Ch 10 Note Sheet L2 Key
Name ___________________________
Note Sheets Chapter 10: Volume
Lesson 10.1 The Geometry of Solids
Study this vocabulary!!
Polyhedron A solid formed by polygonal surfaces that enclose a
single region of space.
D
A
face One of the polygons and its interior forming the surface of a
polyhedron. ΔABD , ΔDBC , ΔDAC , ΔABC
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.
B
Right Rectangular
Prism
Solids:
Lateral edges ⊥ base
right (prism, cylinder or cone)
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).
Axis
⊥
Oblique Hexagonal
Prism
Oblique
Cylinder
Oblique solids are “slanted”
Right Pentagonal Prism
base(s) The congruent parallel polygons that the prism is named after.
base edge The intersection of a lateral face and the base.
lateral edge The intersection of two lateral faces.
Right Cone
base
Prism A polyhedron with two congruent, parallel bases.
lateral face A face of a polyhedron other than a base.
On a right prism, the lateral faces are rectangles.
C
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 8
Ch 10 Note Sheet L2 Key
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.
center (of a sphere) The point from which all points on the
sphere are the same distance.
great circle The intersection of a sphere with a plane that
passes through its center.
Radius
Sphere
Center
Radius
Great
Circle
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.
S. Stirling
Page 2 of 8
Ch 10 Note Sheet L2 Key
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 = ( 2i4 )i3 = 24 cm3
V = ( 3i12 )i8 = 288 cm3
V = (10i30 )i10 = 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 8
Ch 10 Note Sheet L2 Key
Name ___________________________
Read page 533 in the book, if necessary.
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
Find the volume of the
right trapezoidal prism.
Ex B
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)i5 = 30 cm2
2
B = π r2
2
B = π ( 6 ) = 36π in 2
V = Bh note: h = height of the prism = 7
V = 36π i7 = 252π cm3
V = Bh note: h = height of the prism = 10
V = 30i10 = 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
π (8)2 or B = π (8)2
B=
360
8
2
B = 40π cm
V = Bh note: h = height of the prism = 14
V = 40π i14 = 560π 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 cm2
60
30
4
60
2
V = Bh note: h = height of the prism = 10
V = 41.568i10 = 415.68 cm3
V ≈ 416 cm3
S. Stirling
4
a
Page 4 of 8
2
Ch 10 Note Sheet L2 Key
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 haxagon.
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 cm2
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
24π = π ( 3) ih
3
24π = 3π h divide both sides by 3π
24π 3π h
=
3π
3π
8=h
S. Stirling
Page 5 of 8
Ch 10 Note Sheet L2 Key
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 = π (7)2
2
1
B = π 49
2
49π
=
≈ 76.97
2
7
Note: need to find height of
the cone.
Use a 7:24:25 right triangle.
OR
Use Pyth. Thm.
252 = 72 + h2
576 = h2
1
V = Bh
3
1 ⎛ 49π ⎞
3
V= ⎜
⎟ ( 24 ) = 196π cm
3⎝ 2 ⎠
V ≈ 615.75 cm3
h = 576 = 24
S. Stirling
Page 6 of 8
Ch 10 Note Sheet L2 Key
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= π r 3
3
r = radius
Ex A Volume
Plaster cube 12 cm each side.
Find largest possible sphere.
Percentage cut away?
Volume cube:
V = (12 ) = 1728 cm3
3
Radius sphere = 6 cm.
4
V = π r3
3
4
3
V = π ( 6)
3
4
V = π i216 = 288π ≈ 905 cm3
3
Volume cube – volume sphere =
amount cut away
V = 1728 − 905 = 823
Percentage
823
≈ 48%
1728
Ex B Volume
Find volume of the plastic piece.
The outer-hemisphere diameter
is 5.0 in. and the inner-hemisphere
diameter is 4.0 in.
Volume Outer – Volume Inner.
2
3
Vouter = π ( 2.5)
3
2
Vouter = π i15.625
3
Vouter ≈ 32.7
2
3
Vinner = π ( 2 )
3
2
Vinner = π i8
3
Vinner ≈ 16.8
So V ≈ 32.7 − 16.8 ≈ 15.9 in 3
S. Stirling
Page 7 of 8
Ch 10 Note Sheet L2 Key
Name ___________________________
Lesson 10.7 Surface Area of a Sphere
Volume of a Sphere
Read pages 562-563 in the book. Analyze
the Example.
4 3
πr
3
r = radius
V=
Surface Area of a Sphere
SA= 4π r 2
r = radius
Ex A Surface Area
Find surface area of a sphere whose volume is 12,348π m3.
Find Radius:
4
V = π r3
3
4
12348π = π r 3
3
3
9261 = r
r = 21
Surface Area sphere:
SA = 4π r 2
SA = 4π ( 21)
2
= 1764π ≈ 5541.8 m 2
Ex B Surface Area
Find surface area and volume of the cylinder
with the hemisphere taken out of the top.
Volume Hemisphere:
2
V = π r3
3
2
3
V = π ( 4)
3
128
V=
π
3
Volume Cylinder:
V = π r 2h
V = π (4)2 (9)
V = 144π
128
V = 144π −
π ≈ 101.33π
3
S. Stirling
Surface Area
hemisphere:
SA = 2π r 2
SA = 2π ( 4 ) = 32π
2
Surface Area lateral face:
SA = 2π r i H
SA = 2π ( 4 )i( 9 ) = 72π
Area circular base:
A = π r 2 = π ( 4 ) = 16π
2
Total Surface Area:
SA = 32π + 72π + 16π = 120π
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