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# Measurement and Geometry: Area and Volume of Geometric Figures and Objects

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses area and volume of geometric figures and objects. By the end of the module students should know the meaning and notation for area, know the area formulas for some common geometric figures, be able to find the areas of some common geometric figures, know the meaning and notation for volume, know the volume formulas for some common geometric objects and be able to find the volume of some common geometric objects.

## Section Overview

• The Meaning and Notation for Area
• Area Formulas
• Finding Areas of Some Common Geometric Figures
• The Meaning and Notation for Volume
• Volume Formulas
• Finding Volumes of Some Common Geometric Objects

Quite often it is necessary to multiply one denominate number by another. To do so, we multiply the number parts together and the unit parts together. For example,

8 in. 8 in. = 8 8 in. in. = 64 in. 2 8 in. 8 in. = 8 8 in. in. = 64 in. 2

4 mm 4 mm 4 mm = 4 4 4 mm mm mm = 64 mm 3 4 mm 4 mm 4 mm = 4 4 4 mm mm mm = 64 mm 3

Sometimes the product of units has a physical meaning. In this section, we will examine the meaning of the products (length unit)2(length unit)2 size 12{ $$"length unit"$$ rSup { size 8{2} } } {} and (length unit)3(length unit)3 size 12{ $$"length unit"$$ rSup { size 8{3} } } {}.

## The Meaning and Notation for Area

The product (length unit)(length unit)=(length unit)2(length unit)(length unit)=(length unit)2 size 12{ $$"length unit"$$ cdot $$"length unit"$$ = $$"length unit"$$ rSup { size 8{2} } } {}, or, square length unit (sq length unit), can be interpreted physically as the area of a surface.

### Area

The area of a surface is the amount of square length units contained in the surface.

For example, 3 sq in. means that 3 squares, 1 inch on each side, can be placed precisely on some surface. (The squares may have to be cut and rearranged so they match the shape of the surface.)

We will examine the area of the following geometric figures.

## Area Formulas

We can determine the areas of these geometric figures using the following formulas.

 Figure Area Formula Statement Triangle A T = 1 2 ⋅ b ⋅ h A T = 1 2 ⋅ b ⋅ h size 12{A rSub { size 8{T} } = { {1} over {2} } cdot b cdot h} {} Area of a triangle is one half the base times the height. Rectangle A R = l ⋅ w A R = l ⋅ w size 12{A rSub { size 8{R} } =l cdot w} {} Area of a rectangle is the length times the width. Parallelogram A P = b ⋅ h A P = b ⋅ h size 12{A rSub { size 8{P} } =b cdot h} {} Area of a parallelogram is base times the height. Trapezoid A Trap = 1 2 ⋅ ( b 1 + b 2 ) ⋅ h A Trap = 1 2 ⋅ ( b 1 + b 2 ) ⋅ h size 12{A rSub { size 8{ ital "Trap"} } = { {1} over {2} } cdot $$b rSub { size 8{1} } +b rSub { size 8{2} }$$ cdot h} {} Area of a trapezoid is one half the sum of the two bases times the height. Circle A C = π r 2 A C = π r 2 size 12{A rSub { size 8{C} } =πr rSup { size 8{2} } } {} Area of a circle is π π times the square of the radius.

## Finding Areas of Some Common Geometric Figures

### Sample Set A

#### Example 1

Find the area of the triangle.

A T = 1 2 b h = 1 2 20 6   sq ft = 10 6   sq ft = 60 sq ft = 60 ft 2 A T = 1 2 b h = 1 2 20 6   sq ft = 10 6   sq ft = 60 sq ft = 60 ft 2

The area of this triangle is 60 sq ft, which is often written as 60 ft2.

#### Example 2

Find the area of the rectangle.

Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll use the unit fraction 12 in.1 ft12 in.1 ft size 12{ { {"12 in" "." } over {"1 ft"} } } {} since it has inches in the numerator. Then,

4 ft = 4 ft 1 12 in . 1 ft = 4 ft 1 12 in . 1 ft = 48 in . 4 ft = 4 ft 1 12 in . 1 ft = 4 ft 1 12 in . 1 ft = 48 in .

Thus, 4 ft 2 in.= 48 in.+ 2 in.= 50 in.4 ft 2 in.= 48 in.+ 2 in.= 50 in. size 12{"4 ft 2 in" "." =" 48 in" "." +" 2 in" "." =" 50 in" "." } {}

A R = l w = 50   in . 8 in . = 400 sq in . A R = l w = 50   in . 8 in . = 400 sq in .

The area of this rectangle is 400 sq in.

#### Example 3

Find the area of the parallelogram.

A P = b h = 10 . 3   cm 6 . 2 cm = 63 . 86 sq cm A P = b h = 10 . 3   cm 6 . 2 cm = 63 . 86 sq cm

The area of this parallelogram is 63.86 sq cm.

#### Example 4

Find the area of the trapezoid.

A Trap = 1 2 b 1 + b 2 h = 1 2 14.5 mm + 20.4 mm 4.1 mm = 1 2 34.9 mm 4.1 mm = 1 2 143.09 sq mm = 71.545 sq mm A Trap = 1 2 b 1 + b 2 h = 1 2 14.5 mm + 20.4 mm 4.1 mm = 1 2 34.9 mm 4.1 mm = 1 2 143.09 sq mm = 71.545 sq mm

The area of this trapezoid is 71.545 sq mm.

#### Example 5

Find the approximate area of the circle.

A c = π r 2 ( 3.14 ) 16.8 ft 2 3.14 282.24 sq ft 888.23 sq ft A c = π r 2 ( 3.14 ) 16.8 ft 2 3.14 282.24 sq ft 888.23 sq ft

The area of this circle is approximately 886.23 sq ft.

### Practice Set A

Find the area of each of the following geometric figures.

36 sq cm

37.503 sq mm

13.26 sq in.

367.5 sq mi

452.16 sq ft

44.28 sq cm

## The Meaning and Notation for Volume

The product (length unit)(length unit)(length unit)=(length unit)3(length unit)(length unit)(length unit)=(length unit)3 size 12{ $$"length unit"$$ " " $$"length unit"$$ " " $$"length unit"$$ = $$"length unit"$$ rSup { size 8{3} } } {}, or cubic length unit (cu length unit), can be interpreted physically as the volume of a three-dimensional object.

### Volume

The volume of an object is the amount of cubic length units contained in the object.

For example, 4 cu mm means that 4 cubes, 1 mm on each side, would precisely fill some three-dimensional object. (The cubes may have to be cut and rearranged so they match the shape of the object.)

## Volume Formulas

 Figure Volume Formula Statement Rectangular solid V R = l ⋅ w ⋅ h = area of base ⋅ height V R = l ⋅ w ⋅ h = area of base ⋅ height The volume of a rectangular solid is the length times the width times the height. Sphere V S = 4 3 ⋅ π ⋅ r 3 V S = 4 3 ⋅ π ⋅ r 3 size 12{V rSub { size 8{S} } = { {4} over {3} } cdot π cdot r rSup { size 8{3} } } {} The volume of a sphere is 4343 size 12{ { {4} over {3} } } {} times π π times the cube of the radius. Cylinder V Cyl = π ⋅ r 2 ⋅ h = area of base ⋅ height V Cyl = π ⋅ r 2 ⋅ h = area of base ⋅ height The volume of a cylinder is π π times the square of the radius times the height. Cone V c = 1 3 ⋅ π ⋅ r 2 ⋅ h = area of base ⋅ height V c = 1 3 ⋅ π ⋅ r 2 ⋅ h = area of base ⋅ height The volume of a cone is 1313 size 12{ { {1} over {3} } } {} times π π times the square of the radius times the height.

## Finding Volumes of Some Common Geometric Objects

### Sample Set B

#### Example 6

Find the volume of the rectangular solid.

V R = l w h = 9 in. 10 in. 3 in. = 270 cu in. = 270 in. 3 V R = l w h = 9 in. 10 in. 3 in. = 270 cu in. = 270 in. 3

The volume of this rectangular solid is 270 cu in.

#### Example 7

Find the approximate volume of the sphere.

V S = 4 3 π r 3 4 3 3.14 6 cm 3 4 3 3.14 216 cu cm 904.32 cu cm V S = 4 3 π r 3 4 3 3.14 6 cm 3 4 3 3.14 216 cu cm 904.32 cu cm

The approximate volume of this sphere is 904.32 cu cm, which is often written as 904.32 cm3.

#### Example 8

Find the approximate volume of the cylinder.

V Cyl = π r 2 h 3.14 4.9 ft 2 7.8 ft 3.14 24.01 sq ft 7.8 ft 3.14 187.278 cu ft 588.05292 cu ft V Cyl = π r 2 h 3.14 4.9 ft 2 7.8 ft 3.14 24.01 sq ft 7.8 ft 3.14 187.278 cu ft 588.05292 cu ft

The volume of this cylinder is approximately 588.05292 cu ft. The volume is approximate because we approximated π π with 3.14.

#### Example 9

Find the approximate volume of the cone. Round to two decimal places.

V c = 1 3 π r 2 h 1 3 3.14 2 mm 2 5 mm 1 3 3.14 4 sq mm 5 mm 1 3 3.14 20 cu mm 20 . 9 3 ¯   cu mm 20 . 93 cu mm V c = 1 3 π r 2 h 1 3 3.14 2 mm 2 5 mm 1 3 3.14 4 sq mm 5 mm 1 3 3.14 20 cu mm 20 . 9 3 ¯   cu mm 20 . 93 cu mm

The volume of this cone is approximately 20.93 cu mm. The volume is approximate because we approximated π π with 3.14.

### Practice Set B

Find the volume of each geometric object. If π π is required, approximate it with 3.14 and find the approximate volume.

21 cu in.

Sphere

904.32 cu ft

157 cu m

0.00942 cu in.

## Exercises

Find each indicated measurement.

Area

16 sq m

Area

Area

1.21 sq mm

Area

Area

18 sq in.

Area

### Exercise 17

Exact area

#### Solution

60.5 π+132  sq ft60.5 π+132  sq ft size 12{ left ("60" "." 5π+"132" right )"sq ft"} {}

Approximate area

Area

40.8 sq in.

Area

Approximate area

31.0132 sq in.

Exact area

Approximate area

158.2874 sq mm

Exact area

Approximate area

64.2668 sq in.

Area

Approximate area

43.96 sq ft

Volume

Volume

512 cu cm

Exact volume

### Exercise 31

Approximate volume

11.49 cu cm

### Exercise 32

Approximate volume

### Exercise 33

Exact volume

#### Solution

10243π  cu ft10243π  cu ft size 12{ { {"1024"} over {3} } π " cu ft"} {}

### Exercise 34

Approximate volume

### Exercise 35

Approximate volume

22.08 cu in.

### Exercise 36

Approximate volume

### Exercises for Review

#### Exercise 37

((Reference)) In the number 23,426, how many hundreds are there?

4

#### Exercise 38

((Reference)) List all the factors of 32.

#### Exercise 39

((Reference)) Find the value of 434356+123434356+123 size 12{4 { {3} over {4} } - 3 { {5} over {6} } +1 { {2} over {3} } } {}.

##### Solution

3112=2712=2.583112=2712=2.58 size 12{ { {"31"} over {"12"} } =2 { {7} over {"12"} } =2 "." "58"} {}

#### Exercise 40

((Reference)) Find the value of 5+132+2155+132+215 size 12{ { {5+ { {1} over {3} } } over {2+ { {2} over {"15"} } } } } {}.

#### Exercise 41

((Reference)) Find the perimeter.

27.9m

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