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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.

Triangles, a three-sided polygon, have a height, h, measured from bottom to top, and base, b, measured from one end to the other of the bottom side. Rectangles, a four-sided polygon, have a width, w, in this case the vertical side, and a length, l, in this case the horizontal side.
Parallelograms, a four-sided polygon with diagonal sides in the same direction have a height, h, measured as the distance from the bottom to top, and a base, b, measured as the width of the horizontal side. Trapezoids, a four-sided polygon with diagonal sides facing leaning into each other, have a height measured as the distance between the two bases. Trapezoids have two bases of differing lengths, base 1, and base 2.
 Circles. The distance across the circle is the diameter. The distance from the center of the circle to the edge is the radius.

Area Formulas

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

Table 1
  Figure Area Formula Statement
A triangle. 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.
A rectangle. 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.
A parallelogram. 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.
A trapezoid. 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.
A circle. 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 triangle with height 6 feet and length 20 feet.

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.

A rectangle with width 4 feet 2 inches and height 8 inches.

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 parallelogram with base 10.3cm and height 6.2cm

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 trapezoid with height 4.1mm, bottom base 20.4mm, and top base 14.5mm.

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 circle with radius 16.8ft.

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.

Exercise 1

A triangle with base 18cm and height 4cm.

Solution

36 sq cm

Exercise 2

A rectangle with base 9.26mm and height 4.05mm.

Solution

37.503 sq mm

Exercise 3

A parallelogram with base 5.1in and height 2.6in.

Solution

13.26 sq in.

Exercise 4

A trapezoid with height 15mi, bottom base 32mi, and top base 17mi.

Solution

367.5 sq mi

Exercise 5

A circle with radius r = 12ft.

Solution

452.16 sq ft

Exercise 6

A shape composed of a half circle of radius 2cm, a rectangle with base 5cm and height 7cm, and a triangle with base 2cm and height 3cm.

Solution

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.)

A rectangular solid, with length l, width w, and height h. A sphere with radius r.
A cylinder with height h and radius r. A cone with height h and radius r.

Volume Formulas

Table 2
  Figure Volume Formula Statement
A rectangular solid. 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.
A sphere. 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.
A cylinder. 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.
A cone. 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.

A rectangular solid with width 9in, length 10in, and height 3in.

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.

A circle with radius 6cm.

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.

A cylinder with radius 4.9ft and height 7.8ft.

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.

A cone with height 5mm and radius 2mm

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.

Exercise 7

A rectangular solid with width 9in, length 10in, and height 3in.

Solution

21 cu in.

Exercise 8

Sphere
A sphere with radius 6cm.

Solution

904.32 cu ft

Exercise 9

A cylinder with radius 5m and height 2m.

Solution

157 cu m

Exercise 10

A cone with height .9in and radius .1 in.

Solution

0.00942 cu in.

Exercises

Find each indicated measurement.

Exercise 11

Area
 A rectangle with width 8m and height 2m

Solution

16 sq m

Exercise 12

Area
 A rectangle with width 4.1in and height 2.3in.

Exercise 13

Area
graphics40.png

Solution

1.21 sq mm

Exercise 14

Area
A triangle with base 8cm and height 3cm.

Exercise 15

Area
A triangle with base 9in and height 4in.

Solution

18 sq in.

Exercise 16

Area
A parallelogram with base 20cm and height 9cm.

Exercise 17

Exact area
A rectangle with a half-circle on top. The rectangle's width is 22ft, which is also the diameter of the circle, and the rectangle's height is 6ft.

Solution

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

Exercise 18

Approximate area
A triangle with a half-circle on top, like an ice cream cone. The circle's diameter is 18cm, and the height of the triangle is 26cm.

Exercise 19

Area
graphics46.png

Solution

40.8 sq in.

Exercise 20

Area
A trapezoid with bottom base 15 mm, top base 7 mm, and height 8 mm.

Exercise 21

Approximate area
A shape composed of a trapezoid with a half-circle on top. The circle's diameter is the width of the top base. The bottom base is 8.4in, the height of the trapezoid portion is 3.0in, and the radius of the circle is 2.6in.

Solution

31.0132 sq in.

Exercise 22

Exact area
A circle with a diameter of 3ft.

Exercise 23

Approximate area
A circle with a radius of 7.1mm.

Solution

158.2874 sq mm

Exercise 24

Exact area
A shape that looks like an ice rink. A rectangle with a half-circle attached to each end. The radius of the half-circles is 6cm, and the length of the rectangle is 19cm.

Exercise 25

Approximate area
A trapezoid with a half-circle attached to one base. The half-circle's radius is 3.2in. The other base is 9.4in. The height of the trapezoid is 6.1in.

Solution

64.2668 sq in.

Exercise 26

Area
A rectangle with a rectangle cut out of the inside. The inside rectangle has a width of 4.83in and a height of 1.61in. The outside rectangle has a width of 5.21in and a height of 1.74in.

Exercise 27

Approximate area
A tubelike shape in a half circle. The inner circle's radius is 6.0ft. The tube's thickness is 2.0ft.

Solution

43.96 sq ft

Exercise 28

Volume
A rectangular solid with width 4in, length 2in, and height 1in.

Exercise 29

Volume
A rectangular solid with width 8mm, length 8mm, and height 8mm.

Solution

512 cu cm

Exercise 30

Exact volume
A sphere with a radius of 3in.

Exercise 31

Approximate volume
 A sphere with a radius of 1.4cm.

Solution

11.49 cu cm

Exercise 32

Approximate volume
A cylinder with a radius of 2.1ft and a height of 0.9ft.

Exercise 33

Exact volume
Half of a sphere with radius 8ft.

Solution

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

Exercise 34

Approximate volume
A cylinder with a half-sphere on top. The object's radius is 9.2in, and the height of the cylinder is 24.0in.

Exercise 35

Approximate volume
A cone with radius 1.7in and height 7.3in.

Solution

22.08 cu in.

Exercise 36

Approximate volume
 A cylinder with a cone on top. The object has a diameter of 3.0ft. The cone has a height of 3.0ft. The cylinder's height is 8.1ft.

Exercises for Review

Exercise 37

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

Solution

4

Exercise 38

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

Exercise 39

((Reference)) Find the value of 434−356+123434−356+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.
A triangle with sides of the following lengths: 7.2m, 8.3m, and 12.4m.

Solution

27.9m

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