problem 6 : Janice is m aking a gift box. The gift box is a prism with bases that are regular hexagons, and has the dimensions shown in the diagram. a) Find the height PQ of the prism. Area of hexagon = (1/2) x perimeter x Apothem = (1/2) x 6(7) x 6 a)  Volume of figure = base area x height 126 x height = 2835 height = 2835/126 height = 22.5 b)  Surface area of prism = perimeter of the base x height = 6(7) x 22.5 Problem 7 : Container A was filled with water to the brim. Then, some of the water was poured into an empty Container B until the height of the water in both containers became the same. Find the new height of the water in both containers. Both containers are having equal quantity of water. Quantity of water inside the container Let h be the new height of the container. 25 x 30 x 40 = 18 x 25 x h h = (25 x 30 x 40)/(18 x 25) h = 66.6 cm Problem 8 : A fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high. It contains water up to a height of 28 centimeters. How many more cubic centimeters of water are needed to fill the tank to a height of 35 centimeters? Length = 50 cm, width = 30 and height = 28 Capacity of water when its height is 28 cm : = 50 x 30 x 28 = 42000 ----(1) Length = 50 cm, width = 30 and height = 35 Capacity of water when its height is 35 cm : = 50 x 30 x 35 = 52500 ----(2) = 52500 - 42000 = 10500 cm 3 Problem 9 : Find the surface area of a square pyramid given that its base area is 196 square inches and the height of each of its triangular faces is 16 inches Surface area of square base pyramid = perimeter of the base x height Perimeter (4a) = 4(14) = 56 inches Height of the triangular face will be the height of square base prism. Using Pythagorean theorem : height of prism =  √16 2 - 14 2 =  √256-196 =  √60 Surface area = 56√60 Problem 10 : The volume of a rectangular prism is 441 cubic feet. It has a square base with edges that are 7 feet long. a)  Volume of rectangular prism = base area x height 441 = 7 x 7 x h b) surface area = perimeter of base x height Variables and Expressions Variables and Expressions Worksheet Place Value of Number Formation of Large Numbers Laws of Exponents Angle Bisector Theorem Pre Algebra SAT Practice Topic Wise Geometry Topics SMO Past papers Parallel Lines and Angles Properties of Quadrilaterals Circles and Theorems Transformations of 2D Shapes Quadratic Equations and Functions Composition of Functions Polynomials Fractions Decimals and Percentage Customary Unit and Metric Unit Honors Geometry 8th grade worksheets Linear Equations Precalculus Worksheets 7th Grade Math Worksheets Vocabulary of Triangles and Special right triangles 6th Grade Math Topics STAAR Math Practice Math 1 EOC Review Worksheets 11 Plus Math Papers CA Foundation Papers Algebra 1 Worksheets Recent Articles Finding range of values inequality problems. May 21, 24 08:51 PM Solving Two Step Inequality Word Problems May 21, 24 08:51 AM Exponential Function Context and Data Modeling May 20, 24 10:45 PM © All rights reserved. intellectualmath.com High Impact Tutoring Built By Math Experts Personalized standards-aligned one-on-one math tutoring for schools and districts In order to access this I need to be confident with: 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. \begin{aligned} A&=6 \times 6.4 \\ &=38.4 \mathrm{~mm}^2 \end{aligned} \begin{aligned} A&=6 \times 10 \\ &=60 \mathrm{~mm}^2 \end{aligned} \begin{aligned} A&=6 \times 6.4 \\ &=38.4 \mathrm{~mm}^2 \end{aligned} \begin{aligned} A&=\frac{1}{2} \times 10 \times 4.2 \\ &=21 \mathrm{~mm}^2 \end{aligned} 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 Still stuck? At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts. Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. Find out how we can help your students achieve success with our math tutoring programs . [FREE] Common Core Practice Tests (3rd to 8th Grade) Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents. Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers! Privacy Overview 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: 4b b —- —- 2(w+h) w.h —- —- 2(a+b) b.h —- —- a+b+c+d 1/2(a+b).h —- —- 2 π r π r —- —- 2π√(a + b )/2       π a.b —- —- a+b+c 1/2 * b * h —- —- 4(l+b+h) 2(lb+bh+hl) 2h(l+b) l * b * h 6a  6a 4a a —- 2 π r(r+h) 2πrh π r h —- π r(r+l) π r l 1/3π r h —- 4 π r 4π r 4/3π r —- 3 π r 2 π r 2/3π r   Related Articles Surface Area and Volume Class 9 Surface Areas and Volumes Class 10 Notes Also have a look on: 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. Stay tuned to BYJU’S – The Learning App and download the app to explore all Maths-related concepts. 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. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Visit BYJU’S for all Maths related queries and study materials Your result is as below Request OTP on Voice Call MATHS Related Links Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked * Post My Comment You can get all surface area formulas here! https://byjus.com/surface-area-formulas/ its good i uhder stood after seeing this Thanks this helped me a lot. So😇😇😇😇 very thanks, it’s very usefull to mee It’s very useful &helpful for me Thanks byjus💞🙏 Thanks a lots 🙂🙂🙂 it’s very useful for me and everyone. Thank you BYJUS🙏🙏 Thank you so much Byju’s 🤗🤗👍👍it’s so helpful. Thanks a lot. Thanks a lot for sharing this info Thanks a lots 🙂🙂🙂 it’s very useful for me and everyone. Thank you BYJUS🙏🙏 its good you can find surface area formulas here Thiks alot It’s very useful for me and everyone Thinks BYJUS Register with BYJU'S & Download Free PDFs Register with byju's & watch live videos. Surface Area Questions Surface Area of a Cuboid Click here for Questions and Answers Surface Area of a Prism Surface Area of a Cylinder Surface Area of a Sphere Surface Area of a Cone GCSE Revision Cards 5-a-day Workbooks Primary Study Cards Privacy Policy Terms and Conditions Corbettmaths © 2012 – 2024 Surface Area and Volume in the Real World New york state common core math grade 6, module 5, lesson 19. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. 289 IMAGES Surface Area Volume Word Problem Volume and Surface Area Word Problems Problem Solving: Finding Surface Area and Volume Surface Area and Volume Word Problems Volume and surface area word problems Free Printable Surface Area And Volume Worksheets COMMENTS 9.9: Solve Geometry Applications- Volume and Surface Area (Part 1) For a rectangular solid with length 14 cm, height 17 cm, and width 9 cm, find the (a) volume and (b) surface area. Solution. Step 1 is the same for both (a) and (b), so we will show it just once. Step 1. Read the problem. Draw the figure and label it with the given information. PDF 12.4 Real-World Problems: Surface Area and Volume Lesson 12.4 Real-World Problems: Surface Area and Volume 201 L e a r n Solve word problems about surface area and volume of non-rectangular prisms. A block of wood is a prism and has the dimensions shown in the diagram below. a) Find the volume of the block of wood. 3 cm 3 cm 4 cm 4 cm 7 cm 5 cm The base of the prism is a trapezoid. 10.8: Volume and Surface Area Solve application problems involving surface area and volume. ... To study objects in three dimensions, we need to consider the formulas for surface area and volume. For example, suppose you have a box (Figure 10.163) with a hinged lid that you want to use for keeping photos, and you want to cover the box with a decorative paper. ... 9.10: Solve Geometry Applications- Volume and Surface Area (Part 2) 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 and height h, is. S = 2πr2 + 2πrh (9.6.16) (9.6.16) S = 2 π r 2 + 2 π r h. Definition: Volume and Surface Area of a Cylinder. Solving Surface Area and Volume Problems 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. Solution : To know that total cost of silver coating, first we have to know the Surface area of the metal box. Find ... 9.6 Solve Geometry Applications: Volume and Surface Area 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. Volume Problem Solving 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 V cone = 31πr2h and V_ {\text {sphere}} =\frac43 \pi r^3 V sphere = 34πr3. Since the volume of a hemisphere is half the volume of a a sphere of the ... 10.7 Volume and Surface Area Solve application problems involving surface area and volume. Volume and surface area are two measurements that are part of our daily lives. We use volume every day, even though we do not focus on it. When you purchase groceries, volume is the key to pricing. Judging how much paint to buy or how many square feet of siding to purchase is based ... 3.3 Solve Geometry Applications: Volume and Surface Area See Volume and Surface Area of a Cube, below. Substituting, for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get: inches on each side. Find its a) volume and b) surface area. For a cube with side 4.5 metres, find the a) volume and b) surface area of the cube. Khan Academy Khanmigo is now free for all US educators! Plan lessons, develop exit tickets, and so much more with our AI teaching assistant. Math Exercises & Math Problems: Volume and Surface Area of Solids Determine its volume and surface area. Find the volume and surface area of a triangular prism with a right-angled triangle base, if length of the prism base legs are 7.2 cm and 4.7 cm and height of a prism is 24 cm. Find the volume and surface area of a pillar in a shape of a prism with a rhombus base, which diagonals are d 1 = 102 cm, d 2 = 64 ... 9.9: Surface Area and Volume Applications This page titled 9.9: Surface Area and Volume Applications is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform. Use geometric solids to model real world objects and solve problems. Find density by the ratio of mass ... Surface Area and Volume Questions (with Answers) These surface area and volume problems are prepared by our subject experts, as per the NCERT curriculum and latest CBSE syllabus (2022-2023). Learn more: Surface Areas and Volume. Surface Area and Volume Formulas: Total surface area of a cuboid = 2[lb + bh + lh] Total surface area of a cube = 6(side) 2 Surface Area and Volume Word Problems Problem 2 : A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches. a) Find the height of the prism. b) Find the surface area of the prism. Solution : a) Volume of prism = Base area x height. 768 = 8 2 x height. height = 768/64. Surface Area Example 3: surface area of a square-based pyramid in cm. All the lateral faces of the pyramid are congruent. Calculate the surface area. Calculate the area of each face. Show step. The base is a square with the area 6\times {6}=36\text {~cm}^2. 6 × 6 = 36 cm2. Surface Areas and Volume 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. Surface Area Questions Click here for Questions and Answers. Surface Area of a Cylinder. Click here for Questions and Answers. Surface Area of a Sphere. Click here for Questions and Answers. Surface Area of a Cone. Click here for Questions and Answers. Practice Questions. Previous: Surface Area Videos. 3.5: Solve Geometry Applications- Volume and Surface Area Problem Solving Strategy for Geometry Applications Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information. ... 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 ... Surface Area and Volume in the Real World Lesson 19 Student Outcomes. • Students determine the surface area of three-dimensional figures in real-world contexts. • Students choose appropriate formulas to solve real-life volume and surface area problems. Lesson 19 Problem Set. Solve each problem below. 1. Dante built a wooden, cubic toy box for his son. Each side of the box measures ... Khan Academy Khan Academy 6.5: Area, Surface Area and Volume Formulas Surface Area Meaning. SA = 2B + Ph S A = 2 B + P h. Find the area of each face. Add up all areas. SA = B + 1 2sP S A = B + 1 2 s P. Find the area of each face. Add up all areas. SA = 2B + 2πrh S A = 2 B + 2 π r h. Find the area of the base, times 2, then add the areas to the areas of the rectangle, which is the circumference times the height. Khan Academy Math: Pre-K - 8th grade; Pre-K through grade 2 (Khan Kids) Early math review; 2nd grade; 3rd grade; 4th grade; 5th grade; 6th grade; 7th grade; 8th grade; See Pre-K - 8th Math; ... If this problem persists, tell us. Our mission is to provide a free, world-class education to anyone, anywhere. 7.11: Area, Surface Area and Volume Math 27: Number Systems for Educators 7: Geometry 7.11: Area, Surface Area and Volume ... Practice Problems; Definition: Area. The extent or measurement of a surface or piece of land. (2 dimensional) ... Explain the difference between area, surface area, and volume. Estimate the area of the following shapes: essay homework paper phd project resume review speech study thesis