Ch 10 Note Sheets L2 Key V3
Name ___________________________
Note Sheets Chapter 10: Volume
Lesson 10.1 The Geometry of Solids
Study this vocabulary!!
D
Polyhedron A solid formed by polygonal surfaces that enclose a
single region of space.
A
face One of the polygons and its interior forming the surface of a
polyhedron. ABD , DBC , DAC , ABC
edge The intersection of two faces in a polyhedron. AB , AD , DB …
vertex (of a polyhedron) A point of intersection of three or more
edges. Points A, B, C and D.
Solids:
right (prism, cylinder or cone)
Lateral edges

C
B
Right Rectangular
Prism
Right Cone
Axis

base
base
A prism in which the lateral edges are perpendicular to the bases, or
a cylinder or cone in which the axis is perpendicular to the base(s).
oblique (prism, cylinder or cone)
A prism in which the lateral edges are not perpendicular to the bases, or
a cylinder or cone in which the axis is not perpendicular to the base(s).
Oblique Hexagonal
Prism
Oblique
Cylinder
Oblique solids are “slanted”
Prism A polyhedron with two congruent, parallel bases.
Right Pentagonal Prism
base(s) The congruent parallel polygons that the prism is named after.
lateral face A face of a polyhedron other than a base.
On a right prism, the lateral faces are rectangles.
base edge The intersection of a lateral face and the base.
lateral edge The intersection of two lateral faces.
The
lateral
edge is
also the
altitude,
or height
“right”
prism.
Rectangular
Lateral Face
Lateral
Edge
Base Edge
Pentagonal
Base
altitude [of a prism or cylinder] A perpendicular segment from a base to the parallel base or to the
plane containing the parallel base. Height is the length of the altitude.
Oblique Cylinder
Cylinder A solid consisting of two congruent, parallel circles and their
interiors, and the segments having an endpoint on each circle that are
parallel to the segment between the centers of the circles.
Radius
Lateral
Edge
Height
Axis
axis (of a cone or cylinder) The line segment connecting the center of
the base to the vertex or center of the other base.
S. Stirling
Circular Base
Page 1 of 9
Ch 10 Note Sheets L2 Key V3
Name ___________________________
Right Square Pyramid
Pyramid A polyhedron consisting of a polygon base and
Vertex
Triangular
Lateral Face
triangular lateral faces that share a common vertex.
On a right pyramid, the lateral faces are isosceles triangles.
Lateral
Edge
Slant
Height
Height
altitude (of a pyramid or cone) A perpendicular segment from
a vertex to the base or to the plane containing the base.
Height is the length of the altitude.
Base Edge
Oblique Pentagonal Pyramid
Vertex
Lateral Edge
Triangular
Lateral Face
Height
Square
Base
Pentagonal Base Edge
Base
Right Cone
Cone A solid consisting of a circle and its interior, a point
not in the plane of the circle, and all points on line
segments connecting that point to points on the circle.
Slant
Height
Height,
also the
axis.
Circular
Base
Sphere The set of all points in space at a given distance,
radius, from a given point, center.
Radius
Sphere
Center
Radius
center (of a sphere) The point from which all points on the
sphere are the same distance.
Great
Circle
great circle The intersection of a sphere with a plane that
passes through its center.
Hemisphere
Hemisphere Half of a sphere and its great circle base.
Great Circle
Base
Radius
Read pages 520 - 524 in the book for clarification of this vocabulary, if needed.
S. Stirling
Page 2 of 9
Ch 10 Note Sheets L2 Key V3
Name ___________________________
Lesson 10.2 Volume of Prisms and Cylinders
Volume The measure of the amount of space contained in a solid. The volume of an object is the number of unit
cubes that completely fill the space within the object. So, you fill up the 3-d object with cubic units
like in3 or cu. in., m3, etc… Write the units as cm3, or cu. cm, NOT
3
cm!
Read pages 530 – 531 in the book, including the investigation through step 3.
V   2 4  3  24 cm3
V   3 12  8  288 cm3
V  10 30  10  3000 cm3
Read page 532 in the book, through step 4.
Conclusion: The volume of an oblique prism or cylinder is the same as the volume of a
right prism that has the same base area and height.
Complete the formulas below.
Volume of a Right Prism
V= B h
B = base Area
h = height of the prism
Volume of an Oblique
Prism or Cylinder
V= B h
B = base Area
h = height of the prism
S. Stirling
h = the number
of layers
B = the number of
cubes in one layer
h = the number
of layers
B = πr2 gives the cubes in one
layer
Page 3 of 9
Ch 10 Note Sheets L2 Key V3
Read page 533 in the book, if necessary.
Name ___________________________
Complete the Examples below. Notice that a prism (or cylinder) does NOT always “sit” on its base and
the height is NOT always vertical. Pay close attention to how to show your work! Hints: Shade the base
and then mark the height of the figure h. Draw the base flat and find the base area. Now find the volume.
Ex A
Ex B
Find the volume of the
right trapezoidal prism.
Find the volume of the
oblique cylinder. All
All measures are in cm.
measures are in inches.
Base is a trapezoid.
Base is a circle.
1
B   b1  b2  h note: h = height of the trapezoid
2
1
B   4  8  5  30 cm 2
2
B   r2
B    6   36 in 2
2
V  Bh note: h = height of the prism = 7
V  36 7  252 cm3
V  Bh note: h = height of the prism = 10
V  30 10  300 cm3
Or about 791.681
Ex C
Ex D
The solid at right is a right
cylinder with a 135° slice
removed. Find the volume
of the solid. Round your
answer to the nearest cm.
Find the volume
of the solid.
Use Pyth. Thm.
42  22  a 2
12  a 2
Base is a sector. With degree = 360 – 135 = 225
225
5
B
 (8) 2 or B   (8) 2
360
8
2
B  40 cm
V  Bh note: h = height of the prism = 14
V  40 14  560 cm3
V  1759.292 cm3
a  12  3.464
Base is a regular hexagon.
1
B  ap
2
1
B   3.464  6  4 
2
B  41.568 cm 2
60
30
4
60
2
V  Bh note: h = height of the prism = 10
V  41.568 10  415.68 cm3
V  416 cm 3
S. Stirling
4
a
Page 4 of 9
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Ch 10 Note Sheets L2 Key V3
Name ___________________________
Lesson 10.3 Volume of Pyramids and Cones
Read page 538 in the book, include the investigation, which
will be done in class. Complete the conjecture (formula).
Read page 539-340 in the book, if necessary. Pay close
attention to how to show your work!
Volume of a Pyramid or Cone
V=
1
Bh
3
B = base Area
h = height of the pyramid or cone
Ex A
Find the volume of the
a regular hexagonal pyramid.
Base edge = 6 cm and
height = 8 cm
Base is a regular hexagon.
60
30
6
6
a
60
3
3
Use Pyth. Thm.
62  32  a 2
1
B  ap
2
1
B   5.196  6  6 
2
 93.528 cm 2
1
V  Bh note: h = height of the pyramid = 8
3
1
V   93.528  8   249.408 cm3
3
27  a 2
a  27  5.196
Ex B
A cone has a base radius of 3 in and a
volume of 24π in3. Find the height.
1
V  Bh
3
1
2
24    3 h
3
24  3 h divide both sides by 3π
24 3 h

3
3
8h
S. Stirling
Page 5 of 9
Ch 10 Note Sheets L2 Key V3
Name ___________________________
Ex C
Find the volume of this
triangular pyramid.
Base is a
45-45-90
triangle.
45
1
V  Bh note: h = height of the pyramid = 12
3
1
V   25 12   100 cm3
3
10
45
Use Pyth. Thm.
102  x 2  x 2
100  2 x 2
50  x 2
1
B  bh note: h = height of the triangle
2
1
B   7.071 7.071
or
2
3
 24.9995 or 25 cm
x  50  7.071
Ex D
Find the volume of this
half right cone.
25
24
Base is a semi-circle
with r = 7.
1
B   (7) 2
2
1
B   49
2
49

 76.97
2
7
Note: need to find height of
the cone.
Use a 7:24:25 right triangle.
OR
Use Pyth. Thm.
252  7 2  h2
576  h 2
1
V  Bh
3
1  49 
3
V 
  24   196 cm
3 2 
V  615.75 cm3
h  576  24
S. Stirling
Page 6 of 9
Ch 10 Note Sheets L2 Key V3
Name ___________________________
Lesson 10.6 Volume of a Sphere
Read page 558 - 559 in the book. Analyze Example A and Example B.
Volume of a Sphere
4
V=  r 3
3
r = radius
Ex A Volume
Plaster cube 12 cm each side.
Find largest possible sphere.
Radius sphere = 6 cm.
4
V   r3
3
4
3
V    6
3
4
V   216  288 ≈ 905 cm3
3
Ex B Volume
The volume of a hemisphere is 486π. Find the radius.
2
V   r3
3
2
486   r 3
3
3 486
3 2 3



r
2
1
2 3
729  r 3
r  3 729
r 9
S. Stirling
Page 7 of 9
Ch 10 Note Sheets L2 Key V3
Name ___________________________
Lesson 10.7 Surface Area of a Sphere
Read pages 562-563 in the book.
Analyze the Example.
Volume of a Sphere
4 3
r
3
r = radius
V=
Surface Area of a Sphere
SA= 4 r 2
r = radius
Ex A Surface Area
Find surface area of a sphere whose volume is 12,348π m3.
Find Radius:
4
V   r3
3
Surface Area sphere:
4
12348   r 3
3
3
9261  r
SA  4  21
SA  4 r 2
2
2
= 1764  5541.8 m
r  3 9261
r  21
Ex B Surface Area
Find surface area and volume of the cylinder
with the hemisphere taken out of the top.
Volume Hemisphere:
2
V   r3
3
2
3
V    4
3
128
V

3
Volume Cylinder:
V   r 2h
V   (4) 2 (9)
V  144
V  144 
S. Stirling
128
  101.33
3
Surface Area
hemisphere:
SA  2 r 2
SA  2  4  32
2
Surface Area lateral face:
SA  2 r H
SA  2  4   9   72
Area circular base:
A   r 2    4   16
2
Total Surface Area:
SA  32  72 16 = 120π
SA  376.991
Page 8 of 9
Ch 10 Note Sheets L2 Key V3
Name ___________________________
Lesson 10.4 Volume Problems Preparation for your Volume Project.
Read page 547 in the book. Pay close attention to how to show your work as shown in the examples.
Example B
Example A
The volume of this right prism
is 1440 cm3. Find the height
of the prism.
8
h
15
The volume of this sector of
a right cylinder is 2814 m3.
Find the radius of the base of
the cylinder to the nearest m.
r
40
Find area of triangular base.
B
1
815  60
2
V  Bh
1440  60h
24  h
So height of the prism = 24 cm
Base is a sector.
40
 r2
360
1 2
B  r
9
V  Bh
1
2814   r 2 14
9
25326
 r2
14
r  575.8226  23.9963
B
So radius approx. 24 m.
Example C
A swimming pool is in the shape of the prism shown at right. How
many gallons of water can the pool hold? (A cubic foot of water
is about 7.5 gallons.)
The shape is a trapezoidal prism.
1
 b1  b2  h
2
1
B   6  14  30  300 cm 2
2
B
V  Bh note: 16 = height of the prism
V  300 16  4800 cm 3
Since 1 ft3 = 7.5 gallons
4800 ft 3 7.5 gal

 36000 gallons
1
1 ft 3
S. Stirling
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