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6.3: Area, Surface Area and Volume - Mathematics


Definition: Area

The extent or measurement of a surface or piece of land. (2 dimensional)

Definition: Surface Area

The area of such an outer part or uppermost layer. (3 dimensional)

Definition: Volume

The amount of space that a substance or object occupies, or that is enclosed within a container, especially when great. (3 dimensional)

Partner Activity 1

  1. Why is area “squared”? i.e. (15 ext { cm}^{2})
  2. Why is volume “cubed”? i.e. (40 ext { liters}^{3})

Partner Activity 2

Think inside the box and approximate the Shaded Area: (area of a square is base times height)

Partner Activity 3

Think around the box (surface area) and approximate the Shaded Area: (How many sides are not seen in the picture, which must be included in the final answer?)

Partner Activity 4

Think inside the box (volume) and approximate the Shaded Area: Volume is base time’s height times width.


The area of the whole of anything, whether it's an object or a surface, is the sum of the area of its constituent parts. We now know that the surface area of a three-dimensional object is the total area of all of its surfaces. In this section, we will learn about the various formulas used to calculate the surface area of different objects.

Surface Area of Cube

The surface area of the cube is the total area covered by all six faces of the cube. The general formula of the surface area of a cube is given as:

The total surface area of the cube will be the sum of the area of the base and the area of vertical surfaces of the cube. There are a total of 6 surfaces hence, the total surface area = 6s 2
The lateral surface area of a cube is the sum of areas of all side faces of the cube. There are 4 side faces so the sum of areas of all 4 side faces of a cube is its lateral area. LSA = 4a 2 where "a" is the side length.

Surface Area of Cuboid

The surface area of cuboid can be explained in terms of two different categories of area, i.e., lateral surface area and the total surface area. The total surface area of the cuboid is obtained by adding the area of all the 6 faces whereas the lateral surface area of the cuboid is found by finding the area of each face excluding the base and the top. The total surface area and lateral surface area can be expressed in terms of length (l), breadth(b), and height of cuboid(h) as:

  • Total Surface Area of Cuboid, S = 2 (lb + bh + lh) units 2
  • Lateral Surface Area of Cuboid, L = 2h (l + b) units 2

Surface Area of Cone

The surface area of a cone is the amount of area occupied by the surface of a cone. A cone is a 3-D shape that has a circular base. This means the base is made up of a radius and diameter. As a cone has a curved surface, thus we can express its curved surface area as well as total surface area. If the radius of the base of the cone is "r" and the slant height of the cone is "l", the surface area of a cone is given as:

Surface Area of Cylinder

A cylinder is a 3-D solid object which consists of two circular bases connected with a curved face. As a cylinder has a curved surface, thus we can express its curved surface area as well as total surface area. If the radius of the base of the cylinder is "r" and the height of the cylinder is "h", the surface area of a cylinder is given as:

Surface Area of Sphere

A sphere is a three-dimensional solid object which has a round structure, like a circle. The area covered by the outer surface of the sphere is known as the surface area of a sphere. The surface area of a sphere is the total area of the faces surrounding it. The surface area of a sphere is given in square units.

The surface area of a sphere is equal to the lateral surface area of a cylinder. Hence, the relation between the surface area of a sphere and lateral surface area of a cylinder is given as:
Surface Area of Sphere = Lateral Surface Area of Cylinder
⇒ The surface area of Sphere = 2πrh
If a diameter of sphere = 2r
Then the surface area of sphere is 2πrh = 2πr(2r) = 4πr 2 square units.

Surface Area of Hemisphere

Hemisphere is a three-dimensional shape, obtained when a sphere is cut along a plane passing through the center of the sphere. In other words, a hemisphere is half of a sphere. The surface area of a hemisphere is the total area its surface covers. It can be classified into two categories:

  • The curved surface area of a hemisphere(CSA) = ½ (curved surface area of a sphere) = ½ (4 π r 2 ) = 2 π r 2 , where "r" is the radius of the hemisphere.
  • The total surface area of a hemisphere(TSA) = curved surface area + Base Area = 2 π r 2 + π r 2 = 3 π r 2 , where r is the radius of the hemisphere.

Surface Area of Prism

There are two types of areas we read about, first, the lateral surface area of the prism, and second, the total surface area of the prism. Let us learn in detail.

The lateral area of a prism is the sum of the areas of all its lateral faces whereas the total surface area of a prism is the sum of its lateral area and area of its bases.

The lateral surface area of prism = base perimeter × height
The total surface area of a prism = Lateral surface area of prism + area of the two bases = (2 × Base Area) + Lateral surface area or (2 × Base Area) + (Base perimeter × height).

There are seven types of prisms based on the shape of the bases of prisms. The bases of different types of prisms are different so are the formulas to determine the surface area of the prism. See the table below to understand this concept behind the surface area of various prism:

Shape Base Surface Area of Prism = (2 × Base Area) + (Base perimeter × height)
Triangular Prism Triangular Surface area of triangular prism = bh + (s1 + s2 + b)H
Square Prism Square Surface area of square prism = 2a 2 + 4ah
Rectangular Prism Rectangular Surface area of rectangular prism = 2(lb + bh + lh)
Trapezoidal Prism Trapezoidal Surface area of trapezoidal prism = h (b + d) + l (a + b + c + d)
Pentagonal Prism Pentagonal Surface area of pentagonal prism = 5ab + 5bh
Hexagonal Prism Hexagonal Surface area of hexagonal prism = 6b(a + h)
Surface area of regular hexagonal prism = 6ah + 3√3a 2
Octagonal Prism Octagonal Surface area of octagonal prism = 4a 2 (1 + √2) + 8aH


Big Ideas Math Book 6th Grade Answer Key Chapter 7 Area, Surface Area, and Volume

To excel in the exam, candidates of 6th standard have to refer to the important preparatory material of Big Ideas Math Book Answer Key Grade 6 Chapter 7 Area, Surface Area, and Volume. If you facing difficulty in solving or calculating the problems, you can easily understand them with the help of Big Ideas Math Book Answer Key Chapter 7. Check the below sections to know the performance, various chapters involved in this concept. You can find the various chapters like area of parallelograms, area of triangles, areas of trapezoids, three-dimensional figures, and so on.

Performance

Lesson: 1 Areas of Parallelograms

Lesson: 2 Areas of Triangles

Lesson: 3 Areas of Trapezoids and Kites

Lesson: 4 Three-Dimensional Figures

Lesson: 5 Surface Areas of Prisms

Lesson: 6 Surface Areas of Pyramids

Lesson: 7 Volumes of Rectangular Prisms

Chapter 7 – Area, Surface Area, and Volume

Area, Surface Area, and Volume STEAM Video/Performance

Packaging Design

Surface area can be used to determine amounts of materials needed to create objects. Describe another situation in which you need to find the surface area of an object.

Watch the STEAM Video “Packaging Design.” Then answer the following questions. Alex is cutting a design out of paper and folding it to form a box.

Question 1.
Tory says that the length of the design is 47 inches and the width is 16 inches. Show several ways that you can arrange three of the designs on a roll of paper that is 48 inches wide. You can make the length of the paper as long as is needed.

Answer:
We can make the length of the paper = 15.6666 inches

Explanation:
Given that the length of the design = 47 inches
that the width of the design = 16 inches
area of the design = 752 inches
They can arrange the three of the design on a roll of paper that = 48 inches
(752/48)
15.6666 inches
Question 2.
Tory says that the cut-out design has an area of 619 square inches. What is the least possible amount of paper that is wasted when you cut out three of the designs?

Answer:
The least amount of paper that is wasted when you cut out three of the designs = 12.89

Explanation:
The area of the rectangle = breadth x height
619 square inches = 48 in x length
619 = 48l
l = 619/48
l = 12.89

Performance Task

Maximizing the Volumes of Boxes

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the dimensions of a small box and two larger boxes.

You will be asked to determine how many small boxes can be placed in each larger box. When a company is deciding on packaging for a product, why should the surface area of the packaging also be considered?

Area, Surface Area, and Volume Gettig Ready for chapter 7

Chapter Exploration

The formulas for the areas of polygons can be derived from one area formula, the area of a rectangle.

Work with a partner. Find (a) the dimensions of the figures and (b) the areas of the figures. What do you notice?

Question 1.

Answer:
The area of the rectangle = The product of the length and the product of the width
The area of the parallelogram = the product of the width and the product of the height

Explanation:
The area of the rectangle is equal to the area of the parallelogram
The area of the rectangle = the product of the length and the product of the width
The area of the rectangle = length x width
area = l x w
The area of the parallelogram = the product of the base and the product of the width
area = base x height
area = b x h

Question 2.

Answer:
The area of the rectangle = the product of the length and the product of the width
The area of the parallelogram = (1/2) x the product of the base and the product of the height

Explanation:
The area of the rectangle is equal to the area of the parallelogram
The area of the rectangle = the product of the length and the product of the width
The area of the rectangle = length x width
area = l x w
The area of the triangle =(1/2) x the product of the base and the product of the width
area =(1/2) base x height
area =(1/2) b x h

Question 3.

Answer:
The area of the rectangle = the product of the length and the product of the width
The area of the trapezoid = (1/2) x height x (base 1 + base 2)

Explanation:
The area of the rectangle is equal to the area of the parallelogram
The area of the rectangle = the product of the length and the product of the width
The area of the rectangle = length x width
area = l x w
The area of the trapezoid =(1/2) x product of the height and the product of the bases
area =(1/2) x height x bases
area =(1/2) x height (b1 + b2)

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.

Lesson 7.1 Areas of Parallelograms

A polygon is a closed figure in a plane that is made up of three or more line segments that intersect only at their endpoints. Several examples of polygons are parallelograms, rhombuses, triangles, trapezoids, and kites.
The formula for the area of a parallelogram can be derived from the definition of the area of a rectangle. Recall that the area of a rectangle is the product of its length ℓ and its width w. The process you use to derive this and other area formulas in this chapter is called deductive reasoning.

EXPLORATION 1
Deriving the Area Formula of a Parallelogram
Work with a partner.
a. Draw any rectangle on a piece of centimeter grid paper. Cut the rectangle into two pieces that can be arranged to form a parallelogram. What do you notice about the areas of the rectangle and the parallelogram?
b. Copy the parallelogram below on a piece of centimeter grid paper. Cut the parallelogram and rearrange the pieces to find its area.


c. Draw any parallelogram on a piece of centimeter grid paper and find its area. Does the area change when you use a different side as the base? Explain your reasoning.
d. Use your results to write a formula for the area A of a parallelogram.

Answer:
c : The area of the parallelogram = the product of the width and the product of the height
area = width x height
area = w x h
Yes the area changes.
We use the bases as equal length and width as equal length
d : the area of the parallelogram = the product of the width and the product of the height

c : The area of the parallelogram = the product of the width and the product of the height
area = width x height
area = w x h
Yes the area changes.
We use the bases as equal length and width as equal length
d : the area of the parallelogram = the product of the width and the product of the height

The area of a polygon is the amount of surface it covers. You can find the area of a parallelogram in much the same way as you can find the area of a rectangle.

Find the area of the parallelogram.

Question 1.

Explanation:
The area of the parallelogram = the product of the base b and its height h.
area = b x h
b = 20 m, h = 25 m
So the Area of the parallelogram = 500 m

Question 2.

Explanation:
The area of the parallelogram = the product of base b and its height h
area = b x h
b = 7 in , h = 18 in
So area = 126 in

Question 3.

Explanation:
The area of the parallelogram = the product of base b and its height h
area = b x h
b = 30 yd, h = 20.5 yd
So area = 615 yd

When finding areas, you may need to convert square units. The diagrams at the left show that there are 9 square feet per square yard.

1 yd 2 = (1 yd)(1 yd) = (3 ft)(3 ft) = 9 ft 2
You can use a similar procedure to convert other square units.

Question 4.
Find the area of the parallelogram in square centimeters.

Answer:
The area of the parallelogram = the product of the base and the product of the height
Area = 400 cms

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 4 m x 10 m
area = 40 m
1 meter = 100 centimeters
40 meter = 400 centimeters
area = 400 cm

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
WRITING
Explain how to use the area of a rectangle to find the area of a parallelogram

Answer:
Area of the rectangle = length x breadth
Area of the parallelogram = base x height

Explanation:
Area of the rectangle = length x breadth
area = l x b
Area of the parallelogram = base x height
area = b x h
area = bh

FINDING AREA
Find the area of the parallelogram.

Question 6.

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = b x h
b = 16 ft and h = 5 ft given
area = 16 x 5
area = 80 ft

Question 7.

Answer:
The area of the parallelogram = 14 . 4 km

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = b x h
b = 3 km and h = 4.8 km given
area = 3 x 4.8
area = 14.4 km

Question 8.
REASONING
Draw a parallelogram that has an area of 24 square inches.

Answer:
We have to assume that the base = 4 square inches and the height = 6 square inches

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
The side of an office building in Hamburg, Germany, is in the shape of a parallelogram. The area of the side of the building is about 2150 square meters. What is the length x of the portion of the building that extends over the river?

Answer:
The length of the portion of the building that extends over the river = 45 m

Explanation:
The area of the parallelogram = the product of the bases and the product of the bases
area = 2150 square meters given
area = 25 m x (39 +47) given that height = 25 and base = 39
area = 25 x (86)
area = 2150
Thus the length of the portion of the building that extends over the river = 45 m

Question 10.
You make a photo prop for a school fair. You cut a 10-inch square out of a parallelogram-shaped piece of wood. What is the area of the photo prop?

Answer:
The area of the photo prop = 32

Explanation:
the area of the photo prop = the product of the base and the product of the height
area = h x b
area = 8 x 4
area = 32 ft

Question 11.
DIG DEEPER!
A galaxy contains a parallelogram-shaped dust field. The dust field has a base of 150 miles. The height is 14% of the base. What is the area of the dust field?

Answer:
The area of the dust field =3,150

Explanation:
The area of the parallelogram = the product of the height and the product of the base
area = h x b given that height = 14% = (14/100) x 150 = 21, base = 150 miles
area = 3150

Areas of Parallelograms Homework & Practice 7.1

Review & Refresh

Answer:
:

Explanation:
The given equation = y = 4x
for suppose x = 0, then y = 0
x = 1 , then y = 4 that means (1,4) y = 4 x 1 = 4
x = 2 , then y = 8 that means (2,8) y = 4 x 2 = 8
x = 3 , then y = 12 that means (3,12) y = 4 x 3 = 12

Answer:

Explanation:
The given equation = y = x + 3
for suppose x = 0, then y = 3
x = 1 , then y = 4 that means (1,4) y = 1 + 3 = 4
x = 2 , then y = 5 that means (2, 5) y = 2 + 3 = 5
x = 3 , then y = 6 that means (3,16) y = 3 + 3 = 6
x = 4 , then y = 7 that means (4, 7) y = 4 + 3 = 7

Answer:

Explanation:
The given equation = y = 2x + 5
for suppose x = 0, then y = 5
x = 1 , then y = 7 that means (1,7) y = 2 + 5 = 7
x = 2 , then y = 9 that means (2, 9) y = 4 + 5 = 9
x = 3 , then y = 11 that means (3,11) y = 6 + 5 = 11
x = 4 , then y = 13 that means (4, 13) y = 8 + 5 = 13

Represent the ratio relationship using a graph.

Question 4.

Explanation:

Question 5.

Write the prime factorization of the number.

Explanation:
5 x 11 = 55
S0 5,11

Explanation:
2 x 30 = 60
2 x 15 = 30
5 x 3 = 15
3 x 1 = 3

Explanation:
3 x 50 = 150
5 x 10 = 50
5 x 2 = 10
2 x 1 = 2

Explanation:
2 x 63 = 126
3 x 21 = 63
3 x 7 = 21
7 x 1 = 7

Add or subtract.


Question 11.
9.035 – 6.144

Explanation:

Question 12.
28.351 – 19.3518

Answer:
28.351 = 19.3518 = 8.9992

Explanation:

Concepts, Skills, & Problem Solving
USING TOOLS
Rearrange the parallelogram as a rectangle. Then find the area. (See Exploration 1, p. 285.)

Question 13.

Answer:

Question 14.

Answer:

Question 15.

Answer:

FINDING AREA
Find the area of the parallelogram.

Question 16.

Answer:
The area of the parallelogram = 18 ft

Explanation:
The area of the parallelogram = the product of the base and the height
area = b x h given that base = 6 ft height = 3 ft
area = 18 ft
Question 17.

Answer:
The area of the parallelogram = 840 mm

Explanation:
The area of the parallelogram = the product of the base and the height
area = b x h given that base = 20 mm height = 42 ft
area = 840 mm

Question 18.

Answer:
The area of the parallelogram = 187 km

Explanation:
The area of the parallelogram = the product of the base and the height
area = b x h given that base = 17 km height = 11km
area = 187 km

Question 19.

Answer:
The area of the parallelogram = 3750 cm

Explanation:
The area of the parallelogram = the product of the base and the height
area = b x h given that base = 75 cm height = 50 cm
area = 3750 cm

Question 20.

Answer:
The area of the parallelogram = 243 in

Explanation:
The area of the parallelogram = the product of the base and the height
area = b x h given that base = 13.5 in height = 18 in
area = 243 in

Question 21.

Answer:
The area of the parallelogram = 894 mi

Explanation:
The area of the parallelogram = the product of the base and the height
area = b x h given that base = 24 in, height = 37 x (1/4)
area = 24 x 37.25
area = 894 mi

Question 22.
YOU BE THE TEACHER
Your friend finds the area of the parallelogram. Is your friend correct? Explain your reasoning.

Answer:
Yes my friend is correct

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = 8 m x 15 m
Given that base = 8m and height = 15 m
area = base x height
area = 8 m x 15 m
area = 120 m

Question 23.
MODELING REAL LIFE
A ceramic tile in the shape of a parallelogram has a base of 4 inches and a height of 1.5 inches. What is the area of the tile?

Answer:
The area of the tile = 6 inches

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 4 x 1.5
area = 6 inches

FINDING AREA
Find the area of the parallelogram. Round to the nearest hundredth if necessary.

Question 24.

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 6 x 8
area = 48 m
1 meter = 100 centimeter
48 m = 4800 cm

Question 25.

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 1320 yd x 1496 yd
area = 1974720yd
1 yard = 0.000568 mile
1974720 yd = 1,121.64096 miles

Question 26.

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 5 m x 4 m
area = 20 m
1 meter = 3 feet 3.37 inches
20 m = 67 . 4 ft

Question 27.
OPEN-ENDED
Your deck has an area of 128 square feet. After adding a section, the area will be (s 2 + 128) square feet. Draw a diagram of how this can happen.

Answer:

Question 28.
MODELING REAL LIFE
You use the parallelogram-shaped sponge to create the T-shirt design. The area of the design is 66 square inches. How many times do you use the sponge to create the design? Draw a diagram to support your answer.

Answer:
33 times is used to the sponge to create the design = 33
Explanation:
The area of the parallelogram = base x height
area = 3 x 1
area = 3 inches
The area of the design = 66 inches
66/3 = 33

FINDING A MISSING DIMENSION
Find the missing dimension of the parallelogram described.

Question 29.

Answer:
height = 9 ft
Explanation:
area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 6 ft x 9 ft
area = 54 ft
So height = 54

Question 30.

Explanation:
area of the parallelogram = the product of the base and the product of the height
area = base x height
area = 6 . 5 cm x 2.5 cm
area = 16 .25
So height = 2.5 cm

Question 31.

Explanation:
5(x + 4) = 5x + 20

Question 32.
DIG DEEPER!
The staircase has three identical parallelogram-shaped panels. The horizontal distance between each panel is 4.25 inches. The area of each panel is 287 square inches. What is the value of x?

Answer:
The value of x = 14

Explanation:
(1/2)bx = 287
(1/2) x 14 x x = 287
7x = 287
x = 287/7
x = 41
50.5 -8
42.5 x 2 = 8.50

Question 33.
LOGIC
Each dimension of a parallelogram is multiplied by a positive number n. Write an expression for the area of the new parallelogram.

Answer:
The area of the parallelogram = (1/2) x b x h
area = (1/2)bn x hn
(1/2) bhn

Question 34.
CRITICAL THINKING
Rearrange the rhombus shown to write a formula for the area of a rhombus in terms of its diagonals.

Answer:
The area of the rhombus = (1/2) ab
area = (1/2) x ab

Explanation:
The area of the rhombus = (1/2) ab
area = (1/2) x ab

Lesson 7.2 Areas of Triangles

EXPLORATION 1

Deriving the Area Formula of a Triangle
Work with a partner.
a. Draw any parallelogram on a piece of centimeter grid paper. Cut the parallelogram into two identical triangles. How can you use the area of the parallelogram to find the area of each triangle?
b. Copy the triangle below on a piece of centimeter grid paper. Find the area of the triangle. Explain how you found the area.

c. Draw any acute triangle on a piece of centimeter grid paper and find its area. Repeat this process for a right triangle and an obtuse triangle.
d. Do the areas change in part(c) when you use different sides as the base? Explain your reasoning.

Answer:
a : The area of the parallelogram = (1/2) x b x h
The area of the triangle = base x height
b : The area of the triangle = base x height
area = b x h
c : The area of the acute-angled triangle = (1/2) x b x h
The area of the right-angled triangle = (1/2) x base x perpendicular
The area of the obtuse-angled triangle =(1/2) x b x h

Explanation:
a :
b :
c :




e. Use your results to write a formula for the area A of a triangle. Use the formula to find the area of the triangle shown.

Explanation:
The area of the triangle = half x the product of the base b and the product of the height h
area = (1/2) x b x h given that b= 6 m ,h = 5 m
area = (1/2) x 30
area = 15 m

Find the area of the triangle

Question 1.

Explanation:
The area of the triangle = (1/2) x bh
The area of the triangle = (1/2) x the product of base and the product of the height
b = 11 ft , h = 4 ft
area = (1/2) x bh
area = (1/2) x 11 x 4
area = (1/2) x 44
area = 22 ft

Question 2.

Explanation:
The area of the triangle = (1/2) x the product of the base b and the product of the height h
b = 15 cm, h = 8cm
area = (1/2) x 15 x 8
area = (1/2) x 120
area = 60

Question 3.

Explanation:
The area of the triangle = (1/2) x the product of the base b and the product of the height h
So the b = 22m ,h= 10 m
area= (1/2) x 10 x 22
area = (1/2) x 220
area = 110 m

Question 4.

Explanation:
The area of the triangle = (1/2) x the product of base b and the product of the height h
So the b = (13/2) yd h = 2 yd
area = (1/2) x b x h
area = (1/2) x 6.5 x 2
area = (1/2) x 13
area = 6.5

Find the missing dimension of the triangle.

Question 5.

Explanation:
The area of the triangle =(1/2) x the product of the base b and the product of height h
area of the triangle = 24 cm given
So the height = 6cm given
area = (1/2) x b x h
area = (1/2) x 8 x 6
area = (1/2) x 48
area = 24
So base = 8 cm

Question 6.

Explanation:
The area of the triangle =(1/2) x the product of the base b and the product of the height h
area of the triangle = 175 ft given,
So we have to find the height?
area = (1/2) x 20 x 17.5
area = (1/2) x 350
area = 175
So the height = 17.5 ft

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
FINDING AREA
Find the area of the triangle at the left.

Answer:
The area of the triangle = 12

Explanation:
The area of the triangle = (1/2) x the product of base and the product of the height
b = 8 , h = 3
area = (1/2) x bh
area = (1/2) x 8 x 3
area = (1/2) x 24
area = 12

Question 8.
WRITING
Explain how to use the area of a parallelogram to find the area of a triangle.

Answer:
The area of the parallelogram = the product of the base and the product of the height
The area of the triangle = half x the product of the base b and the product of the height h

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = b x h
The area of the triangle = half x the product of the base and the height
area = (1/2) x b x h
FINDING A MISSING DIMENSION
Find the missing dimension of the triangle.

Question 9.

Explanation:
The area of the triangle = (1/2) the product of the base and the product of the height
area = (1/2) x base x height
area = (1/2) x 10 mm x 6 mm
area = (1/2) x 60 mm
area = 30 mm
So base = 10 mm
Question 10.

A composite figure is made up of triangles, squares, rectangles, and other two-dimensional figures. To find the area of a composite figure, separate it into figures with areas you know how to find. This is called decomposition.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
A wildlife conservation group buys the 9 square miles of land shown. What is the distance from Point A to Point B?

Answer:
The distance from point A to point B = 3 miles

Explanation:
The area of the triangle = (1/2) ab
Given that the land shown = 9 square miles
(1/2)bh = triangle ABC + triangle BCD
(3/2)b + (3/2)b = 9
(6/2)b = 9
3 b = 9
b = 3 miles
Question 12.
DIG DEEPER!
Find the area of the side of the chimney. Explain how you found the area.

Answer:
The area of the chimney = 31.5 sq. ft

Explanation:
The area of the chimney = (1/2) x base x height
area = (1/2) x 3 x 21
area = 31.5 sq. ft

Areas of Triangles Homework & Practice 7.2

Review & Refresh

Find the area of the parallelogram.

Question 1.

Answer:
The area of the parallelogram = 72 in

Explanation:
The area of the parallelogram = product of the height h and the product of the base b
b = 12 in, h = 6 in given
area = 12 x 6
area = 72 in

Question 2.

Answer:
The area of the parallelogram = 10.5 km

Explanation:
The area of the parallelogram = the product of the base b and the product of the height h
b = 3.5 km, h = 3km given
area = 3.5 x 3
area = 10.5 km

Question 3.

Answer:
The area of the parallelogram = 255 mi

Explanation:
The area of the parallelogram = the product of the base b and the product of the height h
b = 17 mi, h = 15 mi given
area = 17 x 15
area = 255 mi

Tell which property the statement illustrates.

Answer:
The product of x = the product of y

Explanation:
The product of x = n x 1 = n
the product of y = n
n = n

Answer:
The product of x = the product of y

Explanation:
The product of x = m x 4 = 4 m
the product of y = m x 4 = 4 m
4 . m = m . 4

Question 6.
(x + 2) + 5 = x + (2 + 5)

Answer:
The product of x = the product of y

Explanation:
The product of x = (x + 2) + 5
the product of y = x + (2 +5)
(x +2) +5 = x + (2 +5)

Question 7.
What is the first step when using order of operations?
A. Multiply and divide from left to right.
B. Add and subtract from left to right.
C. Perform operations in grouping symbols.
D. Evaluate numbers with exponents.

Answer:
Evaluate numbers with exponents

Concepts, Skills, & Problem Solving

USING TOOLS
Find the area of the triangle by forming a parallelogram. (See Exploration 1, p. 291.)

Question 8.

Answer:
The area of the right angled triangle = (1/2) x ab

Explanation:

Question 9.

Answer:
obtused angled triangle = (1/2) x b xh

Explanation:

Question 10.

Answer:
acute angled triangle

Explanation:

FINDING AREA
Find the area of the triangle.

Question 11.

Answer:
The area of the triangle = 6 cm

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
So b = 3 cm , h = 4 cm
area = (1/2) x 3 x 4
area = (1/2) x 12
area = 6 cm

Question 12.

Answer:
The area of the triangle = 90 mi

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
So b = 20 mi , h = 9 mi
area = (1/2) x 20 x 9
area = (1/2) x 180
area = 90 sq. miles

Question 13.

Answer:
The area of the triangle =1620 in

Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
So b =60 in , h = 54 in
area = (1/2) x 3240
area = 1620 in

Question 14.

The area of the triangle = 189 mm

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
So b =21 mm , h = 18 mm
area = (1/2) x 378
area = 189 mm

Question 15.

Answer:
The area of the triangle = 1125 cm

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
So b =75 cm , h = 30 cm
area = (1/2) x 2250
area = 1125 cm

Question 16.

Answer:
The area of the triangle = 132 m

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
So b =8 m , h = 33 m
area = (1/2) x 264
area = 132 m

Question 17.
YOU BE THE TEACHER
Your friend finds the area of the triangle. Is your friend correct? Explain your reasoning.

Answer:
Yes my friend is correct

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
Sob =10 m , h = 12 m
area = (1/2) x 120
area = 60 m

Question 18.
MODELING REAL LIFE
Estimate the area of the cottonwood leaf.

Answer:
The area of the cottonwood leaf = 10 in

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
Sob = 5 in , h = 4 in
area = (1/2) x 20
area = 10 in

Question 19.
MODELING REAL LIFE
A shelf has the shape of a triangle. The base of the shelf is 36 centimeters, and the height is 18 centimeters. Find the area of the shelf in square inches.

Answer:
The shape of the triangle = 324

Explanation:
Area of the triangle = (1/2) x the product of the base b and the product of the height h
area = (1/2) x b x h
Sob = 36 cm , h = 18 cm
area = (1/2) x 648
area = 324 sq. cm

FINDING A MISSING DIMENSION
Find the missing dimension of the triangle.

Question 20.

Question 21.

Answer:
Base of the triangle = 4.66 ft

Explanation:
The area of the triangle =(1/2) x the product of the base b and the product of height h
area of the triangle = 14 ft given
So the height = 6 ft given
area = (1/2) x b x h
area = (1/2) x 4.66 x 6
area = (1/2) x 27.96
area = 14
soo base = 4.66 ft

Question 22.

Explanation:
The area of the triangle = (1/2) x the product of the base and the product of the height
area = (1/2) x 1.5 x h given that area = 2 in
2 = (1/2) x 1.5h
1.5 h = 4
h = (4/1.5)
h = 2.66 in
COMPOSITE FIGURES
Find the area of the figure.

Question 23.

Answer:
The area of the figure = 44 ft

Explanation:
The area of the rectangle = 2 x(length + breadth)
area = 2 x (12 +10)
area = 2 x (22)
area = 44 sq. ft

Question 24.

Answer:
The area of the figure = 52 cm

Explanation:
The area of the rectangle = 2 x(length + breadth)
area = 2 x (11 +15)
area = 2 x (26)
area = 52 cm

Question 25.

Answer:
The area of the perimeter = 5/2 x a x b

Explanation:
The area of the perimeter = (5/2) x a x b
where a = middle point
b = sides

Question 26.
WRITING
You know the height and the perimeter of an equilateral triangle. Explain how to find the area of the triangle. Draw a diagram to support your reasoning.

Answer:
The perimeter of the equilateral triangle = a x a x a = 3a

Explanation:
The perimeter of the equilateral triangle = 3a
height of the equilateral triangle = h
the area of the triangle = (1/2) x product of base b and the product of the height h
area = (1/2) x bh

Question 27.
CRITICAL THINKING
The total area of the polygon is 176 square feet. What is the value of x?

Explanation:
The area of the polygon = 176 square feet given
area = 16 x 8 = 128
area = 176 – 128 = 48
The area of the right angle triangle = (a x b )/2
area = (8 x X)/2
48 = 4x + 4x
8x = 48
x = (48/8)
x = 6 ft

Question 28.
REASONING
The base and the height of Triangle A are one-half the base and the height of Triangle B. How many times greater is the area of Triangle B?

Explanation:
The triangle B is 2 times greater than the triangle A

Question 29.
STRUCTURE
Use what you know about finding areas of triangles to write a formula for the area of a rhombus in terms of its diagonals. Compare the formula with your answer to Section 7.1 Exercise 34.

Answer:
The area of the triangle = (1/2) ab
The area of the rhombus = (1/2) ab

Explanation:
The area of the triangle is equal to the area of the rhombus
area of the triangle = (1/2) ab
area of the rhombus = (1/2) ab

Lesson 7.3 Areas of Trapezoids and Kites

EXPLORATION 1
Deriving the Area Formula of a Trapezoid
Work with a partner.
a. Draw any parallelogram on a piece of centimeter grid paper. Cut the parallelogram into two identical trapezoids. How can you use the area of the parallelogram to find the area of each trapezoid?
b. Copy the trapezoid below on a piece of centimeter grid paper. Find the area of the trapezoid. Explain how you found the area.

c. Draw any trapezoid on a piece of centimeter grid paper and find its area.
d. Use your results to write a formula for the area A of a trapezoid. Use the formula to find the area of the trapezoid shown.


You can use decomposition to find areas of trapezoids and kites. A kite is a quadrilateral that has two pairs of adjacent sides with the same length and opposite sides with different lengths.

Try It
Find the area of the figure.

Question 1.

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (5 + 12). Given that b1 = 5 and b2 = 12
area = (1/2) x 9(17) given that h = 9
area = (1/2) x 153
area = 76.5 sq. in

Question 2.

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (3 + 3). Given that b1 = 3 and b2 = 3
area = (1/2) x 7.7(17) given that h = 7.7
area = (1/2) x 130.9
area = 65.45 sq. mi

In Example 1(a), you could have used a copy of the trapezoid to form a parallelogram. As you may have discovered in the exploration, this leads to the following formula for the area of a trapezoid.

Try It
Find the area of the trapezoid.

Question 3.

Answer:
The area of the trapezoid = 22.5 sq. mm

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (5 + 4). Given that b1 = 5 and b2 = 4
area = (1/2) x 8(9) given that h = 8
area = (1/2) x 45
area = 22.5 sq. mm

Question 4.

Answer:
The area of the trapezoid = 30 sq. in

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (7.7 + 2.3). Given that b1 = 7.7 and b2 = 2.3
area = (1/2) x 6(10) given that h = 6 in
area = (1/2) x 60
area = 30 sq. in

Try It
Find the area of the figure.

Question 5.

Answer:
The area of the trapezoid1 = 60 sq. m
The area of the trapezoid2= 88 sq. m

Explanation:
The area of the trapezoid1 = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (5 + 5). Given that b1 = 5 and b2 = 5
area = (1/2) x 12(10) given that h = 6 m
area = (1/2) x 120
area = 60 sq. m
The area of the trapezoid2 = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (8 + 8). Given that b1 = 8 and b2 = 8
area = (1/2) x 11(16) given that h = 6 m
area = 88 sq. m
Question 6.

Answer:
The area of the trapezoid = 112 sq. in

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (6+ 10). Given that b1 = 6 and b2 = 10
area = (1/2) x 14(16) given that h = 14 in
area = (1/2) x 224
area = 112 sq. in

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
WRITING
Explain how to use the area of a parallelogram to find the area of a trapezoid.

Answer:
The area of the parallelogram = product of the height h and the product of the base b
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.

Question 8.
REASONING
What measures do you need to find the area of a kite?

Answer:
Rhombus is used to find the area of the kite

Explanation:
The area of the rhombus = (1/2) x side x height
area = (1/2) x s x h
area = (1/2) sh
kite is also same as the rhombus

FINDING AREA
Find the area of the figure.

Question 9.

Explanation:
The area of the trapezoid = (1/2) x height x the product of the base 1 and the base 2
area = (1/2) x height x (b1 + b2)
area = (1/2) x 5 x (9 + 12)
area = (1/2) x 5 x 21
area = (1/2) x 105
area = 52.5 sq. yd

Question 10.

Explanation:
The area of the trapezoid = (1/2) x height x the product of the base 1 and the base 2
area = (1/2) x height x (b1 + b2)
area = (1/2) x 10 x (7 + 4)
area = (1/2) x 10x 11
area = (1/2) x 110
area = 55 sq. m

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
DIG DEEPER!
An archaeologist estimates that the manuscript shown was originally a rectangle with a length of 20 inches. Estimate the area of the fragment that is missing.

Explanation:
The area of the right angle triangle = a x b where a = height b = base
area = a x b a=12 b = 18 – 6 = 12
area = 12 x 12
area = 144 sq. in

Question 12.
The stained-glass window is made of identical kite-shaped glass panes. The approximate dimensions of one pane are shown. The glass used to make the window costs $12.50 per square foot. Find the total cost of the glass used to make the window.

Answer:
The total cost of the glass used to make the window is 14.4375 $
Explanation:
The area of the rhombus = (1/2) x the product of the side s and the product of height h
area = (1/2) x2 ft x 2.31ft
area = ( 4.62/2)
Area = 2.31 x 12.50$
area = 14.4375$

Areas of Trapezoids and Kites Homework & Practice 7.3

Review & Refresh

Find the area of the triangle.

Question 1.

Explanation:
The area of the triangle = half the product of the base and the product of the height
area = (1/2)bh
area = (1/2) x 18 x 7
area = (126/2)
area = 63 in

Question 2.

Explanation:
The area of the triangle = half the product of the base and the product of the height
area = (1/2)bh
area = (1/2) x 8 x 6.5
area = (52/2)
area = 26 sq. km

Question 3.

Explanation:
The area of the triangle = half the product of the base and the product of the height
area = (1/2)bh
area = (1/2) x 4 x 12.5
area = (50/2)
area = 25 sq. ft

Classify the quadrilateral./

Question 4.

Answer:
The above quadrilateral look similar to the rectangle

Explanation:
Rectangle
the area of the rectangle = length x width
area = l x w

Question 5.

Answer:
The above diagram is similar to the trapezoid

Explanation:
The area of the trapezoid = half the product of the base and the product of the height
area = (1/2) x b x h
Question 6.

Answer:
The above figure is similar to the parallelogram

Explanation:
The area of the parallelogram = the product of the height and the product of the base b
area = b x h

Question 7.
On a normal day, 12 airplanes arrive at an airport every 15 minutes. Which rate does not represent this situation?
A. 24 airplanes every 30 minutes
B. 4 airplanes every 5 minutes
C. 6 airplanes every 5 minutes
D. 48 airplanes each hour

Explanation:
4 airplanes every 5 minutes
Concepts, Skills, & Problem Solving
USING TOOLS
Find the area of the trapezoid by forming a parallelogram. (See Exploration 1, p. 297.)

Question 8.

Answer:
The area of the trapezoid = (1/2) x b x l x h

Explanation:
The area of the trapezoid = (1/2) x b x l x h
area = (1/2) x b x h x l
area = (lxb/2)x h

Question 9.

Answer:
The area of the trapezoid = (1/2) x b x l x h

Explanation:
The area of the trapezoid = (1/2) x b x l x h
area = (1/2) x b x h x l
area = (lxb/2)x h

Question 10.

Answer:
The area of the trapezoid = (1/2) x b x l x h

Explanation:
The area of the trapezoid = (1/2) x b x l x h
area = (1/2) x b x h x l
area = (lxb/2)x h

FINDING AREA
Use decomposition to find the area of the figure.

Question 11.

Explanation:
The area of the triangle = (1/2) b x h
base value = 2, height = 5 given
area = (1/2) x 5 x 2
area = 5
the area of another triangle = 5
the remaining = rectangle
area of rectangle = length x breadth
area = 15 + 5 + 5 = 25 cm

Question 12.

Explanation:
The area of the triangle = (1/2) b x h
base value = 2, height = 5 given
area = (1/2) x 8 x 2=3
area = 12
the remaining = rectangle
area of rectangle = length x breadth
area = 80 + 12 yd
area = 92 yd

Question 13.

Explanation:
The area of the triangle = (1/2) b x h
base value = 2, height = 5 given
area = (1/2) x 8 x 5= 20
area = 20
The area of the triangle = (1/2) b x h
base value = 17, height = 5 given
area = (1/2) x 17 x 5= 20
area = 42.5
area = 20 + 20+ 42.5 + 42.5
area = 125 m

Question 14.

Answer:
The area of the figure = 44 inches

Explanation:
The area of the triangle = (1/2) b x h
base value = 9, height = 4 given
area = (1/2) x 9 x 4= 18
area = 18
the remaining = rectangle
area of rectangle = length x breadth
area = 18 + 18 + 4 +
area = 44 inches
Question 15.

Answer:
The area of the figure = 55 miles

Explanation:
The area of the triangle = (1/2) b x h
base value = 10, height = 7 given
area = (1/2) x 10 x 7= 35
area = 35 mi
The area of the triangle = (1/2) b x h
base value = 10, height = 4 given
area = (1/2) x 10 x 4= 20
area = 20 mi
area = 20 + 35 = 55
area = 55 mi

Question 16.

Answer:
The area of the figure = 17.68 miles

Explanation:
The area of the triangle = (1/2) b x h
base value = 3.2, height = 3.8 given
area = (1/2) x 3.2 x 3.8= 35
area = 6.08
The area of the triangle = (1/2) b x h
base value = 3.2, height = 1.6given
area = (1/2) x 3.2 x 1.6= 20
area = 2.56
area = 2.56 + 2.56 + 6.08 + 6.08 = 17.68 kms
area = 17.68 miles

FINDING AREA
Find the area of the trapezoid.

Question 17.

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (b1 +b2)
area = (1/2) x 4 x (6 + 8)/
area = (1/2) x 4 x 14
area = 28 sq. in

Question 18.

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (b1 +b2)
area = (1/2) x 4 x (3.5 + 1.5)
area = (1/2) x 4 x 5
area = 10 sq. cm

Question 19.

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (b1 +b2)
area = (1/2) x 10 x (13.5 + 7.5)
area = (1/2) x 10 x 21
area = 105 sq. ft

Question 20.
YOU BE THE TEACHER
Your friend finds the area of the trapezoid. Is your friend correct? Explain your reasoning.

Answer:
No my friend is not correct

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (b1 +b2)
area = (1/2) x 8 x (6 +14)
area = (1/2) x 160
area = 80 sq. m

Question 21.
MODELING REAL LIFE
Light shines through a window. What is the area of the trapezoid-shaped region created by the light?

Answer:
16 sq. ft
Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (b1 +b2) given that b1 = 3 and b2 = 5
area = (1/2) x 4 x (5 +3)
area = (1/2) x 32
area = 16 sq. ft

COMPOSITE FIGURES
Find the area of the figure.

Question 22.

Answer:
178.5 sq. ft
Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (14 +7) given that b1 = 14 and b2 = 7
area = (1/2) x 17 x (14 +7)
area = (1/2) x 357
area = 178.5 sq. ft

Question 23.

Question 24.

FINDING A MISSING DIMENSION
Find the height of the trapezoid.

Question 25.

Answer:
The height of the trapezoid = square root of height and bases

Question 26.

Question 27.

FINDING AREA
Find the area (in square feet) of a trapezoid with height hand bases b1 and b2.

Question 28.
h = 6 in.
b1 = 9 in.
b2 = 12 in.

Answer:
The area of the trapezoid = 0.43722 square feet

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (9 +12) given that b1 = 9 and b2 = 12
area = (1/2) x 6 x (9 + 12)
area = (1/2) x 126
area = 63 in
1 inch = 0.00694 sq feet
so 0.43722sq ft

Question 29.
h = 12 yd
b1 = 5 yd
b2 = 7 yd

Answer:
The area of the trapezoid = 648 square feet

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (5 +7) given that b1 = 5 and b2 = 7
area = (1/2) x 12 x (5 + 7)
area = (1/2) x 144
area =72 yd
1 yard = 9 sq feet
so 72 yd = 72 x 9 sq ft
area = 648 sq ft

Answer:
The area of the trapezoid = 355.212 square feet

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (3 +8) given that b1 = 3 and b2 = 8
area = (1/2) x 6 x (3 + 8)
area = (1/2) x 66
area =33 m
1 m = 10.764 sq feet
so 33 m = 33 x 10.764sq ft
area = 355.212 sq ft

Question 31.
OPEN-ENDED
The area of the trapezoidal student election sign is 5 square feet. Find two possible values for each base length.

Answer:
The area of the trapezoid = 4 ft

Explanation:
Area of trapezoid = half the product of the height and the product of the bases
area = (1/2) x h x (4 +4) given that b1 = 4 and b2 = 4
area = (1/2) x 2 x (4 + 4)
area = (1/2) x 8
area = 4 sq. ft

Question 32.
REASONING
How many times greater is the area of the floor covered by the larger speaker than by the smaller speaker?

Explanation:
The larger speaker is 2 times greater than that of the smaller speaker

Question 33.
REASONING
The rectangle and the trapezoid have the same area. What is the length ℓ of the rectangle?

Answer:
The area of the triangle =36
Explanation:
In the above given trapezoid the bases b1 = 12 and b2 = 24 height = 9 ft
So the area of the trapezoid = (1/2) x h x(b1 + b2)
area = (1/2) x 9 x (12 +24)
area = (1/2) x 324
area = 162
The length of the triangle = length X w
162 = 9l
length = 36 ft

CRITICAL THINKING
In the figureshown, the area of the trapezoid is less than twice the area of the triangle. Find the possible values of x. Can the trapezoid have the same area as the triangle? Explain your reasoning.

Answer:
The possible value of x = 15 in

Explanation:
In the above statement they said that the area of the trapezoid is less than twice the area of the triangle
area of the triangle = (1/2) x b x h
area = (1/2) x 15 x 10
area = (150/2) =75
Area of trapezoid = (1/2) x h x (b1 + b2)
area = (1/2) x 10 x (15 +15)
area = (300/2)
area = 150
So the area of the trapezoid is 2 times less than the area of the triangle.

Question 35.
STRUCTURE
In Section 7.1 Exercise 34 and Section 7.2 Exercise 29, you wrote a formula for the area of a rhombus in terms of its diagonals.
a. Use what you know about finding areas of figures to write a formula for the area of a kite in terms of its diagonals.
b. Are there any similarities between your formula in part(a) and the formula you found in Sections 7.1 and 7.2? Explain why or why not.

Answer:
The area of the rhombus = (1/2) x side x height

Explanation:
The kite is looking the same as the rhombus
area of the rhombus = (1/2) x side x height

Lesson 7.4 Three-Dimensional Figures

EXPLORATION 1
Exploring Faces, Edges and Vertices

Work with a partner. Use the rectangular prism shown.
a. Prisms have faces, edges, and vertices. What does each of these terms mean?
b. What does it mean for lines or planes to be parallel or perpendicular in three dimensions? Use drawings to identify one pair of each of the following.

  • parallel faces
  • parallel edges
  • edge parallel to a face
  • perpendicular faces
  • perpendicular edges
  • edge perpendicular to a face

EXPLORATION 2

Drawing Views of a Solid
Work with a partner. Draw the front, side, and top views of each stack of cubes. Then find the number of cubes in the stack. An example is shown at the left.

Answer:
a: the number of cubes = 4
b: the number of cubes = 4
c: the number of cubes = 5
d: the number of cubes = 6

Explanation:
a: />
b:: />
c: />
d: />

A solid is a three-dimensional figure that encloses a space. A polyhedron is a solid whose faces are all polygons.

Question 1.
Find the numbers of faces, edges, and vertices of the solid.

Answer:
faces = 5
edges = 12
vertices = 6

Explanation:
The number of faces = 5
the number of edges = 12
the number of vertices = 6

Draw the solid.

Answer:

Question 3.
pentagonal pyramid

Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
FACES, EDGES, AND VERTICES
Find the numbers of faces, edges, and vertices of the solid at the left.

Answer:
faces : 3
edges : 6
vertices : 3

Explanation:
The number of faces of solid at the left = 3
edges = 6
vertices = 3

Question 5.
DRAWING A SOLID
Draw an octagonal prism.

Answer:

Question 6.
WHICH ONE DOESN’T BELONG?
Which figuredoes not belong with the other three? Explain your reasoning.

Answer:
The 3rd figure is different from the other 3 figures

Explanation:
The 3rd figure is the same as the square prism
the other 3 figures are triangular prisms

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
The Flatiron Building in New York City is in the shape of a triangular prism. Draw a sketch of the building.

Answer:

Question 8.
The Pyramid of the Niches is in El Tajín, an archaeological site in Veracruz, Mexico. Draw the front, side, and top views of the pyramid. Explain.

Answer:
Front = 2
side = 1
top = 1

Explanation:
/>

Question 9.
Use the point-cut diamond shown.
a. Find the numbers of faces, edges, and vertices of the diamond.
b. Draw the front, side, and top views of the diamond.
c. How can a jeweler transform the point-cut diamond into a table-cut diamond as in Example 3?

Answer:
a : the number of faces = 8,edges = 16,vertices = 6
b : front = 2 , side = 2,top = 1

Explanation:
a : faces = 8
edges = 16
vertices = 6
b : />

Three-Dimensional Figures Homework & Practice 7.4

Review & Refresh

Find the area of the figure.

Question 1.

Answer:
The area of the triangle = 15 ft
Explanation:
In the above given trapezoid the bases b1 = 12 and b2 = 24 height = 9 ft
So the area of the trapezoid = (1/2) x h x(b1 + b2)
area = (1/2) x 3 x (4 +6)
area = (1/2) x 30
area = 15
Question 2.

Answer:
The area of the triangle = 18 km
Explanation:
In the above given trapezoid the bases b1 = 12 and b2 = 24 height = 9 ft
So the area of the trapezoid = (1/2) x h x(b1 + b2)
area = (1/2) x 3 x (5 +7)
area = (1/2) x 36
area = 18 km

Question 3.

Answer:
The area of the triangle = 18 km
Explanation:
In the above given trapezoid the bases b1 = 12 and b2 = 24 height = 9 ft
So the area of the trapezoid = (1/2) x h x(b1 + b2)
area = (1/2) x 18 x (6 +6)
area = (1/2) x 216
area = 18 k

Find the LCM of the numbers.

Explanation:
Factors of 8 = 2 x 2 x2
factors of 12 = 2 x 2x 3
l.c.m. = 2 x 2 x 2x 3
l.c.m = 24

Explanation:
Factors of 15 = 5 x 3
factors of 25 = 5 x 5
l.c.m. = 5 x 3 x5
l.c.m = 75

Explanation:
Factors of 32 = 8 x 4
factors of 44 = 11 x 4
l.c.m. = 4 x 11 x 8
l.c.m = 352

Explanation:
Factors of 3 = 3 x 1
factors of 7 = 7 x 1
factors of 10 =5 x 2
l.c.m. = 3 x 7 x 5 x 2
l.c.m = 210

A bucket contains stones and seashells. You are given the number of seashells in the bucket and the ratio of stones to seashells. Find the number of stones in the bucket.

Question 8.
18 seashells 2 to 1

Answer:
No of stones = 12
No of seashells = 6

Explanation:
(18/3) = 6
2 : 1 = 12 : 6

Question 9.
30 seashells 4 : 3

Answer:
No of stones = 17.14
No of seashells = 12.6

Explanation:
(30/7) = 4.2
4 : 3 = 17.4 : 12.6

Question 10.
40 seashells 7 : 4

Answer:
No of stones = 25.45
No of seashells = 14.54

Explanation:
(40/11) = 3.63
7 : 4 = 25.45 : 14.54

Concepts, Skills, & Problem Solving

DRAWING VIEWS OF A SOLID
Draw the front, side, and top views of the stack of cubes. Then find the number of cubes in the stack. (See Exploration 2, p. 305.)

Question 11.

Answer:
front = 10
side = 3
top = 4

Explanation: />

Question 12.

Answer:
front = 9
side = 3
top = 5

Explanation: />

Question 13.

Answer:
front = 5
side = 5
top = 8

Explanation: />

FACES, EDGES, AND VERTICES
Find the numbers of faces, edges, and vertices of the solid.

Question 14.

Answer:
faces = 10
edges = 24
vertices = 9

Explanation:
The number of faces = 10
the number of edges = 24
the number of vertices =9

Question 15.

Answer:
faces = 17
edges = 34
vertices = 13

Explanation:
The number of faces = 17
the number of edges = 34
the number of vertices =13

Question 16.

Answer:
faces = 10
edges = 20
vertices = 7

Explanation:
The number of faces = 10
the number of edges = 20
the number of vertices =7

DRAWING SOLIDS
Draw the solid.

Question 17.
triangular prism

Answer:

Question 18.
pentagonal prism

Answer:

Question 19.
rectangular pyramid

Question 20.
hexagonal pyramid

Answer:

Question 21.
MODELING REAL LIFE
The Pyramid of Cestius in Rome, Italy, is in the shape of a square pyramid. Draw a sketch of the pyramid.

Answer:

Question 22.
RESEARCH
Use the Internet to find a picture of the Washington Monument. Describe its shape.

DRAWING VIEWS OF A SOLID
Draw the front, side, and top views of the solid.

Question 23.

Answer:
front = 1
side = 2
top = 1

Explanation:
/>

Question 24.

Answer:
front =2
side = 2
top = 1

Explanation:
/>

Question 25.

Answer:
front =2
side = 2
top = 1

Explanation:
/>

Question 26.

Answer:
front =2
side = 2
top = 2

Explanation:
/>

Question 27.

Answer:
front = 1
side = 2
top = 1

Explanation:
/>

Question 28.

Answer:
front = 1
side = 1
top = 1

Explanation:
/>

DRAWING SOLIDS
Draw a solid with the following front, side, and top views.

Question 29.

Answer:

Explanation:
front = 2 top = 1 side = 2

Question 30.

Question 31.
MODELING REAL LIFE
Design and draw a house. Name the different solids that you can use to make a model of the house.

Question 32.
DIG DEEPER!
Two of the three views of a solid are shown.
a. What is the greatest number of cubes in the solid?
b. What is the least number of cubes in the solid?
c. Draw the front views of both solids in parts (a) and (b).

Answer:
a: The greatest number of cubes in the solid = 3
b : The least number of cubes in the solid = 2
c : />

Question 33.
OPEN-ENDED
Draw two different solids with five faces.
a. Write the numbers of vertices and edges for each solid.
b. Explain how knowing the numbers of edges and vertices helps you draw a three-dimensional figure.

Question 34.
CRITICAL THINKING
The base of a pyramid has n sides. Find the numbers of faces, edges, and vertices of the pyramid. Explain your reasoning.

Lesson 7.5 Surface Areas of Prisms

EXPLORATION 1
Using Grid Paper to Construct a Solid
Work with a partner. Copy the figure shown below onto grid paper.

a. Cut out and fold the figure to form a solid. What type of solid does the figure form?
b. What is the area of the entire surface of the solid?

EXPLORATION 2
Finding the Area of the Entire Surface
Work with a partner. Find the area of the entire surface of each solid. Explain your reasoning.

The surface area of a solid is the sum of the areas of all of its faces. You can use a two-dimensional representation of a solid, called a net, to find the surface area of the solid. Surface area is measured in square units.

Key Idea
Net of a Rectangular Prism
A rectangular prism is a prism with rectangular bases.

Find the surface area of the rectangular prism.

Question 1.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 9 x 5 + 9 x 5 + 6 x 9+ 6 x 9 + 6 x 5 +6 x 5
surface area = 45 + 45 + 54 + 54 + 30 +30
surface area = 258 sq. mm

Question 2.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 10 x 3.5 + 10 x 3.5 + 10 x 8+ 10 x 8 + 8 x 3.5 + 8 x 3.5
surface area = 35 +35+ 80 + 80 +28 +28
surface area = 286 sq. ft

Key Idea
Net of a Triangular Prism
A triangular prism is a prism with triangular bases.

Find the surface area of the triangular prism.

Question 3.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 3 x 5 = 15+ (1/2) x 3 x 4 = 6 + (1/2) x 3 x 4 = 6 + 4 x 5 = 20 + 5 x 4 = 20
bottom = 3 x 5 , front = (1/2) x 3 x 4, back = (1/2) x 3 x 4, side = 4 x 5 , side = 5 x 4
surface area = 15 +6 +6 +20 + 20
area = 67 sq. yd

Question 4.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 10 x 10 = 100+ (1/2) x 10 x 16=80 + (1/2) x 10 x 16 = 80+ 9 x 10 =90 + 10 x 16=160
bottom = 10 x 10 , front = (1/2) x 10 x 16, back = (1/2) x 10 x 16, side = 9 x 10 , side = 10 x 16
surface area = 100 +80 +80+90 +160
area = 510 sq. m

When all the edges of a rectangular prism have the same length s, the rectangular prism is a cube. The net of a cube shows that each of the 6 identical square faces has an area of s 2 . So, a formula for the surface area of a cube is
S = 6s 2 . Formula for surface area of a cube

Try It
Find the surface area of the cube.

Question 5.

Explanation:
area of the cube = 6 s2
area = 6 x side x side
area = 6 x 4 x 4
area = 96 sq. cm

Question 6.

Explanation:
area of the cube = 6 s2
area = 6 x side x side
area = 6 x 0.5 x 0.5
area = 1.5 sq. in

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
FINDING SURFACE AREA
Find the surface area of a cube with edge lengths of 9 centimeters.

Explanation:
area of the cube = 6 s2
area = 6 x side x side given that s = 9 cm
area = 6 x 9 x 9
area = 486 sq. cm

Question 8.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
Light shines through a glass prism and forms a rainbow. What is the surface area of the prism?

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 6 x 3 = 18+ (1/2) x 5 x 6=30 + (1/2) x 5 x 6 = 30+ 3 x 5 =15+ 5 x 6= 30
bottom = 6 x 3 , front = (1/2) x 5 x 6, back = (1/2) x 5 x 6, side = 3 x 5 , side = 5 x 6
surface area = 18 +15 +15 +15 +30
area = 93 sq. cm

Question 10.
One pint of chalkboard paint covers 60 square feet. What is the least number of pints of paint needed to paint the walls of a room in the shape of a rectangular prism with a length of 15 feet, a width of 13 feet, and a height of 10 feet? Explain.

Question 11.
DIG DEEPER!
A flexible metamaterial is developed for use in robotics and prosthetics. A block of metamaterial is in the shape of a cube with a surface area of 600 square centimeters. What is the edge length of the block of metamaterial?

Answer:
The edge length of the block of metamaterial = 25 square centimeters

Explanation:
The surface area of the cube = 6 s square
area = 6 x 25 x4
area = 6 x100
area = 600 square centimeters

Surface Areas of Prisms Homework & Practice 7.5

Review & Refresh

Draw the front, side, and top views of the solid.

Question 1.

Answer:
front = 1
side = 2
top = 1

Explanation:
/>
Question 2.

Answer:
front = 1
side = 1
top = 1

Explanation:
/>

Question 3.

Answer:
front = 1
side = 2
top = 2

Explanation:
/>
Find the GCF of the numbers.

Explanation:
The factors of 18 are = 3 x 3 x 2
The factors of 72 are = 3 x 3 x 2 x 2 x 2
From the above the greatest common factors are = 3 x 3 x 2 =18

Explanation:
The factors of 44 are = 2 x 2 x 11
The factors of 110 are = 2 x 5 x 11
From the above the greatest common factors are = 2 x 11 = 22

Explanation:
The factors of 78 are = 2 x 3 x 13
The factors of 93 are = 3 x 31
From the above the greatest common factors are = 3

Explanation:
The factors of 60 are = 2 x 2 x 3 x5
The factors of 96 are = 2 x 2 x 2x 2 x 2 x 3
The factors of 156 are =2 x 2 x 3 x 13
From the above the greatest common factors are = 2 x 2 x 3 = 12

Solve the equation.

Explanation:
s = 12 + 5
s = 17

Explanation:
x = 20 -9
x = 11

Explanation:
48 = 6r
r = (48/6)
r = 8

Explanation:
(m/5) = 13
m = 13 x 5
m = 65

Concepts, Skills, & Problem Solving
USING TOOLS
Use a net to find the area of the entire surface of the solid. Explain your reasoning. (See Exploration 2, p. 311.)

Question 15.

Answer:
The surface area of the rectangle = length x breadth
area = l x b

Explanation:
The above figure is same as the rectangle
The surface area of the rectangle = length x breadth
area = l x b

Question 16.

Answer:
The surface area of the triangle = (/2) x length x breadth
area =(1/2) x l x b

Explanation:
The above figure is the same as the triangle
The surface area of the rectangle =(1/2) x length x breadth
area =(1/2) l x b

Question 17.

Answer:
The surface area of the square = (/2) x length x breadth
area =(1/2) x l x b

Explanation:
The above figure is the same as the triangle
The surface area of the rectangle =(1/2) x length x breadth
area =(1/2) l x b

FINDING SURFACE AREA
Find the surface area of the rectangular prism.

Question 18.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 5 x 1 + 5 x 1 + 10 x 5 + 10 x 5 + 10 x 1 +10 x 1
surface area = 5 + 5 + 50 + 50 +10 +10
surface area = 130

Question 19.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 6 x 9 + 6 x 9 + 9 x 3 + 9 x 3 + 6 x 3 +6 x 3
surface area = 54 + 54 + 27 + 27 +18 +18
surface area = 198 sq. cm

Question 20.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 2 x 4+ 2 x 4 + 4 x 5 + 4 x 5 + 2 x 5 +2 x 5
surface area = 8 + 8 + 20 + 20+10+10
surface area = 76

Question 21.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 2 x 4+ 2 x 4 + 3 x 2 + 3 x 2 + 4 x 3 +4 x 3
surface area = 8 + 8 + 6 + 6 +12 +12
surface area = 52

Question 22.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 5.5 x 3 + 5.5 x 3 + 3 x 7.25 + 3 x 7.25 + 5.5 x 7.25 + 5.5 x 7.25
surface area = 16.5+ 16.5 + 21.75 + 21.75 +39.875 +39.875
surface area = 116.375 m

Question 23.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 6 x2 .33 + 6 x 2.33 + 2.33 x 1.25 + 2.33 x 1.25 + 6 x 1.25 + 6 x 1.25
surface area = 13.98+ 13.98 + 2.9125 + 2.9125 +7.5 +7.5
surface area = 48.785 mi

FINDING SURFACE AREA
Find the surface area of the triangular prism.

Question 24.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 3 x 4 = 12 + (1/2) x 3 x 10 = 15 + (1/2) x 3 x 10 = 15 + 4 x 5 = 20 + 4 x 10 = 40
bottom = 3 x 4 , front = (1/2) x 3 x 10, back = (1/2) x 3 x 10, side = 4 x 5 , side = 4 x 10
surface area = 12 +15 +15+20 +40
area = 102 cm

Question 25.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 16 x 10 = 160+ (1/2) x 16 x 17 = 136 + (1/2) x 16 x 17 = 136+ 15 x 10=150 + 15 x 16 = 240
bottom = 16 x 10 , front = (1/2) x16 x 17, back = (1/2) x 16 x 17, side = 15 x 10 , side =15 x 16
surface = 160 +136+ 136 + 150 + 240
area = 822 m

Question 26.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 12 x 9 = 108+ (1/2) x 12 x 10 = 60 + (1/2) x 12 x 10 = 60 + 15 x 9 = 135 + 9 x 20 = 180
bottom = 12 x 9 , front = (1/2) x12 x 10, back = (1/2) x 12 x 10, side = 115 x 9 , side = 9 x 20
surface area =108 +60 +60+ 135 +180
area = 543 in

Question 27.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 1 x 3 = 3+ (1/2) x 1 x 2.2 = 1.1 + (1/2) x 1 x 2.2 = 1.1 + 2 x 3 = 6+ 3 x 2.2 = 6.6
bottom = 1 x 3 , front = (1/2) x1 x 2.2, back = (1/2) x 1 x 2.2, side = 2 x 3 , side = 3 x 2.2
surface area =3 + 1.1 + 1.1+ 6 + 6.6
area = 17. 8 ft

Question 28.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 4 x 4 = 16+ (1/2) x 4 x 3 = 6 + (1/2) x 4 x 3 = 6 + 5.7 x 4=22.8+ 3 x 4 = 12
bottom = 4 x 4 , front = (1/2) x 4 x 3, back = (1/2) x 4 x 3, side = 5.7 x 4 , side = 3 x 4
surface area = 16 + 6 + 6 +22.8 + 12
area = 62. 8 mm

Question 29.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 10 x (17/3) = 56.6+ (1/2) x 10 x 13 = 65 + (1/2) x 10 x 13 = 65 + 12 x (17/3) = 68+ (17/3) x 13 = 73.6
bottom = 10 x (17/3) , front = (1/2) x 10 x 13, back = (1/2) x 10 x 13, side = 12 x (17/3) , side = (17/3) x 13
surface area = 56.6 + 65 + 65 + 68 + 73.6
area = 3228. 26 m

Question 30.
MODELING REAL LIFE
A gift box in the shape of a rectangular prism measures 8 inches by 8 inches by 10 inches. What is the least amount of wrapping paper needed to wrap the gift box? Explain

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 8 x 8 + 8 x 8 + 10 x 8 + 10 x 8+ 8 x 10 + 8 x 10
surface area = 64 + 64+ 80 + 80 + 80 + 80
surface area = 440 in
The least amount of wrapping paper needed to wrap the gift box = 440 in

Question 31.
MODELING REAL LIFE
What is the least amount of fabric needed to make the tent?

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 6 x 4 = 24+ (1/2) x 6 x 5 = 15 + (1/2) x 6 x 5 = 15+ 7 x 4 = 28+ 5 x 4 = 20
bottom = 6 x 4 , front = (1/2) x 6 x 5, back = (1/2) x 6 x 5, side = 7 x 4 , side = 5 x 4
surface area = 24 + 15 + 15 + 28 + 20
area = 20 sq. ft

FINDING SURFACE AREA

Find the surface area of the cube.

Question 32.

Explanation:
area of the cube = 6 s2
area = 6 x side x side
area = 6 x 6 x 6
area = 216 sq. km

Question 33.

Explanation:
area of the cube = 6 s²
area = 6 x side x side
area = 6 x (1/2) x (1/2)
area = 6 x 0.11111
area = 0.6666 sq. ft

Question 34.

Explanation:
area of the cube = 6 s²
area = 6 x side x side
area = 6 x 5.5 x 5.5
area = 6 x 30.25
area = 181 .5 sq. yd

Question 35.
MODELING REAL LIFE
A piece of dry ice is in the shape of a cube with edge lengths of 7 centimeters. Find the surface area of the dry ice.

Answer:
294 centimeters
Explanation:
area of the cube = 6 s2
area = 6 x side x side
area = 6 x 7 x 7
area = 6 x 49
area = 294 sq. cm

Question 36.
YOU BE THE TEACHER
Your friend finds the surface area of the prism. Is your friend correct? Explain your reasoning.

Answer:
No, my friend is not correct because my friend did the surface area of the cube

Explanation:
My friend did the surface area of the cube
cube = 6s²

Question 37.
CRITICAL THINKING
A public library has the aquarium shown. The front piece of glass has an area of 24 square feet. How many square feet of glass were used to build the aquarium?

Answer:
The square feet of glass were used to build the aquarium = 12 ft

Explanation:
The square feet of glass were used to build the aquarium = l x b
square = 6 x 2.5
square = 12

Question 38.
PROBLEM SOLVING
A cereal box has the dimensions shown.
a. Find the surface area of the cereal box.
b. The manufacturer decides to decrease the size of the box by reducing each of the dimensions by 1 inch. Find the decrease in surface area.

Answer:
a : 216 in
b : 144 in

Explanation:
Surface area of a = 6 x s x s
area = 6 x 36
area = 216
surface area of b = 6 x s x s
area = 6 x 24
area = 144 because they said that cereal box is decreased by 1 inch

Question 39.
REASONING
The material used to make a storage box costs $1.25 per square foot. The boxes have the same volume. Which box might a company prefer to make? Explain your reasoning.

Question 40.
LOGIC
Which of the following are nets of a cube? Select all that apply.

Question 41.
MODELING REAL LIFE
A quart of stain covers 100 square feet. How many quarts should you buy to stain the wheelchair ramp? (Assume you do not have to stain the bottom of the ramp.)

Question 42.
DIG DEEPER!
A cube is removed from a rectangular prism. Find the surface area of the figure after removing the cube.

Explanation:
Surface area of the rectangular prism = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 9 x 8= 72 + (1/2) x 9 x 5=94.5 + (1/2) x 9 x 5 = 94.5 + 5 x 8 = 40+ 8 x 5 =40
bottom = 9 x 8 , front = (1/2) x 9 x 5, back = (1/2) x 9 x 5, side = 5 x 8 , side = 8 x 5
surface area = 72 + 94.5 + 94.5 +40 +40
area = 341 sq. ft

Lesson 7.6 Surface Areas of Pyramids

EXPLORATION 1
Using a Net to Construct a Solid
Work with a partner. Copy the net shown below onto grid paper.

a. Cut out and fold the net to form a solid. What type of solid does the net form?
b. What is the surface area of the solid?

EXPLORATION 2
Finding Surface Areas of Solids
Work with a partner. Find the surface area of each solid. Explain your reasoning.

Key Idea

In this book, the base of every pyramid is either a square or an equilateral triangle. So, the lateral faces are identical triangles.

Try It
Find the surface area of the square pyramid.

Question 1.

Explanation:
surface area of the square pyramid = bottom + side + side + side + side
area = 4 + 3 + 3 + 3
bottom = 2 x 2= 4, side = (1/2) x 2 x 3 = 3,
area = 16 sq. ft

Question 2.

Answer:
62.5 sq. cm
Explanation:
surface area of the square pyramid = bottom + side + side + side + side
area = 25 + 12.5 + 12.5 + 12.5
bottom = 5 x 5 = 25, side = (1/2) x 5 x 5 = 12.5,
area = 62.5 sq. cm

Try It
Find the surface area of the triangular pyramid.

Question 3.

Explanation:
surface area of the triangular pyramid = bottom + side + side + side
area = 17 + 3 + 3 + 3
bottom = (1/2 ) x 3 x 4 = 17 , side =(1/2) x 3 x 2 = 3
area = 26 sq. cm

Question 4.

Explanation:
surface area of the triangular pyramid = bottom + side + side + side
area = 17 + 3 + 3 + 3
bottom = (1/2 ) x 3 x 4 = 17 , side =(1/2) x 3 x 2 = 3
area = 26 sq. cm

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
PRECISION
Explain how to find the surface area of a pyramid.

Answer:
Surface area of the pyramid = area +(1/2) x p x s

Explanation:,
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height

FINDING SURFACE AREA
Find the surface area of the pyramid.

Question 6.

Explanation:,
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height
area = 7 +(1/2) x 4 x 7
area = 7 + 14
area = 21 sq. yd

Question 7.

Explanation:,
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height
area = 8 +(1/2) x 7 x 6.9
area = 7 + (48.3/2)
area = 7 + 24.15
area = 31.15 sq. m

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
A salt lamp is shaped like a triangular pyramid. Find the surface area of each triangular face.

Answer:
The surface area of the triangular pyramid = 115.5 sq. in

The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 3 in
area of the face2 = 3 in
area of the face3 = 3 in
area of the face4 = 3in
area of the base = 3.5
surface area = 12 + 3.5 = 15.5 sq. in

Question 9.
DIG DEEPER!
Originally, each triangular face of the Great Pyramid of Giza had a height of 612 feet and a base of 756 feet. Today, the height of each triangular face of the square pyramid is 592 feet. Find the change in the total surface area of the four triangular faces of the Great Pyramid of Giza.

Answer:
The surface area of the four triangular faces = 612 sq. ft

The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 612 sq. ft
area of the face2 = 612 sq. ftft
area of the face3 =612 sq. ft
area of the face4 = 612 sq. ft
area of the base = 756 sq. ft
surface area = 3204 sq. ft

Surface Areas of Pyramids Homework & Practice 7.6

Review & Refresh

Find the surface area of the prism.

Question 1.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 7 x 2 + 7 x 2 + 7 x 3+ 7 x 3 + 3 x 2 + 3 x 2
surface area = 14 +14+ 21 + 21 +6+6
surface area = 82 sq. ft

Question 2.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = + 7 x 2 + 7 x 3+ 7 x 3 + 3 x 2 + 3 x 2
surface area = 14 +14+ 21 + 21 +6+6
surface area = 82 sq. ft

Question 3.

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area =8 x 8 + 8 x 8 + 8 x 8+ 8 x 8 + 8 x 8 + 8 x 8
surface area = 64 +64 + 64 + 64+ 64+64
surface area = 384 sq. in

Match the expression with an equivalent expression.

Question 4.

Explanation:
3 x (4n + 2 )
12n + 6
A. 2 x (6n + 6)
12n + 12

Question 5.

Explanation:
6 x (2n + 3 )
12n + 18
A. 6 x (2n + 3)
12n + 18

Question 6.

Explanation:
4 x (3n + 4 )
12n + 16
C. 2 x (6n + 3)
12n + 6

Question 7.

Answer:
12n + 12
D. 2 x (6n +8)

Explanation:
4 x (3n + 3)
12n + 12
C. 2 x (6n + 8)
12n + 16

Write the fraction or mixed number as a percent.

Explanation:
(17/25) = (17/25 x 100)
(17 x 4)/(25 x 4)
(68/100)
(34/50)
68%

Explanation:
(19/20) = (19/20 x 100)
(19 x 5)/(20 x 5)
(95/100)
95%
Question 10.
6(frac<7><8>)

Explanation:
(7/8) = (7/8 x 100)
6 x (7/8) = 6 x (7/8 x 100)
(55/8)

Explanation:
(3/400) = (3/400 x 100)
(3/4) = 0.75%

Concepts, Skills, &Problem Solving
USING TOOLS
Use a net to find the surface area of the solid. Explain your reasoning. (See Exploration 2, p. 319.)

Question 12.

Answer:
The surface area of the triangular pyramid = a+(1/2) x p x s

Explanation:
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height

Question 13.

Answer:
The surface area of the triangular pyramid = a+(1/2) x p x s

Explanation:
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height

Question 14.

Answer:
The surface area of the triangular pyramid = a+(1/2) x p x s

Explanation:
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height

FINDING SURFACE AREA
Find the surface area of the pyramid.

Question 15.

Answer:
The surface area of the pyramid = 27 in

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 5 sq. in
area of the face2 = 5 sq. in
area of the face3 = 5 sq. in
area of the face4 = 5 sq. in
area of the base = 7 sq. in
surface area = 27 sq. in

Question 16.

Answer:
The surface area of the pyramid = 51 .6 sq. yd

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 11.4 sq. yd
area of the face2 = 11.4 sq. yd
area of the face3 = 11.4 sq. yd
area of the face4 = 11.4 sq. yd
area of the base = 6 sq. yd
surface area = 51.6 sq. yd

Question 17.

Answer:
The surface area of the pyramid = 80 sq. cm

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 17 sq. cm
area of the face2 = 17 sq. cm
area of the face3 = 17 sq. cm
area of the face4 = 17 sq. cm
area of the base = 12 sq. cm
surface area = 80 sq. cm

Question 18.

Answer:
The surface area of the pyramid = 50.8 sq. ft

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 10.4 sq. ft
area of the face2 = 10.4 sq. ft
area of the face3 = 9 sq. ft
area of the face4 = 9 sq. ft
area of the base = 12 sq. ft
surface area = 50.8 sq. ft

Question 19.

Answer:
The surface area of the pyramid = 49.8 sq. in

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 6.9 sq. in
area of the face2 = 6.9 sq. in
area of the face3 = 14 sq. in
area of the face4 = 14 sq. in
area of the base = 8 sq. in
surface area = 49.8 sq. in

Question 20.

Answer:
The surface area of the pyramid = 27 sq. m

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 8 sq. m
area of the face2 = 8 sq. m
area of the face3 = 3.5 sq. m
area of the face4 = 3.5 sq. m
area of the base = 4 sq. m
surface area = 27 sq. m

Question 21.
MODELING REAL LIFE
A paperweight is shaped like a triangular pyramid. Find the surface area of the paperweight.

Answer:
The surface area of the pyramid = 9.8 sq. in

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 2.2 sq. in
area of the face2 = 2.2 sq. in
area of the face3 = 1.7 sq. in
area of the face4 = 1.7 sq. in
area of the base = 2 sq. in
surface area = 9.8 sq. in

Question 22.
PROBLEM SOLVING
The entrance to the Louvre Museum in Paris, France, is a square pyramid. The side length of the base is 116 feet, and the height of one of the triangular faces is 91.7 feet. Find the surface area of the four triangular faces of the entrance to the LouvreMuseum.

Answer:
The surface area of the pyramid = 48.2 sq. ft

Explanation:
The surface area of the triangular pyramid = area of the faces + area of the base
area of the face1 = 91.7 sq. ft
area of the face2 = 91.7 sq. ft
area of the face3 = 91.7 sq. ft
area of the face4 = 91.7 sq. ft
area of the base = 116 sq. ft
surface area = 482.8 sq. ft

Question 23.
MODELING REAL LIFE
A silicon wafer is textured to minimize light reflection. This results in a surface made up of square pyramids. Each triangular face of one of the pyramids has a base of 5 micrometers and a height of 5.6 micrometers.Find the surface area of the pyramid, including the base.

Answer:
Area = 42 sq. micrometers

Explanation:,
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height
area = 28 +(1/2) x 5 x 5.6
area = 28+ (28/2)
area = 28+ 14
area = 42 sq. micrometers

Question 24.
REASONING
A hanging light cover made of glass is shaped like a square pyramid. The cover does not have a bottom. One square foot of the glass weighs 2.45 pounds. The chain can support 35 pounds. Will the chain support the light cover? Explain.

Answer:
Yes the chain support the light cover

Explanation:
The surface area of the pyramid = area + (1/2) x p x s where p = perimeter of the base, s = slant height
area = 4+(1/2) x 2 x 2
area = 4+ 4
area = 8
area = 8 sq. ft

Question 25.
GEOMETRY
The surface area of the square pyramid shown is 84 square inches. What is the value of x?

Answer:
The value of x = 14 in

Explanation:
The surface area of the square pyramid = length x breadth
area = l x b
84 =6x
x =(84/6)
x =14 in

Question 26.
STRUCTURE
In the diagram of the base of the hexagonal pyramid, all the triangles are the same. Find the surface area of the hexagonal pyramid.

Answer:
The surface area of the hexagonal pyramid = 478.32 cm

Explanation:
The surface area of the hexagonal pyramid = 3ab +3bs
area = 3 x 8 x 6.93 +3 x 8 x 13 where a = 8,b =8, s = 13
area = 166.32+312
area = 478.32 sq. cm

Question 27.
CRITICAL THINKING
Can you form a square pyramid using a square with side lengths of 14 inches and four of the triangles shown? Explain your reasoning.

Answer:
yes
Explanation:
given that square has side lengths of 14 inches and four of the triangles.
so we can form the square pyramid

Lesson 7.7 Volumes of Rectangular Prisms

Recall that the volume of a three-dimensional figureis a measure of the amount of space that it occupies. Volume is measured in cubic units.

EXPLORATION 1

Using a Unit Cube
Work with a partner. A unit cube is a cube with an edge length of 1 unit. The parallel edges of the unit cube have been divided into 2, 3, and 4 equal parts to create smaller rectangular prisms that are identical.

a. The volumes of the identical prisms are equal. What else can you determine about the volumes of the prisms? Explain.

Answer:
The volume of the prism = 1 unit

Explanation:
The volume of the prism = Bh
where b = area of the base
volume = 1 x 1
volume = 1
b. Use the identical prisms in part(a) to find the volume of the prism below. Explain your reasoning.

Answer:
The volume of the prism = 4.3125 cu. unit

Explanation:
The volume of the prism = Bh
where b = area of the base
volume = (3/4) x (2/3)
volume = 0.75 x 5.75
volume = 4.3125 cu. units

c. How can you use a unit cube to find the volume of the prism below? Explain.

Answer:
The volume of the prism = 0.375 cu. m

Explanation:
The volume of the prism = Bh
where b = area of the base
volume = (1/2) x (3/4)
volume = 0.5x 0.75
volume = 0.375 cu. m


d. Do the formulas V = Bh and V = ℓwh work for rectangular prisms with fractional edge lengths? Give examples to support your answer.

Key Idea
Volume of a Rectangular Prism
Words
The volume V of a rectangular prism is the product of the area of the base and the height of the prism.
Algebra
V = Bh or V = lwh

When a rectangular prism is a cube with an edge length of s, you can also use the formula V = s 3 to find the volume V of the cube.

Try It
Find the volume of the prism.

Question 1.

Answer:
The volume of the prism = 0.6665 cu. ft

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (1.333) x (0.5)
volume = 1.33 x 0.5
volume = 0.6665 cu. ft

Question 2.

Answer:
The volume of the prism = 0.6665 cu. ft

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (1.333) x (0.5)
volume = 1.33 x 0.5
volume = 0.6665 cu. ft

Try It
Find the missing dimension of the prism.

Question 3.

Answer:
The missing dimension of the prism = 6 in

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = b x l x h
volume = 6x 6 x 2
volume = 72 in
so length = 6 in

Question 4.

Answer:
The width of the prism = 12.5 cm

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = 5 x(1/2) x 12.5 x20
volume = (11/2)x 12.5 x 20
volume = 1375 cm
so width = 12.5 cm

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
CRITICAL THINKING
Explain how volume and surface area are different.

Answer:
Volume grows exponentially and the surface area grows with the volume
the surface area grows with the rate of volume

Question 6.
FINDING A MISSING DIMENSION
The base of a rectangular prism has an area of 24 square millimeters. The volume of the prism is 144 cubic millimeters. Make a sketch of the prism. Then find the height of the prism.

Answer:
The height of the prism = 6 millimeters

Explanation:
The volume of the rectangular prism = Bh
where b = base area and h= height
so volume = 144 cubic millimeters given
B = 24
volume = Bh
144 = 24h
h = (144/24)
h = 6 mm

FINDING VOLUME
Find the volume of the prism.

Question 7.

Answer:
The volume of the prism = 0.2343 cu. m

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (3/4) x(1/2) x (5/8)
volume = 0.75 x 0.5x 0.625
volume = 0.2343 cu. m

Question 8.

Answer:
The volume of the prism = 0.343 cu. in

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (7/10) x(7/10) x (7/10)
volume = 0.7 x 0.7 x 0.7
volume = 0.343 cu. in

When finding volumes, you may need to convert cubic units. The diagrams at the left show that there are 27 cubic feet per cubic yard.

1 yd 3 = (1 yd)(1 yd)(1 yd) = (3 ft)(3 ft)(3 ft) = 27 ft 3
You can use a similar procedure to convert other cubic units.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
DIG DEEPER!
The shark cage is in the shape of a rectangular prism and has a volume of 315 cubic feet. Find a set of reasonable dimensions for the base of the cage. Justify your answer.

Answer:
The base of the prism = 4.906 ft

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
314 = (8) x(8 ) x h
314 = 64h
h = (314/64)
h = 4.906 ft

Question 10.
The hot tub is in the shape of a rectangular prism. How many pounds of water can the hot tub hold? One cubic foot of water weighs about 62.4 pounds.

Answer:
The hot tub can hold = 312 pounds

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
hot tub = (2) x(2) x 1.25
hot tub = 4 x 1.25
hottub = 5
hot tub = 5
5 x 62.4 = 312 pounds

Volumes of Rectangular Prisms Homework & Practice 7.7

Review & Refresh

Find the surface area of the pyramid.

Question 1.

Answer:
The surface area of the pyramid = 220.5 sq. ft

Explanation:
The surface area of the pyramid = area + (1/2) x p x s
where area = l x w
surface area =21 x (1/2) x 7 x3
surface area = 21x 10.5
surface area = 220.5 sq. ft

Question 2.

Answer:
The surface area of the pyramid = 364.5 sq. m

Explanation:
The surface area of the pyramid = area + (1/2) x p x s
where area = l x w
surface area =27 x (1/2) x 4.5 x6
surface area = 27 x (27/2)=13.5
surface area = 364.5 sq. m

Question 3.

Answer:
The surface area of the pyramid = 20,760 sq. yd

Explanation:
The surface area of the pyramid = area + (1/2) x p x s
where area = l x w
surface area =346 x (1/2) x 12 x 20
surface area = 346 x (120/2)=60
surface area = 20,760 sq. yd

Write the phrase as an expression. Then evaluate the expression when x = 2 and y = 12.

Question 4.
8 more than a number x

Answer:
Yes
Explanation:
8 is more than 2
where x = 2 given

Question 5.
the difference of a number y and 9

Answer:
The difference of a number y and 9 = 3

Explanation:
difference = 12 – 9
diff = 3
where y = 12 given

Concepts, Skills, & Problem Solving
STRUCTURE
The unit cube is divided into identical rectangular prisms. What is the volume of one of the identical prisms? (See Exploration 1, p. 325.)

Question 6.

Explanation:
In the above figure the base = (2/2) units
base = (2/2) units
height = (4/2) units

Question 7.

Explanation:
In the above figure the base = (3/3) units
base = (3/2) units
height = (2/3) units

Question 8.

Explanation:
In the above figure the base = (5/5) units
base = (5/3) units
height = (3/5) units

FINDING VOLUME
Find the volume of the prism.

Question 9.

Answer:
The volume of the prism = 0.297 Cu. in

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (0.66) x(0.6) x (0.75)
volume = 0.297 Cu. in

Question 10.

Answer:
The volume of the prism = 1.3125 Cu. cm

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (7/4) x(3/2) x (1/2)
volume = 1.75 x 1.5 x 0.5
volume = 1.3125 Cu. cm

Question 11.

Answer:
The volume of the prism = 0.064 Cu. ft

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (2/5) x(2/5) x (2/5)
volume = 0.4 x 0.4 x 0.4
volume = 0.064 Cu. ft

Question 12.

Answer:
The volume of the prism =0.75 Cu. m

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (5/8) x(3/4) x (2)
volume = 0.625 x 0.6 x 2
volume = 0.75 Cu. m

Question 13.

Answer:
The volume of the prism =3.111255 Cu. cm

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (5/3) x(5/6) x (9/4)
volume = 1.66 x 0.833 x 2.25
volume = 3.111255 Cu. cm

Question 14.

Answer:
The volume of the prism = 12.425 Cu. m

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (14/5) x(10/7) x (25/8)
volume = 2.8 x 1.42 x 3.125
volume = 12.425 Cu. m

FINDING A MISSING DIMENSION
Find the missing dimension of the prism.

Question 15.
Volume = 1620 cm 3

Answer:
The missing dimension of the prism = 20 cm

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
1620 = (9) x(9) x h
1620 = 81 h
h = (1620/8)
h = 8
Question 16.
Volume = 220.5 cm 3

Answer:
The missing dimension of the prism = 4.5 cm

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
220.5 = (7) x(7) x w
220.5 = 49 w
w = (220.5/49)
w = 4.5 cm

Question 17.
Volume = 532 in 3

Answer:
The missing dimension of the prism = 16 in

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
532 = (7/4) x(19) x w
532 = 33.25
h = (532/33.25)
h = 16 in

Question 18.
MODELING REAL LIFE
An FBI agent orders a block of ballistics gel. The gel weighs 54 pounds per cubic foot. What is the weight of the block of gel?

Answer:
The volume of the prism = 720 in

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (6) x(20) x 6
volume = 720 in
Question 19.
MODELING REAL LIFE
a. Estimate the amount of casserole left in the dish.
b. Will the casserole fit in the storage container? Explain your reasoning.

Answer:
a : The amount of casserole left in the dish = 396 in

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (2.75) x(12) x 12
volume = 396

b : The casserole will not fit in the container

Explanation:
The length of the container is less than the casserole

Question 20.
GEOMETRY
How many (frac<3><4>)-centimeter cubes do you need to create a cube with an edge length of 12 centimeters?

Explanation:
16 cubes are used to create a cube with an edge length of 12 centimeters
in the above-given question (3/4) = 0.75
0.75 x 16 = 12

Question 21.
REASONING
How many one-millimeter cubes do you need to fill a cube that has an edge length of 1 centimeter? How can this result help you convert a volume from cubic millimeters to cubic centimeters? from cubic centimeters to cubic millimeters?

Answer:
10 mm cubes are needed to fill a cube that has an edge length of 1 cm

Explanation:
1 mm =0.1 cm
0.1 x 10 = 1 cm
1 cubic mm = 0.001 cubic cm
1 cubic cm = 1000 cubic mm

Question 22.
LOGIC
The container is partially filled with unit cubes. How many unit cubes fit in the container? Explain your reasoning.

Answer:
6 cubes fit in the container

Explanation:
In the above diagram,6 units are filled with color
so 6 unit cubes fit in the container.

Question 23.
PROBLEM SOLVING
The area of the shaded face is 96 square centimeters. What is the volume of the rectangular prism?

Answer:
The volume of the rectangular prism = 4 x 3 x 8

Explanation:
The volume of the rectangular prism = length x width x height
volume = 4 x 3 x 8

Question 24.
DIG DEEPER!
Is the combined volume of a 4-foot cube and a 6-foot cube equal to the volume of a 10-foot cube? Use a diagram to justify your answer.

Question 25.
PROJECT
You have 1400 square feet of boards to use for a new tree house.
a. Design a tree house that has a volume of at least 250 cubic feet. Include sketches of your tree house.
b. Are your dimensions reasonable? Explain your reasoning.

Area, Surface Area, and Volume Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A sports complex has two swimming pools that are shaped like rectangular prisms. The amount of water in the smaller pool is what percent of the amount of water in the larger pool?

Understand the problem.
You know the shape and the dimensions of the two swimming pools. You are asked to find the amount of water in the smaller pool as a percent of the amount of water in the larger pool.
Make a plan.
First, find the volume of each pool. Then represent the amount of water in the smaller pool as a fraction of the amount of water in the larger pool. Find an equivalent fraction whose denominator is 100 to find the percent.

Answer:
28% of water in the 1st pool is greater than that of the smaller pool

Explanation:
volume of the first pool = 25 x 3 x 50=3750
volume of the second pool = 25 x 21 x 2= 1050
1050 : 3750 = 0.28
0.28 = 28 %
Solve and Check
Use the plan to solve the problem. Then check your solution.

Question 2.
Use a graph to represent the relationship between the surface area S(in square meters) and the height h(in meters) of the triangular prism. Then find the height when the surface area is 260 square meters.

Answer:
The height of the triangular prism = 43.33 meters

Explanation:
The surface area of the triangular prism = (1/2) x b x h
area = (1/2) x 12h
260 = 6h
h =(260/6)
h = 43.33 m

Question 3.
A toy company sells two different toy chests. The toy chests have different dimensions, but the same volume. What is the width w of Toy Chest 2?

Answer:
The width of the toy chest2 = 384 in

Explanation:
In the above figure the toy chest 2 has length = 24 in and volume = 16 in
width = 24 x 16
width = 384 in

Performance Task
Maximizing the Volumes of Boxes

At the beginning of this chapter, you watched a STEAM Video called “Packaging Design.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.

Area, Surface Area, and Volume Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.

Answer:
polygon = a polygon is a closed figure in a plane that is made up of 3 or more line segments that intersect only at their endpoints.
composite figure = composite figure is made up of triangles, squares, rectangles and other two-dimensional figures.
kite = kite is a quadrilateral that has two pairs of adjacent sides with the same length and opposite sides with different lengths.

Graphic Organizers
You can use a Four Square to organize information about a concept. Each of the four squares can be a category, such as definition, vocabulary, example, non-example, words, algebra, table, numbers, visual, graph, or equation. Here is an example of a Four Square for the area of a parallelogram.

Choose and complete a graphic organizer to help you study the concept.

  1. area of a triangle
  2. area of a trapezoid
  3. area of a composite figure
  4. polyhedron
  5. surface area of a prism
  6. surface area of a pyramid
  7. volume of a rectangular prism

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.

7.1 Areas of Parallelograms (pp. 285–290)

Find the area of the parallelogram.

Question 1.

Explanation:
area of the parallelogram = b x h
area = 25 x 20
area = 500 yd

Question 2.

Explanation:
area of the parallelogram = b x h
area = 22 x 11
area = 242 mm

Question 3.

Explanation:
area of the parallelogram = b x h
area = 9 x 5
area = 45 cm

Question 4.
Find the area (in square inches) of the parallelogram.

Answer:
area = 72 square in

Explanation:
area of the parallelogram = b x h
area = 2 x 3
area = 6 ft
1 feet = 12 inches
area = 72 square inches

Question 5.
The billboard shown is in the shape of a parallelogram with a base of 48 feet. What is the height of the billboard?

Answer:
The height of the billboard = 14 ft

Explanation:
The area of the parallelogram = b x h
672 = 48 h
h = (672/48)
h = 14 ft

Question 6.
The freeway noise barrier shown is made of identical parallelogram-shaped sections. The area of each section is 7.5 square meters, and the height of the barrier is 5 meters. How many meters wide is each section of the noise barrier?

Answer:
Each section of the noise barrier is 1.5 m

Explanation:
The area of the parallelogram = b x h
area = 1.5 x5
area = 7.5
so b = 1.5 m

Question 7.
Draw a parallelogram that has an area between 58 and 60 square centimeters.

Answer:
59 square centimeters

Explanation:

7.2 Areas of Triangles (pp. 291–296)

Find the area of the triangle.

Question 8.

Answer:
The area of the triangle = 80 sq. km

Explanation:
The area of the triangle = half the product of the base and the product of the height
area = (1/2) x 16 x 10
area = (1/2) x 160
area = 80 sq. km

Question 9.

Answer:
The area of the triangle = 175 sq. cm

Explanation:
The area of the triangle = half the product of the base and the product of the height
area = (1/2) x 14 x 25
area = (1/2) x 350
area = 175 sq. cm

Find the missing dimension of the triangle.

Question 10.

Explanation:
The area of the tringle = b x h
given that h = 7 mi and area = 35 mi
base = 5 mi

Question 11.

Explanation:
The area of the tringle = b x h
given that b = 4 cm and area = 5 mi
height = 5 cm

Find the area of the figure.

Question 12.

Answer:
The area of the trapezoid = 112.5 sq. yd

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (12 + 3). Given that b1 = 12 and b2 = 3
area = (1/2) x 15(15) given that h = 15 yd
area = (1/2) x 225
area = 112.5 sq. yd

Question 13.

Answer:
The area of the trapezoid = 4 sq. mm

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (8 + 8). Given that b1 = 8 and b2 = 8
area = (1/2) x 2(16) given that h = 2 yd
area = (1/2) x 8
area = 4 sq. mm

Question 14.
Draw a composite figure that has an area of less than 35 square inches.

Answer:
The area of the triangle = b x h
area = 6 x 4
area = 24 square inches

Explanation:

Question 15.
Te triangle-shaped entrance to the cavern is 2([frac<1><2>/latex] feet tall and 4 feet wide. What is the area of the entrance?

Answer:
The area of the entrance = 10

Explanation:
2 x(1/2) = 5/2 = 2.5
2.5 x 4 = 10
7.3 Areas of Trapezoids and Kites (pp. 297 – 304)

Question 16.
Use decomposition to find the area of the kite.

Answer:
The area of the kite = 255 sq. in

Explanation:
The area of the trapezoid = (1/2) x 10 x 9 + 10 x 21
area = (1/2) x 90 +210
area = 45 + 210
255 sq. in
Find the area of the trapezoid.

Question 17.

Answer:
The area of the trapezoid = 105 sq. m

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (15 + 6). Given that b1 = 15 and b2 = 6
area = (1/2) x 10(21) given that h = 10 in
area = (1/2) x 210
area = 105 sq. m

Question 18.

Answer:
The area of the trapezoid = 6 sq. in

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (1.5 + 2.5). Given that b1 = 1(1/2) = 1.5 and b2 = 2(1/2) = 2.5
area = (1/2) x 3(4) given that h = 10 in
area = (1/2) x 12
area = 6 sq. in

Question 19.

Answer:
The area of the trapezoid =49 sq. mi

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (6 + 8). Given that b1 = 6 and b2 = 8
area = (1/2) x 7(14) given that h = 7 mi
area = (1/2) x 98
area = 49 sq. mi

Find the area of the figure.

Question 20.

Answer:
The area of the trapezoid = 56 sq. ft

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (6 + 7). Given that b1 = 6 and b2 = 7
area = (1/2) x 8(14) given that h = 8 ft
area = (1/2) x 112
area = 56 sq ft

Question 21.

Answer:
The area of the trapezoid = 48 sq. cm

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (4 + 8). Given that b1 = 4 and b2 = 8
area = (1/2) x 8(12) given that h = 8 ft
area = (1/2) x 96
area = 48 sq. cm

Question 22.

Answer:
The area of the trapezoid = 40 sq. in

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (6 +10). Given that b1 = 6 and b2 = 10
area = (1/2) x 5(16) given that h = 80
area = (1/2) x 80
area = 40 sq. in

Question 23.
You are creating a design for the side of the soapbox car. How much area do you have for the design?

Answer:
The area of the trapezoid =240 sq. in

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (6 +14). Given that b1 = 6 and b2 = 14
area = (1/2) x 24(20) given that h = 24
area = (1/2) x 480
area = 240 sq. in

Question 24.
Find the area (in square centimeters) of a trapezoid with a height of 2 meters and base lengths of 3 meters and 5 meters.

Answer:
The area of the trapezoid =240 sq. in

Explanation:
The area of the trapezoid = one half the product of its height h and the sum of its bases b1 and b2.
area = (1/2) x h x( b1 + b2)
area =(1/2) x h (3 +5). Given that b1 = 3 and b2 = 5
area = (1/2) x 2(8) given that h = 2
area = (1/2) x 16
area = 8 sq. m
1 meter = 100 cm
area = 8 x 100 = 800 sq. cm

7.4 Three-Dimensional Figures (pp. 305–310)

Find the numbers of faces, edges, and vertices of the solid.

Question 25.

Answer:
faces = 6
edges = 12
vertices =7

Explanation:
The number of faces = 6
The number of edges = 12
The number of vertices =7

Question 26.

Answer:
faces = 5
edges = 10
vertices =6

Explanation:
The number of faces = 5
The number of edges = 10
The number of vertices =6

Question 27.

Answer:
faces = 9
edges = 20
vertices = 12

Explanation:
The number of faces = 9
The number of edges = 20
The number of vertices =12

Question 28.
square pyramid

Question 29.
hexagonal prism

Draw the front, side, and top views of the solid.

Question 30.

Answer:
front = 1
side =2
top = 1

Explanation:
/>

Question 31.

Answer:
front = 4
side =2
top = 2

Explanation:
/>

Question 32.

Answer:
front = 1
side = 1 />
top = 1

7.5 Surface Areas of Prisms (pp. 311–318)

Find the surface area of the prism.

Question 33.

Answer:
The surface area of the prism = 100 sq. in

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 7 x 2 + 7 x 2 + 4x 7+ 4 x 7 + 4 x 2 + 4 x 2
surface area = 14 +14+ 28 + 28 +8+8
surface area = 100 sq. in

Question 34.

Answer:
The surface area of the prism = 243 sq. m

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 6 x 9 + 6 x 9 + 9 x 4.5+ 9 x 4.5 + 6 x 4.5 + 6 x 4.5
surface area = 54 +54+ 40.5 + 40.5 +27+ 27
surface area = 243 sq. m

Question 35.

Answer:
The surface area of the prism = 590 cm

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 8 x 7 + 8 x 7 + 15 x 8+ 15 x 8 + 17 x 7 + 17 x 7
surface area = 56 +56+ 120 + 120 +119+ 119
surface area = 590 sq. cm

Question 36.

Answer:
The surface area of the triangular prism = 49 sq. ft

Explanation:
The surface area of the triangular prism = 2 b + p s
surface area = 2 x 5 +6 x 6.5
surface area = 10 + 39
surface area = 49 sq. ft

Question 37.

Answer:
The volume of the prism = 1,17,649 cu. yd

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (7) x(7) x (7)
volume = 49 x 49 x 49
volume = 1,17,649 cu. yd

Question 38.

Answer:
The volume of the prism = 614.125 cu. mi

Explanation:
The volume of the prism = Bh
where b = l x w
where b = area of the base
volume = (17/2) x(17/2) x (17/2)
volume = 8.5 x 8.5 x 8.5
volume = 614.125 cu. mi

Question 39.
One quart of water-resistant paint covers 75 square feet. A swimming pool is in the shape of a rectangular prism with a length of 20 feet, a width of 10 feet, and a height of 5 feet. How many quarts should you buy to paint the swimming pool with two coats of paint?

Answer:
The volume of rectangular prism = 1000 cu. ft

Explanation:
The volume of the rectangular prism = b ase x height
volume = b x h where b = l x w
volume = 20 x 10 x 5
volume = 1000 cu. ft
7.6 Surface Areas of Pyramids (pp. 319–324)

Find the surface area of the pyramid.

Question 41.

Answer:
The surface area of the pyramid = 95.2 m

Explanation:
The surface area of the square pymarid = area + (1/2) x ps
surface area = 55.2 + (1/2) x 10 x 8
surface area = 55.2 + 40
surface area = 95.2 m

Question 42.

Answer:
The surface area of the pyramid = 88.1 cm

Explanation:
The surface area of the square pymarid = area + (1/2) x ps
surface area = 65.8 + (1/2) x 7 x 9.4
surface area = 55.2 + 32.9
surface area = 88.1 cm

Question 43.
You make a square pyramid for a school project. Find the surface area of the pyramid.

Answer:
The surface area of the pyramid = 41.25

Explanation:
The surface area of the square pyramid = area + (1/2) x ps
surface area = 27.5 + (1/2) x 5.5 x 5
surface area = 27.5 + 13.75
surface area = 41.25 in

7.7 Volumes of Rectangular Prisms (pp. 325–330)

Find volumes and missing dimensions of rectangular prisms.

Question 44.

Answer:
The volume of rectangular prism = 4.875 ft

Explanation:
The volume of the rectangular prism = b ase x height
volume = b x h where b = l x w
volume = (5/2) x (3/2) x (4/3)
volume = 2.5 x 1.5 x 1.33
volume = 4.875 ft

Question 45.

Answer:
The volume of rectangular prism = 0.605 cm

Explanation:
The volume of the rectangular prism = b ase x height
volume = b x h where b = l x w
volume = (1/2) x (2/3) x (11/6)
volume = 0.5 x 0.66 x 1.83
volume = 0.605 cm

Question 46.

Answer:
The volume of rectangular prism = 0.05272 cu. in

Explanation:
The volume of the rectangular prism = b ase x height
volume = b x h where b = l x w
volume = (3/8) x (3/8) x (3/8)
volume = 0.375 x 0.375 x 0.375
volume = 0.05273 cu. in

Question 47.
The prism has a volume of 150 cubic feet. Find the length of the prism.

Answer:
The length of rectangular prism = 7 ft

Explanation:
The volume of the rectangular prism = base x height
volume = b x h where b = l x w
volume = (5) x (4) x (7)
volume = 150
length = 7 ft

Question 48.
How many cubic inches of tissues can the box hold?

Answer:
The length of rectangular prism = 162 . 5 in

Explanation:
The volume of the rectangular prism = b ase x height
volume = b x h where b = l x w
volume = (5) x (5) x (6.5)
volume = 162.5 in

Question 49.
Draw a rectangular prism that has a volume less than 1 cubic inch.

Area, Surface Area, and Volume Practice Test

7 Practice Test

Find the area of the figure.

Question 1.

Answer:
The area of the parallelogram = 13000 cm

Explanation:
The area of the parallelogram = the product of the base and the product of the height
area = 130 x 100
area = 13000 cm

Question 2.

Answer:
The area of the triangle = 154 in

Explanation:
The area of the triangle = (1/2 ) x b x h
area = (1/2) x 14 x 22
area = (1/2) x 308
area = 154 in

Question 3.

Find the surface area of the solid.

Question 4.

Explanation:
Surface area = area of the bottom + area of the front +area of the back +area of a side + area of a side
surface area = 12 x 7 = 84+ (1/2) x 7 x 5 = 17.5+ (1/2) x 7 x 5 = 17.5+ 13 x 5 = 65 + 12 x 13 = 156
bottom = 12 x 7 , front = (1/2) x 7 x 5, back = (1/2) x 7 x 5, side = 13 x 5 , side = 12 x 13
surface area = 84 + 17.5 +17.5 + 65+156
area = 340 ft

Question 5.

Answer:
The volume of the prism =1567.5 m

Explanation:
The volume of the prism = Bh where b = l x w
where b = area of the base
volume = (9.5) x(11) x (15)
volume = 1567.5 m

Find the volume of the prism.

Question 6.

Answer:
The volume of the prism = 3.5 cm

Explanation:
The volume of the prism = b x h
volume = (3/2) x (7/3)
volume = 1.5 x 2.33
volume = 3.5 cm

Question 7.

Answer:
The volume of the prism = 0.64 yd

Explanation:
The volume of the prism = b x h
volume = (4/5) x (4/5)
volume = 0.8 x 0.8
volume = 0.64 yd

Question 8.
Draw an octagonal prism.

Answer:

Question 9.
Find the numbers of faces, edges, and vertices of the solid.

Answer:
faces = 8
edges = 26
vertices = 10

Explanation:
The number of faces = 8
the number of edges = 26
the number of vertices = 10

Question 10.
The area of a parallelogram is 156 square meters. What is the height of the parallelogram when the base is 13 meters?

Answer:
The base of the parallelogram = 12 square meters

Explanation:
The area of the parallelogram = b x h
area = 13 x 12
area = 156
so base = 12 square meters

Question 11.
A candle is shaped like a square pyramid. Find the surface area of the candle.

Answer:
The surface area of the candle = 21.6 in

Explanation:
The area of the square pyramid = area +(1/2) x p x s
surface area = 14.4 +(1/2) x 4 x 3.6
surface area = 14.4 + 7.2
surface area = 21.6 in

Question 12.
You are wrapping the boxed DVD collection as a present. What is the least amount of wrapping paper needed to wrap the box?

Question 13.
A cube has an edge length of 4 inches. You double the edge lengths. How many times greater is the volume of the new cube?

Explanation:
Given that the cube has an edge length of 4 inches
They said to double the cube
4 x 2 = 8
the volume of the new cube is 2 times greater than that of the old cube

Question 14.
The Pentagon in Arlington, Virginia, is the headquarters of the U.S. Department of Defense. The building’s center contains a pentagon-shaped courtyard with an area of about 5 acres. Find the land areas (in square feet) of the courtyard and the building.

Area, Surface Area, and Volume Cumulative Practice

7 Cumulative Practice

Question 1.
A cruise ship is carrying a total of 4971 people. Each life boat can hold a maximum of 150 people. What is the minimum number of lifeboats needed to evacuate everyone on the cruise ship?
A. 33 lifeboats
B. 34 lifeboats
C. 54 lifeboats
D. 332 lifeboats

Answer:
option A is correct

Explanation:
150/33 = 4971 boats
In the above question, the lifeboat can hold a maximum of 150 people
The minimum no of lifeboats needed = 33

Question 2.
Which number is equivalent to the expression?
3 . 4 2 + 6 ÷ 2
F. 27
G. 33
H. 51
I. 75

Answer:
option H is correct

Explanation:
3 . 16 + 3
48 + 3
51
Question 3.
What is the volume of the package?

Question 4.
A housing community started with 60 homes. In each of the following years, 8 more homes were built. Let represent the number of years that have passed since the first year, and let n represent the number of homes. Which equation describes the relationship between n and y?
F. n = 8y + 60
G. n = 68y
H. n = 60y + 8
I. n = 60 + 8 + y

Answer:
option I is the correct

Explanation:
n = 60 + 8 + y
where n = no of homes
y = number of years

Question 5.
What is the value of m that makes the equation true?

Answer:
the value of m = (6/4)

Explanation:
m = (6/4)
4m = 6
4 x (6/4) = 6
6 = 6

Question 6.
What is the surface area of the square pyramid?

Answer:
none of the options are correct

Explanation:
surface area of the pyramid = area +(1/2)x(5x 3)
area = 15+ (1/2) (15)
area = 15 + 7.5
area = 22.5 in
Question 7.
A wooden box has a length of 12 inches, a width of 6 inches, and a height of 8 inches.

Part A
Draw and label a rectangular prism with the dimensions of the wooden box.
Part B
What is the surface area, in square inches, of the wooden box? Show your work.
Part C
You have a two-fluid ounce sample of wood stain that covers 900 square inches. Is this enough to give the entire box two coats of stain? Show your work and explain your reasoning.

Answer:
The surface area of the rectangular prism = 432 in

Explanation:
The surface area of the rectangular prism = Area of the top + area of the bottom +area of front + area of back + area of side + area of a side
surface area = 12 x 6 + 12 x 6 + 6 x 8 + 6 x 8 + 12 x 8+ 12 x 8
surface area = 72+ 72 + 48 + 48 + 96 +96
surface area = 432 in
a :

Question 8.
On Saturday, you earned $35 mowing lawns. This was dollars more than you earned on Thursday. Which expression represents the amount, in dollars, you earned mowing lawns on Thursday?
F. 35x
G. x + 35
H. x – 35
I. 35 – x

Answer:
Option G is correct

Explanation:
x + 35 is correct

Question 9.
What is the area, in square yards, of the triangle?

Answer:
The area of the triangle = 20 yd

Explanation:
Area of the triangle = (1/2) x b x h
area = (1/2) x 5 x 8 where b = 5 and h = 8 given
area = (1/2) x 40
area = 20yd

Question 10.
Which expression is equivalent to [latex]frac<12><35>) ?

Answer:
option b is correct

Explanation:
(2 x 6)/(7 x 5)
12 / 35
so option b is correct
Question 11.
The description below represents the area of which polygon?

F. rectangle
G. parallelogram
H. trapezoid
I. triangle

Answer:
option h is correct

Explanation:
The formula of trapezoid = 0ne half the product of its height and the sum of its bases.

Question 12.
What is the missing quantity in the double number line?

A. 25 ounces
B. 165 ounces
C. 525 ounces
D. 600 ounces

Answer:
Option b is correct

Explanation:
6 + 15 = 21
150 + 15 = 165
so option b is correct

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6.3: Area, Surface Area and Volume - Mathematics

In this section we will start looking at the volume of a solid of revolution. We should first define just what a solid of revolution is. To get a solid of revolution we start out with a function, (y = fleft( x ight)), on an interval (left[ ight]).

We then rotate this curve about a given axis to get the surface of the solid of revolution. For purposes of this discussion let’s rotate the curve about the (x)-axis, although it could be any vertical or horizontal axis. Doing this for the curve above gives the following three dimensional region.

What we want to do over the course of the next two sections is to determine the volume of this object.

In the final the Area and Volume Formulas section of the Extras chapter we derived the following formulas for the volume of this solid.

where, (Aleft( x ight)) and (Aleft( y ight)) is the cross-sectional area of the solid. There are many ways to get the cross-sectional area and we’ll see two (or three depending on how you look at it) over the next two sections. Whether we will use (Aleft( x ight)) or (Aleft( y ight)) will depend upon the method and the axis of rotation used for each problem.

One of the easier methods for getting the cross-sectional area is to cut the object perpendicular to the axis of rotation. Doing this the cross section will be either a solid disk if the object is solid (as our above example is) or a ring if we’ve hollowed out a portion of the solid (we will see this eventually).

In the case that we get a solid disk the area is,

where the radius will depend upon the function and the axis of rotation.

In the case that we get a ring the area is,

where again both of the radii will depend on the functions given and the axis of rotation. Note as well that in the case of a solid disk we can think of the inner radius as zero and we’ll arrive at the correct formula for a solid disk and so this is a much more general formula to use.

Also, in both cases, whether the area is a function of (x) or a function of (y) will depend upon the axis of rotation as we will see.

This method is often called the method of disks or the method of rings.

The first thing to do is get a sketch of the bounding region and the solid obtained by rotating the region about the (x)-axis. We don’t need a picture perfect sketch of the curves we just need something that will allow us to get a feel for what the bounded region looks like so we can get a quick sketch of the solid. With that in mind we can note that the first equation is just a parabola with vertex (left( <2,1> ight)) (you do remember how to get the vertex of a parabola right?) and opens upward and so we don’t really need to put a lot of time into sketching it.

Here are both of these sketches.

Okay, to get a cross section we cut the solid at any (x). Below are a couple of sketches showing a typical cross section. The sketch on the right shows a cut away of the object with a typical cross section without the caps. The sketch on the left shows just the curve we’re rotating as well as its mirror image along the bottom of the solid.

In this case the radius is simply the distance from the (x)-axis to the curve and this is nothing more than the function value at that particular (x) as shown above. The cross-sectional area is then,

[Aleft( x ight) = pi - 4x + 5> ight)^2> = pi left( <- 8 + 26 - 40x + 25> ight)]

Next, we need to determine the limits of integration. Working from left to right the first cross section will occur at (x = 1) and the last cross section will occur at (x = 4). These are the limits of integration.

The volume of this solid is then,

In the above example the object was a solid object, but the more interesting objects are those that are not solid so let’s take a look at one of those.

First, let’s get a graph of the bounding region and a graph of the object. Remember that we only want the portion of the bounding region that lies in the first quadrant. There is a portion of the bounding region that is in the third quadrant as well, but we don't want that for this problem.

There are a couple of things to note with this problem. First, we are only looking for the volume of the “walls” of this solid, not the complete interior as we did in the last example.

Next, we will get our cross section by cutting the object perpendicular to the axis of rotation. The cross section will be a ring (remember we are only looking at the walls) for this example and it will be horizontal at some (y). This means that the inner and outer radius for the ring will be (x) values and so we will need to rewrite our functions into the form (x = fleft( y ight)). Here are the functions written in the correct form for this example.

[eginy & = sqrt[3]hspace<0.5in> Rightarrow hspace<0.5in>x = y & = frac<4>hspace <0.65in>Rightarrow hspace<0.5in>x = 4yend]

Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right.

The inner radius in this case is the distance from the (y)-axis to the inner curve while the outer radius is the distance from the (y)-axis to the outer curve. Both of these are then (x) distances and so are given by the equations of the curves as shown above.

The cross-sectional area is then,

Working from the bottom of the solid to the top we can see that the first cross-section will occur at (y = 0) and the last cross-section will occur at (y = 2). These will be the limits of integration. The volume is then,

With these two examples out of the way we can now make a generalization about this method. If we rotate about a horizontal axis (the (x)-axis for example) then the cross-sectional area will be a function of (x). Likewise, if we rotate about a vertical axis (the (y)‑axis for example) then the cross-sectional area will be a function of (y).

The remaining two examples in this section will make sure that we don’t get too used to the idea of always rotating about the (x) or (y)-axis.

First let’s get the bounding region and the solid graphed.

Again, we are going to be looking for the volume of the walls of this object. Also, since we are rotating about a horizontal axis we know that the cross-sectional area will be a function of (x).

Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right.

Now, we’re going to have to be careful here in determining the inner and outer radius as they aren’t going to be quite as simple they were in the previous two examples.

Let’s start with the inner radius as this one is a little clearer. First, the inner radius is NOT (x). The distance from the (x)-axis to the inner edge of the ring is (x), but we want the radius and that is the distance from the axis of rotation to the inner edge of the ring. So, we know that the distance from the axis of rotation to the (x)-axis is 4 and the distance from the (x)-axis to the inner ring is (x). The inner radius must then be the difference between these two. Or,

The outer radius works the same way. The outer radius is,

Note that given the location of the typical ring in the sketch above the formula for the outer radius may not look quite right but it is in fact correct. As sketched the outer edge of the ring is below the (x)-axis and at this point the value of the function will be negative and so when we do the subtraction in the formula for the outer radius we’ll actually be subtracting off a negative number which has the net effect of adding this distance onto 4 and that gives the correct outer radius. Likewise, if the outer edge is above the (x)-axis, the function value will be positive and so we’ll be doing an honest subtraction here and again we’ll get the correct radius in this case.

The cross-sectional area for this case is,

[Aleft( x ight) = pi left( <<+ 2x + 4> ight)>^2> - < ight)>^2>> ight) = pi left( <- 4 - 5 + 24x> ight)]

The first ring will occur at (x = 0) and the last ring will occur at (x = 3) and so these are our limits of integration. The volume is then,

As with the previous examples, let’s first graph the bounded region and the solid.

Now, let’s notice that since we are rotating about a vertical axis and so the cross-sectional area will be a function of (y). This also means that we are going to have to rewrite the functions to also get them in terms of (y).

[eginy &= 2sqrt hspace <0.5in>Rightarrow hspace<0.5in>x = frac<<>> <4>+ 1 y & = x - 1hspace <0.75in>Rightarrow hspace<0.5in>x = y + 1end]

Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right.

The inner and outer radius for this case is both similar and different from the previous example. This example is similar in the sense that the radii are not just the functions. In this example the functions are the distances from the (y)-axis to the edges of the rings. The center of the ring however is a distance of 1 from the (y)-axis. This means that the distance from the center to the edges is a distance from the axis of rotation to the (y)-axis (a distance of 1) and then from the (y)-axis to the edge of the rings.

So, the radii are then the functions plus 1 and that is what makes this example different from the previous example. Here we had to add the distance to the function value whereas in the previous example we needed to subtract the function from this distance. Note that without sketches the radii on these problems can be difficult to get.

So, in summary, we’ve got the following for the inner and outer radius for this example.

The cross-sectional area is then,

The first ring will occur at (y = 0) and the final ring will occur at (y = 4) and so these will be our limits of integration.


Surface area of a composite figure

3D composite figures are figures that are made up of two or more types of figures. Their surface areas can be calculated by breaking them down into their components, calculating the surface area of each component, then summing the surface areas.

Find the surface area of the perfect ice-cream cone composed of a right cone and a hemisphere.

Let S be the surface area of the composite figure. The slant height of the right cone is 10 and its base radius is 3. The hemisphere also has a radius of 3. Let S1 be the surface area of the right cone's lateral surface and S2 be the surface area of the hemisphere:


Area, Surface Area and Volume Reference Sheet

The graphic on this page, is designed to be a quick reference for calculating the area, surface area and volume of common shapes.

For more information and examples of these calculations see our pages:
Calculating Area, Three-Dimensional Shapes and Calculating Volume.

Definitions

Apothem: The line connecting the centre of a regular polygon with one of its sides. The line is perpendicular (at a right angle) to the side.

Axis: A line of reference about which an object, point or line is drawn, rotated or measured. In a symmetrical shape, an axis is usually a line of symmetry.

Radius: The distance from the geometric centre of a curved shape to its circumference (edge).

Further Reading from Skills You Need

This four-part guide takes you through the basics of numeracy from arithmetic to algebra, with stops in between at fractions, decimals, geometry and statistics.

Whether you want to brush up on your basics, or help your children with their learning, this is the book for you.


Mathematical Practices

Mathematical Practice 1: Make sense of problems and persevere in solving them.

Perseverance may be called for because the solutions to the problems may not seem immediately obvious. For example, some students, distracted by the different shapes of the four-sided, partial rectangular prisms, may not immediately realize that they must have the same surface areas, as they are formed from paper of the same area. It is not so easy or obvious without numerical calculation to determine which rectangular prism has the greater volume.


Strategies for Teaching Elementary Mathematics

Surface Area : In general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object.

In order for students to understand surface area, they first need to understand concepts such as base, height, diameter, radius, pi, etc.

Surface area can be a challenging concept for students because of the different formulas that are used when dealing with different geometric shapes.

http://www.math.com/tables/geometry/surfareas.htm -This is a link to a website that lists the formulas to find the surface area of specific shapes.

Here are the links to the Utah State Core Curriculum for teaching surface area.

http://www.uen.org/core/core.do?courseNum=5060 -5 th grade core for surface area. It is found in Standard 4.

http://www.uen.org/core/core.do?courseNum=5050 -6 th grade core for surface area. It is also found in Standard 4.

The standard from the NCTM for 5 th grade is listed like this:

Geometry and Measurement and Algebra: Describing three-dimensional shapes and analyzing their properties, including volume and surface area.
Students relate two-dimensional shapes to three-dimensional shapes and analyze properties of polyhedral solids, describing them by the number of edges, faces, or vertices as well as the types of faces. Students recognize volume as an attribute of three-dimensional space. They understand that they can quantify volume by finding the total number of same-sized units of volume that they need to fill the space without gaps or overlaps. They understand that a cube that is 1 unit on an edge is the standard unit for measuring volume. They select appropriate units, strategies, and tools for solving problems that involve estimating or measuring volume. They decompose three-dimensional shapes and find surface areas and volumes of prisms. As they work with surface area, they find and justify relationships among the formulas for the areas of different polygons. They measure necessary attributes of shapes to use area formulas to solve problems. ( http://www.nctm.org/standards/focalpoints.aspx?id=334 )

Lesson Plans -Here are some links to some websites that have great ideas for surface area lesson plans:

Activities -Here are some links to some websites that have great activities that can be used to teach surface area. The first two links are especially good ideas for teaching surface area of cylinders.


Lateral & Surface Areas, Volumes

The lateral area of a regular pyramid or right cone is similar to that of prisms, but since each face is a triangle (or triangle-like), there is a factor of one half. The lateral area is thus half the slant height times the perimeter. The slant height is the distance from the vertex to the edge of the base where it is halfway between the base's vertices. If the pyramid is irregular and certainly if the cone is oblique, the surface area might not be calculatable using elementary techniques (which is a fancy way to say you may need calculus). It depends on if you can obtain the altitude (slant height) of each triangular face.

Surface Area = Lateral Area + n × Bases
n = 2 for prisms/cylinders n = 1 for pyramids/cones n = 0 for spheres.

The surface area of a pyramid or cone is the lateral area plus the area of the single base.

The surface area of a sphere is equal to 4 r 2 . Analogous to the unit circle is the unit sphere. Similarly, just as there are 2 radians of angle in one revolution, there are 4 steradians of solid angle in all directions.

Example: Consider a right pyramid A-BCDE with vertex A and square base BCDE of length 20" on each side and a slant height of 26". What are its lateral and surface areas?

Answer: We don't need the height for this calculation, but we will calculate it anyway to stress the difference between slant height and height. The slant height is the hypotenuse of a right triangle where the height is one leg and 20"/2 = 10" is the other leg. Thus 10 2 + h 2 = 26 2 or 100 + h 2 = 676. Thus h 2 = 576 or h = 24". The lateral surfaces are all triangles with a base of 20" and a height (the slant height) of 26". There are four of them. Thus the lateral area is 4×½吐"吖" = 1040 in 2 . The base is 20" square or 400 in 2 . Thus the [total] surface area is 1440 in 2 .

Understanding surface area may be clearer if you refer back to the net associated with the object. At left is a net for a cube and at right a portion of a net for a sphere. Each of these portions of a sphere is called a gore .

Now is a good time to review something learned in algebra, namely ( x + y ) 2 = x 2 + xy + xy + y 2 = x 2 + 2 xy + y 2 . The diagram at the right should clarify this further, help you remember the FOIL method , as well as give a physical basis for this relationship. (Remember also, the square root of ( x 2 + y 2 ) does NOT equal x + y .) Consider extending the FOIL method first into trinomials: ( a + b + c )( d + e + f ) = ad + ae + af + bd + be + bf + cd + ce + cf . The distributive property is another way to consider this situation. Here the box method is useful.

  d e f
a ad ae af
b bd be bf
c cd ce cf

Now extend the method into three dimensions to find: V = ( a + b )( c + d )( e + f ) = ace + acf + ade + adf + bce + bcf + bde + bdf . This would be useful in finding the volume, which is why it is difficult to display in two dimensions.

  • Every polyhedral region has an unique volume , dependent only on your unit cube .
  • A box has a volume of length × width × height ( V = lwh ).
  • Congruent figures have equivalent volume.
  • Total volume is the sum of all nonoverlapping regions.

By knowing the volume, one can determine the dimensions of a polyhedron. Specifically for a cube with edge s and volume s 3 , given a cube with volume 1000 cubic centimeters (1 liter), you can take the cube root to determine each side had length 10 centimeters or about 3.937 inches. Since one gallon is 231 cubic inches, it is thus about 3.785 liters. Other unit conversions can be expected and are summarized in Numbers lesson 9. Cube roots and volume are at the heart of an ancient impossible geometric construction from antiquity, the Delian Cube Doubling problem. Another important concept is that if you double the dimensions of a cube, the volume goes up by a factor of 8=2 3 , just like area went up by a factor of 4=2 2 . This is a problem commonly encountered when converting from cubic feet into cubic yards!

Example: Suppose you wish to pour concrete 4" deep in your driveway which is 90' long and 9' wide.

Answer: You quickly discover that there are 90࡯঩=270 feet 3 . However, there are only 10 yards 3 since each yard is 3 feet and 3 3 =27.

Calculating in the "native unit" of yards: 30ࡩয can help prevent such an error. By "native" we mean here that the final results are expected in cubic yards. If the initial units are converted to yards fewer mistakes will be made. It is EXTREMELY common to erroneously divide by 3 or 9 and not 27 when converting cubic feet into cubic yards.

Example: Suppose you wish to find the volume of the square based right pyramid A-BCDE given in an earlier example with slant height 26" and base 20" on each side.

Answer: The height is 24" as calculated in the previous example. Thus the volume is (1/3)× B × h = (1/3) × 20" × 20" × 24" = 3200 in 3 .

Volume Formulae

Prism or cylinder: V = base area × height
Pyramid or cone: V = (1/3) × base area × height
Sphere: V = (4/3) × (radius) 3

Typically these formula are written as V = Bh (prism or cylinder), V =(1/3) Bh (pyramid or cone), or V =(4/3) r 3 (sphere). Note how a big B is used to signify that this is a two dimensional base or area and not the same (linear) b we use in triangles.

Oblique Prisms and cylinders have the same volume as a right prism or cylinder with the same height and base area. Think of a stack of paper whose top has been pushed to one side. The stack is no longer vertical. However, the volume of paper hasn't changed. In the formula for finding the volume of an oblique prism please note that the height is the perpendicular segment between the top and bottom bases. When you learn calculus you will discover the surface area of a sphere to be the derivative with respect to r of the sphere's volume formula. A similar thing happens between area of a circle and its circumference. This may be happenstance or there may be a deep reason which I'd like to know.

Example: A favorite volume/surface area problem is as follows. A swimming pool is 24' long, 20' wide, 3' deep at the shallow end, and 10' deep at the deep end. The floor slopes evenly. What is the inside surface of the swimming pool and what is the volume (in gallons)?

Answer: The swimming pool is a trapezoidal prism. The floor is 25 feet long since 10' - 3' = 7' and 7 2 +24 2 = 49 + 576 = 625 = 25 2 . The surface area is the sum of 5 surfaces: 2 congruent trapezoidal sides (½(3+10)㩌), 2 rectangular ends (3㩈 + 10㩈), and the bottom (20㩍). This is 2𤚤 + 60 + 200 + 500 = 1072 feet 2 . The volume is Base × height, where Base is the area of one side (½(3+10)㩌), and the height is the width of the pool (20). Thus the volume is 3120 feet 3 or 23339 gallons (multiply by 12 3 cubic inches per cubic foot and divide by 231 cubic inches per gallon).

A gedanken experiment (thought experiment) used to justify the volume formula for a sphere is as follows. First, remember the circle area activity where we cut the circle into 16 wedges, then rearranged the wedges into a r × r parallelogram. Along the same lines, cut a sphere into pyramids. The total area of the bases of these pyramids is 4 r 2 . The height of each is r . Hence the formula is derived. Along the same lines, some have suggested remembering the 1/3 in conal volume formulae by correlating it to the analoguous two dimensional trianglular area formula which has a 1/2 in it.

Given two solids included between parallel planes. If every plane cross section parallel to the given planes has the same area in both solids, then the volumes of the solids are equal. This is know as Cavalieri's Principle .

The Greek Archimedes is one of the three greatest mathematicians of all time. Among his important discoveries is the relationship between the volumes of the cone, sphere, and cylinder. In fact, this discovery was so much his favorite that he requested it to be inscribed on his tombstone. Specifically, consider a sphere of radius r , two cones each with the same radius and height ( r ), and a cylinder with the same radius and height (2 r ). The cylinder will contain either the two cones or the sphere. Their volumes can easily be seen to be (4/3) r 3 , 2(1/3) r 3 , and 2 r 3 . Thus the cones plus the sphere equals the cylinder exactly. (Actually, Archimedes is more commonly credited with showing the sphere's volume to be 2/3's that of the cylinder.) See the corresponding diagrams in the textbook related to the proof of Cavalieri's Principle.

Example: Question 10.2#24 in our text asked the students about cones made from circles (radius 4") with central angles of 45°, 60°, and 120° removed (which got taped to the board amid Madonna jokes). Perform the following. Find the volume of each cone. Find the central angle which maximizes volume. Find the central angle which maximized volume to surface ratio.

Answer: For bonus points hand in your solution by the time the chapter reviews are due.


Key Facts & Information

AREA OF A RIGHT TRIANGLE

  • In this section, we will solve for the area of a right triangle.
  • As a refresher, remember that in order to find the area of a rectangle, we multiply its width by its length.
    • w x l = a
    • area of right triangle = (l x w)/2

    AREA OF TRIANGLES

    • Now that we know how to compute for the area of a right triangle, we can derive the equation that we can use to compute for the area of other triangles.
    • Note that any two triangles will form a parallelogram.
    • And we know that to solve for an area of any parallelogram, we just multiply its base and its height.
    • Therefore, we can write the formula for the area of any triangle as:
      • area of triangle = (b x h)/2

      AREA OF SPECIAL QUADRILATERALS

      • In this section, we will learn how to solve for the area of special quadrilaterals.
      • We will take a trapezoid as our special quadrilateral example.
      • In this case, we would not be able to use the same equation that we used in parallelograms since this is not a parallelogram.
      • However, we can transform this to create a parallelogram.
      • First, we can duplicate this trapezoid.
      • Now that we have two trapezoids, we have to flip the other one vertically and connect them in order to have a parallelogram.
      • After connecting them both, we have a parallelogram.
      • Remember that for us to be able to find the area of a parallelogram, we have to know its height and base.
      • For us to identify the height and the base, we have to first label them.
      • Based on the diagram above, the height of the parallelogram is already given, however, for the base, we still have to compute for it.
        • base = a + b
        • area = base x height
        • area = (a + b) x height
        • area = ((a + b) x height)/2

        SURFACE AREA OF A CUBE

        • If area is the measurement of the size of a flat surface in a two-dimensional plane, then surface area is the is the measurement of the exposed surface of a shade in a three-dimensional plane.
        • Let us start with the most simple three-dimensional shape – a cube.
        • We know that to find the area of a square, we just need to multiply one side with another side.
        • On the other hand, a cube has 6 faces and each face can be represented by a square.
        • Therefore, if we want to get the surface area of a cube, we can first get the area of one face (one square).
          • a = s x s
          • Wherein s represents the length of the side.

          SURFACE AREA OF A RECTANGULAR PRISM

          • To find the area of a rectangle, we just need to multiply the length and the width.
          • Now, a rectangular prism is composed of 6 faces. However, we cannot use the same method we used for computing for the surface area of a cube since the faces of a rectangular prism are not equal.
          • However, we know that the top and the bottom faces are the same, the left and the right faces are also the same, and the front and back faces are also the same.
          • Therefore, we only need to identify 3 rectangular faces.
          • Now, we have to identify 3 face combinations: (1) top and bottom, (2) front and back, and (3) right and left.
          • Let us first identify the top and bottom face combination, to get the area of it, the sides that we have to multiply are side a and side c.
            • top/bottom = a x c
            • front/back = b x c
            • right/left = a x b
            • surface area = 2(a x b) + 2(b x c) + 2(a x c)

            SURFACE AREA OF A PYRAMID

            • Now, we will try to get the surface area of pyramid.
            • If the triangular faces of a rectangular pyramid are the same, then we can just use the formula for getting the area of a triangle.
            • With that, we can compute for the area of a pyramid by first computing for the perimeter of the base.
            • Since the base is a square, we just need to multiply the length of the side or edge by 4.
              • perimeter = 4s
              • base area = s x s
              • area of triangle = (b x h)/2
              • SA of a pyramid = ((p x h)/2) + ba

              SURFACE AREA OF ANY PRISM

              • To find the area of any prism, there are just 3 things that we have to remember: (1) base perimeter, (2) base area, and (3) height of the prism.
                • surface area = (p x h) + 2b

                VOLUME OF A CUBE

                • Now that we know how to get the area of a square, and the surface area of a cube, we will now move on to finding the volume of a cube.
                • But first, let us identify what a volume is. Volume is the measurement of how much space a three-dimensional shape occupies.
                • We computed for the area of a square by multiplying its side by itself. Then, we multiplied this by 6 to compute for the surface area of a cube.
                • This time, to find the volume of the cube, we will need to follow this formula:
                  • volume = s x s x s

                  VOLUME OF A RECTANGULAR PRISM

                  • Moving forward, let us not compute for the volume of a rectangular prism.
                  • Since we were able to compute for the volume of a cube by multiplying the side to itself twice, we just need to apply this concept to find the volume of a rectangular prism.
                  • Therefore, to find the volume of a rectangular prism we just need to follow the formula:
                    • volume = l x w x h

                    VOLUME OF A PYRAMID

                    • Where “l” is the length of the prism, “w” is the width of the prism, and “h” is the height of the prism.
                    • If for the area of a triangle we used base times height then divide it by 2. This time, for the volume of a pyramid, we will use:
                      • volume of a pyramid = (b x h)/3

                      Area, Surface Area, and Volume Worksheets

                      This is a fantastic bundle which includes everything you need to know about the Area, Surface Area, and Volume across 35 in-depth pages. These are ready-to-use Area, Surface Area, and Volume worksheets that are perfect for teaching students how we can find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes.

                      Complete List Of Included Worksheets

                      • Lesson Plan
                      • Area, Surface Area, and Volume
                      • Find A
                      • D&C
                      • Cubes
                      • Pyramid
                      • Prisms
                      • Words
                      • VC
                      • Find
                      • Blank
                      • Versus

                      Link/cite this page

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                      Use With Any Curriculum

                      These worksheets have been specifically designed for use with any international curriculum. You can use these worksheets as-is, or edit them using Google Slides to make them more specific to your own student ability levels and curriculum standards.


                      Watch the video: Area and Volume Q1 f (December 2021).