Transcript MATH 2160 1st Exam Review - Valdosta State University

Perimeter, Area,
Surface Area, and
Volume Examples
Geometry
and
Measurement
Geometry

Polyhedron: V + F – E = 2
Vertices
 Edges
 Faces


Should be able to draw ALL of the
following:
Sphere
 Prisms – Cube, Rectangular,
Triangular
 Cylinder
 Cone
 Pyramids – Triangular, Square

Measurement

5 ft
Rectangle
3 ft
Perimeter
 P = 2l + 2w, where l = length and
w = width
 Example: l = 5 ft and w = 3 ft


P rectangle
=

P
=
2(5 ft) + 2(3 ft)
P
P
=
=
10 ft + 6 ft
16 ft

2l + 2w
Measurement

5 ft
Rectangle
3 ft
Area
 A = lw where l = length and w =
width
 Example: l = 5 ft and w = 3 ft


A rectangle = lw

A
=
(5 ft)(3 ft)

A
=
15 ft2
Measurement

Square
3 ft
Perimeter
 P = 4s, where s = length of a side
 Example: s = 3 ft


P square
=

P
=
4(3 ft)

P
=
12 ft
4s
Measurement

Square
3 ft
Area
 A = s2 where s = length of a side
 Example: s = 3 ft


A square =
s2

A
=
(3 ft)2

A
=
9 ft2
Measurement

Triangle
Perimeter
 P = a + b + c, where a, b, and c
are the lengths of the sides of the
triangle
 Example: a = 3 m; b = 4 m; c = 5
m

P triangle
P
=
P
=

=
a+b+c
3m+4m+5m
12 m
Measurement

Triangle
4m
5m
3m
Area
 A = ½ bh, where b is the base and
h is the height of the triangle
 Example: b = 3 m; h = 4 m

A triangle =
A
=
A
=

½ bh
½ (3 m) (4 m)
6 m2
Measurement

3 cm
Circle
Circumference
 C circle = d or C = 2r, where d =
diameter and r = radius


Example: r = 3 cm

C circle =
2r

C
=
2(3 cm)

C
=
6 cm
Measurement

3 cm
Circle
Area
 A = r2, where r = radius


Example: r = 3 cm
r2

A circle
=

A
=
(3 cm)2

A
=
9 cm2
7 cm
Measurement

Rectangular Prism


5 cm
6 cm
Surface Area: sum of the areas of all of the
faces
Example: There are 4 lateral faces: 2 lateral
faces are 6 cm by 7 cm (A1= wh) and 2
lateral faces are 5 cm by 7 cm (A2 = lh).
There are 2 bases 6 cm by 5 cm (A3 = lw)



A1 = (6 cm)(7 cm) = 42 cm2
A2 = (5 cm)(7 cm) = 35 cm2
A3 = (6 cm)(5 cm) = 30 cm2

SA rectangular prism = 2wh + 2lh + 2lw

SA = 2(42 cm2) + 2(35 cm2) + 2(30 cm2)
SA = 84 cm2 + 70 cm2 + 60 cm2
SA = 214 cm2


Measurement

Rectangular Prism
7 cm
5 cm
Volume:
6 cm
 V = lwh where l is length; w is width;
and h is height
 Example: l = 6 cm; w = 5 cm; h = 7 cm


V rectangular prism = Bh = lwh

V
=
(6 cm)(5 cm)(7 cm)

V
=
210 cm3
Measurement

Cube
5 cm
Surface Area: sum of the areas of all
6 congruent faces
 Example: There are 6 faces: 5 cm by
5 cm (A = s2)


SA cube = 6A = 6s2

SA = 6(5 cm)2

SA = 6(25 cm2)

SA = 150 cm2
Measurement

Cube
5 cm
Volume:
 V = s3 where s is the length of a side
 Example: s = 5 cm


V cube = Bh = s3

V
=
(5 cm)3

V
=
125 cm3
Measurement
7m
5m

Triangular Prism


Surface Area: sum of the areas of all of the
faces
Example: There are 3 lateral faces: 6 m by
7 m (A1= bl). There are 2 bases: 6 m for the
base and 5 m for the height (2A2 = bh).


A1 = (6 m)(7 m) = 42 m2
2A2 = (6 m)(5 m) = 30 m2

SA triangular prism = bh + 3bl

SA = 30 m2 + 3(42 m2)

SA = 30 m2 + 126 m2
SA = 156 m2

6m
Measurement
7m
5m

Triangular Prism
6m
Volume:
 V = ½ bhl where b is the base; h is
height of the triangle; and l is length of
the prism
 Example: b = 6 m; h = 5 m; l = 7 m


V triangular prism = Bh = ½ bhl

V
=
½ (6 m)(5 m)(7 m)

V
=
105 m3
3 ft
Measurement
12 ft

Cylinder
Surface Area: area of the circles plus
the area of the lateral face
 Example: r = 3 ft; h = 12 ft

2rh +2r2

SA cylinder=

SA = 2 (3 ft)(12 ft) + 2 (3 ft)2
SA
 SA
 SA

=
=
=
72 ft2 + 2 (9 ft2)
72 ft2 + 18 ft2
90 ft2
3 ft
Measurement
12 ft

Cylinder
Volume of a Cylinder: V = r2h
where r is the radius of the base
(circle) and h is the height.
 Example: r = 3 ft and h = 12 ft.
 V cylinder =
Bh = r2h
 V
=
(3 ft)2  (12 ft)
 V
=
(9 ft2)(12 ft)
 V
=
108 ft3

13 ft
12 ft
Measurement
5 ft

Cone
Surface Area: area of the circle plus
the area of the lateral face
 Example: r = 5 ft; t = 13 ft

rt +r2

SA cone=

SA = (5 ft)(13 ft) +  (5 ft)2
SA
 SA
 SA

=
=
=
65 ft2 +  (25 ft2)
65 ft2 + 25 ft2
90 ft2
13 ft
12 ft
Measurement
5 ft

Cone
Volume: V = r2h/3 where r is the
radius of the base (circle) and h is the
height.
 Example: r = 5 ft; h = 12 ft
 V cone=
r2h/3
 V
=
[(5 ft)2  12 ft ]/ 3
 V
=
[(25 ft2)(12 ft)]/3
 V
=
(25 ft2)(4 ft)
 V
=
100 ft3

8 mm
Measurement

Sphere
Surface Area: 4r2 where r is the
radius
 Example: r = 8 mm
 SA sphere =
4r2
 SA
=
4(8 mm)2
 SA
=
4(64 mm2)
 SA
=
256 mm2

6 mm
Measurement

Sphere
Volume of a Sphere: V = (4/3) r3
where r is the radius
 Example: r = 6 mm
 V sphere =
4r3/3
 V
=
[4 x (6 mm)3]/3
 V
=
[4 x 216 mm3]/3
 V
=
[864 mm3]/3
 V
=
288 mm3

Measurement
V

Triangular Pyramid

Square Pyramid
Bh
3