Ch 10 Note sheet L1 Key
Name ___________________________
Lesson 10.1 The Geometry of Solids
Study this vocabulary!!
Polyhedron A solid formed by polygonal surfaces that enclose a
single region of space.
tetrahedron A polyhedron with four faces.
face One of the polygons and its interior forming the surface of a
polyhedron. ABD , DBC , DAC , ABC
D
A
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 sheet L1 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 prism, the lateral faces are isosceles triangles.
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.
Note: slant heights are on the
surface of pyramids (and
cones), so they are used in
calculating surface area, not
volume.
Lateral
Edge
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
Radius
Sphere
Sphere The set of all points in space at a given distance,
Center
radius, from a given point, 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 of your book.
S. Stirling
Page 2 of 9
Ch 10 Note sheet L1 Key
10.1 Page 524 Exercise #2-9
Name ___________________________
Prism
2. Bases: ∆PQR, ∆TUS
3. Lateral Faces: Parallelogram TUQP, TSRP, USRQ
4. Lateral Edges: QU , RS , TP
5. Height: 6
Pyramid
6. Base: GYPTAN
7. Vertex: point E
8. Lateral Edges: GE , YE , PE , TE , AE , NE
9. Height: 13
Lesson 10.2 Volume of Prisms and Cylinders
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. Complete the conjectures.
Volume of an Oblique
Prism or Cylinder
V= B h
Volume of a Right Prism
V= B h
h
B = base Area
h = height of the prism
S. Stirling
h
B = base Area
h = height of the prism
Page 3 of 9
Ch 10 Note sheet L1 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! If you get confused, draw
the base flat.
Example A
Example B
4
Find the volume of the
right trapezoidal prism.
7
All measures are in cm.
5
9
measures are in inches.
B
6
6
Find the volume of the
oblique cylinder. All
7h
10 h
B
8
Base is a circle.
Base is a trapezoid!!
1
 b1  b2  h note: h = height of the trapezoid
2
1
B   4  8 5  30 cm2
2
B
V  Bh note: h = height of the prism
V  30 10  300 cm3
Example C
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.
h
Base is a sector. With degree = 360 – 135 = 225
225
5
 (8)2 or B   (8)2
360
8
2
B  40 cm
V  Bh note: h = height of the prism
V  40 14  560 cm3
V  1759 cm3
V  Bh note: h = height of the prism
V  36 7  252 cm3
Or about 791.68
B
B
B   r2
2
B    6   36 in 2
Example D
Find the volume
of the solid.
B
Use 30-60-90
triangle for
apothem.
h
long  sh 3
a2 3
Base is a regular hexagon.
1
B  ap
2
1
B  (2 3)(4 6)
2
B  24 3 cm2
60
30
4
60
2
V  Bh note: h = height of the prism
V  24 3 10  240 3 cm3
V  416 cm3
S. Stirling
4
a
Page 4 of 9
2
Ch 10 Note sheet L1 Key
Lesson 10.3 Volume of Pyramids and Cones
Name ___________________________
Read page 538 in the book, include the investigation.
Complete the conjecture.
Read page 539-340 in the book, if necessary. Pay close
attention to how to show your work as shown.
Example A
Find the volume of the
a regular hexagonal pyramid.
Base edge = 6 cm and
height = 8 cm
6 a
3
Base is a regular hexagon.
1
B  ap
2
1
B  3 3  6 6
2
 54 3 cm 2
 
Use 30-60-90
triangle for
apothem.
long  sh 3
a3 3
1
V  Bh note: h = height of the pyramid
3
1
V  54 3 8  144 3 cm3
3

B = base Area
h = height of the pyramid
or cone
A cone has a base radius of 3 in and a volume of
24π in3. Find the height.
6
60
3
V= 1/3 B h
Example B
60
30
Volume of a Pyramid or
Cone
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

Example C
Example D
Find the volume of this
triangular pyramid.
Find the volume of this
half right cone.
Base is a 45-45-90 triangle.
h
10
10  leg 2 so leg 
2
1
B  bh note: h = height of the triangle
2
1  10  10  100
45
10
B 

 25


2  2  2  4
B
2
Base is a semi-circle.
1
V  Bh note: h = height of the pyramid
3
1
V   2512   100 cm3
3
S. Stirling
B
1
B   (7) 2
2
1
49
B   49 
2
2
h
25
24
10
45
10
2
h
note: need to find height of
the cone
1
V  Bh
3
1  49 
3
V 
  24   196 cm
3 2 
7
Use a 7:24:25
right triangle.
Page 5 of 9
Ch 10 Note sheet L1 Key
Name ___________________________
Lesson 10.4 Volume Problems
Intro:
How many feet in a yard?
3 feet  1 yard
How many square feet in a square yard?
 3 feet 
So, how many cubic feet in a cubic yard?
3
3
 3 feet   1 yard  so
So, how many cubic inches in a cubic foot?
3
3
12 in   1 ft  so 1728 in3  1 ft3
2
 1 yard  so 9 ft 2  1 yd 2
2
27 ft 3  1 yd3
Review Conversions:
Water weighs 63 pounds per cubic foot.
A cubic foot = 7.5 gallons.
If a can has a diameter of 16 inches and a height of 20 inches, how many gallons can it
hold?
Since 1 ft 3  7.5 gallons , need ft3
V   (8)2 20  1280 in 3  4021.24
3
3
12 in   1 ft  so 1728 in3  1 ft3
1280 in 3 1 ft 3 7.5 gal
 17.45 gal
1
1728 in 3 1 ft 3
Boxes are to be shipped to a drought ridden community. The boxes are 1 foot by 8 inches
by 1.25 feet. How would you figure out the weight of one box?
Since 1 ft 3  63 pounds , need ft3
8 in 1 ft
2
 ft
1 12 in 3
2 5 5 3
V 1
 ft
3 4 6
Convert
5 ft 3 63 pounds
 52.5 pounds
6
1 ft 3
Continued…
S. Stirling
Page 6 of 9
Ch 10 Note sheet L1 Key
Name ___________________________
Read page 547 in the book. Pay close attention to how to show your work as shown in the examples.
Example A
17
The volume of this right prism
is 1440 cm3. Find the height
h
of the prism.
B
8
Example B
15
h
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
B
14 h
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.
B
40
1
 r 2 so B   r 2
360
9
V  Bh
1
2814   r 2 14
9
9
14 2 9
2814 
r
14
9
14
25326
 r2
14
r  575.8226  23.9963
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.)
30
The shape is a trapezoidal prism.
B
1
 b1  b2  h
2
1
B   6  14  30  300 ft 2
2
B
V  Bh note: 16 = height of the prism
V  300 16  4800 ft 3
Since 1 ft3 = 7.5 gallons
4800 ft 3 7.5 gal

 36000 gallons
1
1 ft 3
S. Stirling
Page 7 of 9
14
h
Ch 10 Note sheet L1 Key
Name ___________________________
Lesson 10.6 & 10.7 Volume & Surface Area of a Sphere
Read page 558 - 559 in the book.
Volume of a Sphere
Surface Area of a Sphere
Analyze Example A and Example B.
4 3
V=  r
Read pages 562-563 in the book.
SA= 4 πr2
3
Analyze the Example.
r = radius
r = radius
Example A Volume
Plaster cube 12 cm each side.
Find largest possible sphere.
Percentage cut away?
12
6
Volume cube:
V  12   1728 cm3
3
Radius sphere = 6 cm.
12
12
Volume cube – volume sphere =
amount cut away
4
V   r3
3
4
3
V    6
3
4
V   216  288 ≈ 905 cm3
3
Example 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.
V  1728  905  823
Percentage
5.0
4.0
Volume Outer – Volume Inner.
823
 48%
1728
2
3
Vouter    2.5
3
2
Vouter   15.625
3
Vouter  32.7
2
3
Vinner    2 
3
2
Vinner   8
3
Vinner  16.8
So V  32.7  16.8  15.9 in3
Example 1 Surface Area
Find surface area of a sphere whose volume is 12,348π m3.
Find Radius:
4
V   r3
3
4
12348   r 3
3
3
9261  r
r  21
S. Stirling
Surface Area sphere:
SA  4 r 2
2
SA  4  21
2
= 1764  5541.8 m
Page 8 of 9
Ch 10 Note sheet L1 Key
Name ___________________________
Example 2 Surface Area
4
Find surface area and volume of the cylinder
with the hemisphere taken out of the top.
4
h
Volume Hemisphere:
Surface Area
hemisphere:
2
V   r3
3
2
3
V    4
3
128
V

3
SA  2 r 2
2
SA  2  4   32
Surface Area lateral face:
SA  2 r H
SA  2  4   9   72
Volume Cylinder:
V   r 2h
V   (4)2 (9)
V  144
Area circular base:
A   r 2    4   16
2
Total Surface Area:
SA  32  72  16 = 120π
128
V  144 
  101.33
3
Example 3 Volume
A giant scoop in the shape of a hemisphere of radius 5 cm is used to transfer liquid from a large vat into a storage
tank. The cylindrical storage tank has a diameter of 12 inches and a height of 15 inches. How high will the liquid
rise if 9 scoops are poured into it?
12
Volume Hemisphere:
2
V   r3
3
2
3
V    5
3
125
V

3
Volume Cylinder:
V   r 2h
375    6  h
15
2
375 36  h

36
36
h  10.4167
h
9 Scoops:
V  9
S. Stirling
125
  375
3
Page 9 of 9