Area: Parallelograms

Area: ParallelogramsArea: Parallelograms

10-1

PRE-ALGEBRA LESSON 10-1PRE-ALGEBRA LESSON 10-1

It takes 126 ft of fence to enclose a field that is twice as long as it iswide. What is the area of the field?

882 ft2

Area: ParallelogramsArea: ParallelogramsPRE-ALGEBRA LESSON 10-1PRE-ALGEBRA LESSON 10-1

(For help, go to Lesson 3-4.)

Use A = w and find the third value.

1. A = 54 in.2, w = 6 in. 2. = 35 m, w = 7 m

3. A = 25 cm2, = 2.5 cm 4. = 7.2 ft, w = 7.2 ft

Check Skills You’ll Need

10-1


Solutions

1. A = w 2. A = w54 = 6 A = 35 • 7 = 9 in. A = 245 m2

3. A = w 4. A = w25 = 2.5w A = 7.2 • 7.2w = 10 cm A = 51.84 ft2

10-1


Find the area of the rectangle.


Step 1 Change the units so that they are the same.

150 cm = 1.5 m Change 150 centimeters to meters.

Step 2 Find the area.

A = bh Use the formula for area of a rectangle.

= (4)(1.5) Replace b and h with the dimensions 4 and 1.5.

= 6 Simplify.

The area of the rectangle is 6 m2. Quick Check

10-1


Find the area of each parallelogram.


a. b.

A = bh area formula

= (8)(2) Substitute.

= 16 Simplify.

The area is 16 m2. The area is 15 in.2.

A = bh

= (2.5)(6)

= 15

Quick Check

10-1


Find each area.

1. rectangle ABFE

2. parallelogram ACFD

3. a rectangle with a base of 50 cm and a height of 5 cm

250 cm2

40 cm2

48 cm2

10-1

Area: Triangles and TrapezoidsArea: Triangles and Trapezoids

10-2


Find the area of a rectangle that is 3 ft wide and twice as high.12

24 ft212

Area: Triangles and TrapezoidsArea: Triangles and TrapezoidsPRE-ALGEBRA LESSON 10-2PRE-ALGEBRA LESSON 10-2


Find each product.

1. • 16 2. • 14 • 6

3. • 5 • 15 4. • 2 • 812

12

12

12

12


10-2


Solutions

1. • 16 2. • 14 • 6 • 168 = 8 • 147 • 6

7 • 6 = 42

3. • 5 • 15 4. • 2 • 8 • 75 = 37 • • 8

• • 82

5 • 2 = 10

12

121

12

121

12

12

12

12

12

12

52

121

521

10-2


Find the area of the triangle.


The area is 39 in.2.

A = bh Use the formula for area of a triangle.12

= • 13 • 6 Replace b with 13 and h with 6.12

= 39 Simplify.

Quick Check

10-2


Find the area of the figure.


Add to find the total: 450 + 1,350 = 1,800.

Area of triangle

A = bh12

= • 45 • 2012

= 450

Area of rectangle

A = bh

= 45 • 30

= 1,350

The area of the figure is 1,800 cm2.Quick Check

10-2


Suppose that, through the years, a layer of silt and mud

settled in the bottom of the Erie Canal. Below is the resulting cross

section of the canal. Find the area of the trapezoidal cross section.


The area of the cross section is 106.5 ft2.

A = h(b1 + b2) Use the formula for the area of a trapezoid.12

A = • 3(31 + 40) Replace h with 3, b1 with 31, and b2 with 40.12

= • 3(71) Simplify.12

= • 21312

= 106.5Quick Check

10-2


Find each area.

1. trapezoid PQRU 2. triangle PTU

3. triangle QRS 4. trapezoid PQSU

28 ft2

192 ft2 20 ft2

164 ft2

10-2

Area: CirclesArea: Circles

10-3


Explain how this pattern works.

(15,873 7) 2 = (222,222)

(15,873 7) 3 = (333,333)

(15,873 7) 4 = (444,444)

because (15,873 7) 1 = (111,111)

Area: CirclesArea: CirclesPRE-ALGEBRA LESSON 10-3PRE-ALGEBRA LESSON 10-3


Simplify each expression.

1. 3.14 • 42 2. 3.14 • 52

3. 3.14 • 92 4. 3.14 • 0.52


10-3


Solutions

1. 3.14 • 42 2. 3.14 • 52

= 3.14 • 4 • 4 = 3.14 • 5 • 5= 50.24 = 78.5

3. 3.14 • 92 4. 3.14 • 0.52

= 3.14 • 9 • 9 = 3.14 • 0.5 • 0.5= 254.34 = 0.785

10-3


Find the exact area of a circle with diameter 20 in.


A = r 2

= (10)2 r = d; r = 1012

= 100 Simplify.

The area is 100 in.2.

Quick Check

10-3


A TV station’s weather radar can detect precipitation in a circular region having a diameter of 100 mi. Find the area of the region.


A = r 2

The area of the region is about 7,850 mi2.

= (50)2 r = d; r = 5012

= 2,500 exact area

(2,500)(3.14) Use 3.14 for .

= 7,850 approximate area

Quick Check

10-3


A pound of grass seed covers approximately 675 ft2. Find the

area of the lawn below. Then find the number of bags of grass seed

you need to buy to cover the lawn. Grass seed comes in 3-lb bags.


Area of region that is one fourth of a circle:

area of circle = r 2

area of quarter circle = r 2

A (3.14)(15)2 Replace with 3.14 and r with 15.

= 176.625 ft2

14

14

10-3


(continued)


Area of region that is a rectangle:

area of rectangle = bh

A = 45 • 25 Replace b with 45 and h with 25.

= 1,125 ft2

The area of the lawn is about 177 ft2 + 1,125 ft2 = 1,302 ft2.

You need to buy one 3-lb bag of grass seed.

1,302 ÷ 675 1.93 Divide to find the number of pounds of seed.

Quick Check

10-3


Find the area.

1. Find the exact area of a circle with diameter 32 in.

2. A 5-ft-diameter round table is in a 12 ft-by-15 ft room.

a. What is the area covered by the table? Round to the nearest unit.

b. What is the area of the rest of the room? Round to the nearest unit.

160 ft2

20 ft2

256 in.2

10-3

Space FiguresSpace Figures

10-4


Find the area of a square 15.7 mm on each side.

246.49 mm2

Space FiguresSpace FiguresPRE-ALGEBRA LESSON 10-4PRE-ALGEBRA LESSON 10-4


Judging by appearance, classify each polygon.

1. 2.

3. 4.


10-4


Solutions

1. triangle 2. square

3. rectangle 4. hexagon

10-4


For each figure describe the bases and name

the figure.


The bases are circles.

a. b.

The figure is a cylinder.

The bases are rectangles.

The figure is a rectangular prism.

Quick Check

10-4


Name the space figure you can form from each net.


With two hexagonal bases and rectangular sides, you can form a hexagonal prism.

a. b.

With a rectangular base and triangular sides, you can form a rectangular pyramid.

Quick Check

10-4


Describe the base(s) and name each solid.

1. 2.

3. Name the solid you can form from this net.

octagonal pyramid

circular base; cone triangular bases; triangular prism

10-4

Surface Area: Prisms and CylindersSurface Area: Prisms and Cylinders

10-5


A prism has 7 faces and 10 vertices. How many edges does theprism have?

15

Surface Area: Prisms and CylindersSurface Area: Prisms and CylindersPRE-ALGEBRA LESSON 10-5PRE-ALGEBRA LESSON 10-5


Find the circumference of each circle with the given radius or diameter.

1. r = 5 in. 2. r = 4.2 cm

3. d = 8 ft 4. d = 6.8 in.


10-5


Solutions

1. C = 2 r 2. C = 2 rC = 2(3.14)(5 in.) C = 2(3.14)(4.2 cm)C = 31.4 in. C = 26.4 cm

3. C = d 4. C = dC = (3.14)(8 ft) C = (3.14)(6.8 in.)C = 25.1 ft C = 21.3 in.

10-5


Find the surface area of the rectangular prism

using a net.


60 + 60 + 150 + 90 + 150 + 90 = 600 Add the areas.

The surface area is 600 cm2.

Draw and label a net.Find the area of each rectangle in the net.

Quick Check

10-5


Find the surface area of the rectangular prism.


The surface area of the rectangular prism is 500 in.2.

Step 1 Find the lateral area.

L.A. = ph Use the formula for lateral area.

= (5 + 6 + 5 + 6)20 p = 5 + 6 + 5 + 6 and h = 20

= 440

Step 2 Find the surface area.S.A. = L.A. + 2B = 440 + 2(5 • 6) L.A. = 440 and B = 5 • 6 = 440 + 60 = 500

Quick Check

10-5


Find the surface area of the cylindrical water tank.


The surface area of the water tank is about 1,156 ft2.

Step 1 Find the lateral area.

L.A. = 2 rh Use the formula for lateral area.

2(3.14)(8)(15)

754

Step 2 Find the surface area. S.A. = L.A. + 2B Use the formula for surface area. = L.A. + 2( r 2) 754 + 2(3.14)(8)2

1,156

Quick Check

10-5


Find the surface area of each figure rounded to the nearest whole unit.

1. triangular prism with base perimeter 24 cm, base area 24 cm2, and height 15 cm

2. rectangular prism with base perimeter 30 cm, base area 50 cm2, and height 150 cm

3. cylindrical candle with radius 2 cm and height 16 cm

about 226 cm2

408 cm2

4,600 cm2

10-5

Surface Area: Pyramids, Cones, and SpheresSurface Area: Pyramids, Cones, and SpheresPRE-ALGEBRA LESSON 10-6PRE-ALGEBRA LESSON 10-6

10-6

Give the number of faces, edges, and vertices of a rectangularprism.

6 faces, 12 edges, 8 vertices



Use the Order of Operations to simplify each expression.

1. (9 ) + (8 ) 2. (12 ) + (15 )

3. (24 ) + (3 ) 4. (32 ) + (14 )

23

12

34

25

16

13

58

17


10-6


Solutions

1. • 93 + • 84 2. • + •

= 6 + 4 = 9 + 6 = 10 = 15

3. • 244 + • 31 4. • 324 + • = 4 + = 20 + 2= 5 = 22

231

121

341

251

161

131

581

171

10-6

153 123

142

Surface Area: Pyramids, Cones, and SpheresSurface Area: Pyramids, Cones, and Spheres

Find the surface area of the square pyramid.


The surface area of the pyramid is 105 m2.

Step 1 L.A. = p Use the formula for lateral area.12

= • 20 • 8 = 80 p = 4(5) and = 8.12

Step 2 S.A. = L.A. + B = 80 + 52 Lateral area = 80 and B = 52. = 80 + 25 = 105

Quick Check

10-6


Find the surface area of the cone.


The surface area of the cone is about 94 m2.

Step 1 L.A. = r Use the formula for lateral area.

3.14(3)(7) r = 3 and = 7.

= 65.94

Step 2 S.A. = L.A. + B Use the formula for surface area.

65.94 + 3.14(3)2 L.A. 65.94 and B = (3)2.

= 65.94 + 28.26

= 94.2Quick Check

10-6


Earth has an average radius of 3,963 mi. What is Earth’s approximate surface area to the nearest 1,000 mi2? Assume that Earth is a sphere.


The surface area of Earth is about 197,259,000 mi2.

S.A. = 4 r 2 Use the formula for surface area.

= 197,259,434.64 Multiply.

197,259,000 Round to nearest 1,000.

4(3.14)(3,963)2 r 3,963

Quick Check

10-6


Find the surface area of each space figure. Round to the nearest whole unit.

1. a square pyramid with base edge 80 m and slant height 100 m

2. a cone with slant height 22 cm and radius 7 cm

3. a sphere with radius 12 cm

about 1,809 cm2

22,400 m2

about 637 cm2

10-6

Volume: Prisms and CylindersVolume: Prisms and Cylinders

10-7


Use graph paper. Design and draw a diagram to determine whichhas the greater area—a square with sides 10 cm or a circle with adiameter 10 cm?

Draw a circle within the square to prove that the square has the greater area.

Volume: Prisms and CylindersVolume: Prisms and CylindersPRE-ALGEBRA LESSON 10-7PRE-ALGEBRA LESSON 10-7


Find the area of each circle.

1. 2. 3.


10-7


Solutions

= (82)( ) = (122)( )

= 64(3.14) = 144(3.14)

201 cm2 452.2 cm2

12= (20)

= 10A = r2

= (102)( )

= 100(3.14)

314 cm2

10-7

1. A = r2 2. A = r2 3. r = d12


Find the volume of the triangular prism.


The volume is 1,260 cm3.

V = Bh Use the formula for volume.

= 63 • 20 B = • 9 • 14 = 63 cm212

= 1,260 Simplify.

Quick Check

10-7


Find the volume of the juice can, to the nearest cubic centimeter.


The volume is about 581 cm3.

V = Bh Use the formula for volume.

V = r 2h B = r 2

= 580.7744 Simplify.

3.14 • 3.42 • 16 Replace r with 3.4, and h with 16.

Quick Check

10-7


Find the volume of each space figure.

1. rectangular prism with base 12 m by 14 m and height 50 m

2. cylindrical pool with diameter 24 ft and height 4 ft

3. right triangular prism with base legs 8 cm and 10 cm and height 20 cm

800 cm3

8,400 m3

about 1,808.64 ft3

10-7

Problem Solving Strategy: Make a ModelProblem Solving Strategy: Make a Model

10-8


A roll of wallpaper is 24 in. wide. You used all but 4 ft of one roll.Another roll has 7 ft of wallpaper on it. What is the total area you can cover with the remaining wallpaper?

12

23 ft223

13

Problem Solving Strategy: Make a ModelProblem Solving Strategy: Make a ModelPRE-ALGEBRA LESSON 10-8PRE-ALGEBRA LESSON 10-8


Draw each figure described below.

1. a rectangle with small squares drawn in each corner

2. a rectangle divided into eight congruent rectangles

3. two parallelograms that have different shapes but the same perimeter


10-8


SolutionsAnswers may vary. Samples:

1. 2.

3.

10-8


A can company rolls rectangular pieces of metal that measure 8 in. by 10 in. to make the sides of cans. Which height, 8 in. or 10 in., will make the can with the greater volume?


Build two cans using 8 in.-by-10 in. pieces of paper. You do not need to make the bases, just the sides.

not to scale

10-8


(continued)


The volume of the can with height 8 in. is greater.

V = r 2h

(3.14)(1.32)(10)

53.1 in.3

V = r 2h

(3.14)(1.62)(8)

64.3 in.3

Find the volumes.

not to scale

Measure your models to find approximate radii.Radius of 10-in. high can 1.3 in.Radius of 8-in. high can 1.6 in.

Quick Check

10-8


Solve.

1. You cut square corners from a piece of cardboard that has dimensions 32 cm by 40 cm. You then fold the cardboard to create

a box with no lid. To the nearest centimeter, what are the dimensions of the box that will have the greatest volume?

20 cm by 28 cm by 6 cm

10-8

Volume: Pyramids, Cones, and SpheresVolume: Pyramids, Cones, and Spheres

10-9


Multiply. Write each answer in simplest form.

a. 3

b. 1 2

c.

1 78

34

12

17

25

2 716

2 35

18

16

Volume: Pyramids, Cones, and SpheresVolume: Pyramids, Cones, and SpheresPRE-ALGEBRA LESSON 10-9PRE-ALGEBRA LESSON 10-9


Multiply.

1. (3.14)(2)2(5) 2. (4)2(6)

3. (3.14)(2)3 4. (3.14)(0.5)3

13

13

43

43


10-9


Solutions

1. (3.14)(22)(5) = (3.14)(4)(5) 2. (42)(6) = (16)(6)

= (62.8) = (96)

= 20.93 = 32

3. (3.14)(23) = (3.14)(8) 4. (3.14)(0.5)3 = (3.14)(0.125)

= (25.12) = (0.3925)

= 33.493 = 0.523

13

1313

13

1313

43

4343

13

4343

10-9


Find the volume of the cone.


V = Bh Use the formula for volume.13

V = r 2h B = r 213

(3.14)(2)2(12) Replace r with 2 and h with 12.13

= 50.24 Simplify.

The volume of the cone is about 50 in.3.Quick Check

10-9


Find the volume of the square pyramid.


The volume of the pyramid is 256 in.3.

V = Bh Use the formula for volume.13

V = s 2h B = s213

= (8)2(12) Replace s with 8 and h with 12.13

= 256 Simplify.Quick Check

10-9


Earth has an average radius of 3,963 mi. What is Earth’s approximate volume to the nearest 1,000,000 mi3? Assume that Earth is a sphere.


The volume of the Earth is about 260,580,000,000 mi3.

V = r 3 Use the volume formula.43

(3.14)(3,963)3 Replace r with 3,963.43

260,579,713,159 Simplify.

Quick Check

10-9


Find the volume of each space figure to the nearest unit.

1. a cone with diameter 9 cm and height 12 cm

2. a square pyramid with base edges 12 m and height 18 m

3. a basketball with diameter 10 in.

about 523 in.3

about 254 cm2

864 m3

10-9

Area: Parallelograms

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