Bell Ringer
2/19/15
13.2 ft
8.7 ft
Find the Area: ______
Volume and Surface Area
(Three Dimensional)
Mr. Turner’s 7th Grade
Math Class
Recap on Unit 7
The test from Unit 7 included the following:
• Perimeter = Sum of all sides; 2(L+W) or 2L+2W or
L+L+W+W
• Circumference = πd or 2πr
• Area of a Rectangle = b x h
• Area of a Triangle =
• Area of a Circle = π r²
1
bh
2
or
𝑏ℎ
2
Unit 7
vs.
Unit 8
1. Volume of a Triangular Prism
V = Bh
1
1. Area of a Rectangle
or Square
A = bh
V = 2bh x h
2. Volume of a Rectangular Prism
V = Bh
V = bh x h
3. Volume of a Triangular Pyramid
V=
V=
2. Area of a Triangle
A=
1
bh
2
1
Bh
3
1 1
( bh)
3 2
xh
4. Volume of a Rectangular Pyramid
1
3
1
3
V = Bh
V = (bh) x h
What is
?
Volume is the amount of space occupied by a threedimensional object; and it’s measured in cubic units.
Ex. cm³, ft³, in³
Face, Edge, and Vertex
• Face: A face is any of the individual surfaces of a
solid object (Think Two-Dimensional).
• Edge: An edge is a line segment that joins two
vertices.
• Vertices: A vertex (plural: vertices) is a point
where two or more straight lines (Edges) meet.
Cube
Faces of a Cube
Faces are also known as Surfaces
• A cube has 6 faces.
• All the faces of a cube are
squares.
Face
Vertices of a Cube
Vertices are also known as Corners
vertex
• A cube has 8 vertices.
Edges of a Cube
Edges are also known as Straight Lines
Edge
• A cube has 12 edges.
Cuboid
How many faces
does a cuboid
have?
Faces of a Cuboid
Faces are also known as Surfaces
• A cuboid has 6 faces.
• A cuboid can have 6 faces
which are rectangles.
• A cuboid can have 4 faces
which are rectangles and 2
faces which are squares.
How many vertices
does a cuboid
have?
Vertices of a Cuboid
Vertices are also known as Corners
• A cuboid has 8 vertices.
How many edges
does a cuboid
have?
Edges of a Cuboid
Edges are also known as Straight Lines
• A cuboid has 12 edges.
A Cube has 6 square faces.
The length of each square face is equal.
height2
height1
base
Volume of CUBE = Bh
= b X h1 X h2
= units³
V = Bh
Where B = Area
A rectangular prism that has 6
congruent faces is called a cube.
A rectangular prism that has 6
congruent faces is called a cube.
A rectangular prism that has 6
congruent faces is called a cube.
A rectangular prism that has 6
congruent faces is called a cube.
A rectangular prism that has 6
congruent faces is called a cube.
A rectangular prism that has 6
congruent faces is called a cube.
The solid shown below is made up of twenty-seven 1-cm
cubes. Find the volume of the solid.
V = Bh
Where B = Area
Find the Volume: bh x h
3 cm
3 cm
3 cm
Volume of cube = 3 x 3 x 3 cm³
= 27 cm³
The volume of the cube is 27 cm³
Example 1
Find the volume of the cube.
5 cm
V = Bh
5 cm
Where B = Area
5 cm
Volume of cube = b x h x h
=5 x 5 x 5
= 125cm³
Example 2
10 cm
The box shown on the
left is a present given
to class the Spirit
Squad for winning the
Cheer Competition. Find
the volume of the
V
=
present if it is a cube.
Volume of present = b x h x h
= 10 x 10 x 10
= 1000 cm³
Bh
CUBOID (Rectangular Prism)
A cuboid also has 6 faces but
NOT all the faces are equal
height2
base
height1
V = Bh
Where B = Area
Volume of CUBOID = base X height1 X height2
= b X h
X h
Example 3
Find the volume of the rectangular prism shown
below.
4cm
V = Bh
4cm
10cm
Volume = b x h x h
= 10 x 4 x 4
= 160 cm³
Example 4
V = Bh
12.5 m
5m
7m
Volume of a Rectangular
Prism
=b x h x h
= 7 x 5 x 12.5
= 437.5 m³
Practice
Try these questions yourself. Make
sure you solve each question in your
journal.
Question 1
Find the volume of the
following solid.
V = Bh
Volume of cube
= b x h x
h
=
2
2cm
2cm
x
2
2
2cm
=
8
cm³
x
Question 2
V = Bh
Volume of cube
= b x h x h
=
6
= 216
x
6
x
Find the volume
of the cube
shown below.
6
cm³
6 cm
Question 3
V = Bh
Volume of a = 4
Rectangular Prism
x
3
=
x 10.5
126
10.5m
m³
3m
4m
Question 4
6.5 cm
V = Bh
11cm
7.5 cm
Volume of = 11
x 7.5
a Rectangular Prism
= 536.25 cm³
x
6.5
Bell Ringer
2/20/15
Area of a Rectangle =
Volume of a Rectangle =
A = bh
V = Bh
5 cm
5 cm
5 cm
12 cm
A = 12 x 5
A = 60 cm²
12 cm
V = 12 x 5 x 5
V = 300 cm³
Area of Triangles
What is the formula?
Area =
1
b
2
x h or
Area =
h
b
𝑏ℎ
2
Area of Triangles
Example: Find the area of the triangle
1
A=
bh
2
8 in
1
A = 8(7)
2
7 in
A = 28 in²
Area of Triangles
Example: Find the area of the triangle
1
A=
bh
2
11 in
1
A = 11(3)
2
3 in
A = 16.5 in²
Unit 7
Area of a Triangle =
𝑏ℎ
2
or
1
bh
2
Unit 8
Volume of a Triangular Prism =
V = Bh (STAAR Reference Sheet)
or
V=
𝑏ℎ
xh
2
or
V=
1
bh x h
2
Volume of a Triangular Prism
Volume of a
h
b
To find the volume of the
triangular prism, we must
first find the area of the
triangular base (shaded in
yellow).
Volume of a
• To find the area of the Base…
Area (triangle) = b x h
2
h
b
This gives us the Area of
the Base (B).
Volume of a
• Now to find the volume…
B
h
We must then multiply
the area of the base (B) by
the height (h) of the
prism.
This will give us the
Volume of the Prism.
Volume of a
Volume
(triangular prism)
B
h
V =
B x h
Volume of a
Volume =
V = B x h
V = (8 x 4) x 12
2
V = 16
x 12
V =
192 cm3
Practice
Try these questions yourself. Make
sure you solve each question in your
journal.
Volume of a
Volume =
V = B x h
V = (6 x 4) x 12
2
V = 12
x 12
V =
144 cm3
Volume of a
Volume =
V = B x h
V = (7 x 3)
2
V = 10.5
V =
x 10
x 10
105 m3
Volume of a
Volume =
V = B x h
V = (8 x 10) x 60
2
V = 40
x 60
V =
2,400 cm3
Volume of a
Volume =
V = B x h
9.5 in
17 in
67 in
V = (17 x 9.5) x 67
2
V = 80.75 x 67
V = 5,410.25 in3
Bell Ringer
Area of a Triangle:
2/26/15
Volume of a Triangular Prism:
Round to the nearest hundredth.
A=
𝑏ℎ
2
or
1
bh
2
V = Bh
3 ft
8 in
3.5 ft
3.5 ft
12.8 in
A = 12.8 x 8 ÷ 2
A = 51.2 in²
V=
3.5 x 3.5
2
V = 300 cm³
x2
Vs.
Prisms Have:
• 2 Congruent Bases
• Rectangular Side Faces
Pyramids Have:
• 1 Base
• Triangular Side Faces That
Meet at One Vertex
Volume of a
Pyramid
Base
Faces:
5 total faces
• The 4 Side Faces are Triangles
• The Base is a Square
It has 5 Vertices (corner points)
It has 8 Edges
Net
1
3
Volume = Bh
B = Base area
B = b xh
Height
1
3
Volume = BH
1
3
V = (8 x 7) x 6
1
3
V = (56) x 6
1
3
V = (336)
V=
336
3
= 112in³
1
3
Volume = BH
1
3
V = (24 x 8) x 10
1
3
V = (192) x 10
1
3
V = (1,920)
V=
1,920
3
= 640in³
1
3
Volume = Bh
1
3
V = (10 x 8) x 6
1
3
V = (80) x 6
1
3
V = (480)
V=
480
3
= 160in³
Practice
Try these questions yourself. Make
sure you solve each question in your
journal.
1
3
Volume = Bh
1
3
V = (6 x 6) x 15
1
3
V = (36) x 15
1
3
V = (540)
V=
540
3
= 180m³
1
3
Volume = Bh
1
3
V = (8 x 8) x 9
1
3
V = (64) x 9
1
3
V = (576)
V=
576
3
= 192cm³
Bell Ringer
2/27/15
Volume of a Rectangular Pyramid:
1
V = Bh
3
1
B = bh so V = bhh
3
6.8 ft
4.5 ft
4.5 ft
1
V = 4.5 x 4.5 x 6.8
3
V = 45.9 ft³
Pyramid
Base
Faces:
4 total faces
• The 3 Side Faces are Triangles
• The Base is also a Triangle
It has 4 Vertices (corner points)
It has 6 Edges
Net
1
Volume = 3 Bh
B = Base area
1
𝑏ℎ
B = bh or
2
2
height
1
Volume = Bh
3
1 8𝑥6
V= (
) x 12
3
2
1
3
V = (24) x 12
1
3
V = (288)
V=
288
3
= 96cm³
1
Volume = Bh
3
1 8 𝑥 12
V= (
) x 16
3
2
1
3
V = (48) x 17
1
3
V = (816)
V=
816
3
= 272cm³
1
Volume = Bh
3
1 14 𝑥 8
V= (
) x 18
3
2
1
3
V = (56) x 18
1
3
V = (1,008)
V=
1,008
3
= 336cm³
18
Practice
Try these questions yourself. Make
sure you solve each question in your
journal.
1
Volume = Bh
3
1 4 𝑥 10
V= (
) x 12
3
2
1
3
V = (20) x 12
4
1
3
V = (240)
V=
240
3
= 80cm³
10
1
Volume = Bh
3
1 8.5 𝑥 16.5
V= (
)
3
2
1
3
V = (70.125) x 12
1
3
V = (841.5)
V=
841.5
3
m
x 12
= 280.5 m³
8.5 m
16.5 m
1
Volume = Bh
3
1 16 𝑥 8
V= (
) x 24
3
2
1
3
24.6 cm
V = (64) x 24
16 cm
1
3
V = (1,536)
V=
1,536
3
= 512cm³
8 cm
Bell Ringer
3/2/15
Volume of a Triangular Pyramid:
1
V = Bh
3
1 1
B = bh so V = ( bh) x h
3 2
6.8 ft
4.5 ft
4.5 ft
1
4.5 x 4.5
3
2
V= x
x 6.8
V = 4.5 x 4.5 x 6.8 x 0.5 ÷ 3
V = 22.95 ft³
What is
surface area?
The number of square units needed to cover all of the
surfaces (bases and lateral faces)
In order to find the
total surface area all
you have to do is find
the area of each
shape and add them
together.
base
Lat. Face
Lat
F.
Base
Lat. Face
Lat. Face
Base
Lat. Face
Lat. Face
The answer is always
in square units.
Ex. units²
Remember: In order to find the total surface area all you have to
do is find the area of each shape and add them together. Find the
total surface area.
9 yd
6.4 yd
4.6 yd
6.4 yd
Base
TSA = 2(9 x 6.4) + 2(4.6 x 6.4) + 2(9 x 4.6)
TSA = 115.2 + 58.88 + 82.8
TSA = 256.88 yd²
Base
Remember: In order to find the total surface area all you have to
do is find the area of each shape and add them together. Find the
total surface area.
TSA =
7.6 𝑥 5.2
2
+
6.4 𝑥 5.2
3( 2 )
TSA = 19.76 + 49.92
7.6 in.
TSA = 19.76 + 3(16.64)
TSA = 69.68 in²
5.2 in.
Find the
for the following shapes.
9 yd
9 in.
6 yd
6 in.
5 yd
5 yd
Base
Base
T.S.A. = ___________
T.S.A. = ___________
What is
surface area?
The number of square units needed to cover the lateral view
(area excluding the base(s) of a three-dimensional figure)
Find the triangular pyramids lateral surface area.
Remember: The Lateral Surface Area is
everything except the base(s).
LSA =
9.3 𝑥 7
2(
)
2
+
9.3 𝑥 4
2(
)
2
LSA = 65.1 + 37.2
LSA = 102.3 ft²
4 ft
7 ft
9.3 ft
Find the rectangular pyramids lateral surface area. (Paint the 4 walls in
your bedroom, but not the floor or the ceiling)
7 yd
LSA = 2(4.8 x 7) + 2(3.6 x 4.8)
LSA = 2(33.6) + 2(17.28)
4.8 yd
4.8 yd
LSA = 67.2 + 34.56
3.6 yd
Floor
LSA = 101.76 yd²
Ceiling
Find the
for the following shapes.
10 yd
6 ft
10 ft
8 yd
15 ft
6 yd
6 yd
Base
Base
L.S.A. = ___________
L.S.A. = ___________