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Lesson 12.1Exploring Solids
Today, we will learn to…> use vocabulary associated with solids> use Euler’s Formula
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Polyhedron
~ a solid formed by polygons
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Prismsare polyhedra named by their basesname the base
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Prisms have 2 parallel bases, Pyramids have 1 base.
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Cylinders, Cones, & Spheres
with curved surfaces are
NOT polyhedra
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Parts of Solids FacesEdgesVertex
Bases vs Lateral FacesHeight
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F + V = E =
+ = 6 8 1412
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F + V = E =
+ = 5 5 108
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Euler’s Theorem
# Faces + # Vertices = # Edges + 2
F + V = E + 2
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1. In a polyhedron,F = 8 V = 10 E = ?find the number of edges.
F + V = E + 28 + 10 = E + 2
18 = E + 2___ edges16
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2. In a polyhedron,F = 5 V = ? E = 8find the number of vertices.
F + V = E + 25 + V = 8 + 25 + V = 10
___ vertices5
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3. In a polyhedron,F = ? V = 6 E = 9find the number of faces.
F + V = E + 2F + 6 = 9 + 2F + 6 = 11
___ faces5
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A soccer ball is a polyhedron with 32 faces (20 hexagons & 12 pentagons). How many vertices does this polyhedron have?
32 + V = 90 + 2# Edges =
hexagons pentagons20(6) 12(5)
½ (120 + 60) = 90
V = 60 vertices
= 120 = 60
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A polyhedron can be convex or concave.
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4. Describe the cross section shown.
square
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5. Describe the cross section shown.
pentagon
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6. Describe the cross section shown.
triangle
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A polyhedron is regular if all of its faces are congruent
regular polygons.
Platonic Solids
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Lesson 11.1Angle Measures
in Polygons
Today, we are going to…> find angle measures in polygons
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# of sides 3 4 5 6 8 n# triangles
sum of all angles
measure of each angle in regular polygon
180˚
2 3 4 6 n-2
360˚
540˚
720˚
1
1080
˚
(n-2)180˚60
˚
90˚
108˚
120˚
135˚ (n-2)180
n
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Theorem 11.1Polygon Interior Angles
Theorem
The sum of the measures of the interior angles is
____________(n - 2)180°
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The measure of one interior angle in a regular
polygon is…
(n - 2) 180°n
Corollary to Theorem 11.1
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1. Find the sum of the measures of the interior angles of a 30-gon.
(30 - 2) 180
(28) 180
5040˚
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2. How many sides does a polygon have if the sum of the interior angles is 3240˚?
n = 20 sides
180(n-2) = 3240
n – 2 = 18
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3. What is the measure of each interior angle of a regular
nonagon?
1260˚ 9 =
(9 - 2)180 9
(7)180 9
140
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4. Find x.
180(7 - 2)180(5)900˚
How many angles? 7
900° – given angles = x
900 – 783 = 117
Sum?
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5. Find x.
180(5 - 2)180(3)540˚
How many angles? 5
sum of all angles = 54043x – 19 = 540
13
Sum?
x =
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1 2
34
5
Exploring Exterior Angles
GSP
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Theorem 11.2Polygon Exterior Angles
Theorem
The sum of the measures of one set of exterior
angles in any polygon is
_________360°
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The measure of one exterior angle in any
regular polygon is
360°n
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The number of sides in any regular polygon is…
360°ext. # sides =
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One exterior angle and its interior angle are always
________________.supplementary
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6. What is the measure of each exterior angle in a regular
decagon?
360˚ 10
36˚
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7. How many sides does a regular polygon have if each exterior angle measures 40˚?
360˚ 40° = 9 sides
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8. How many sides does a regular polygon have if each
interior angle measures 165.6˚?Don’t write this down, yet.
180(n – 2) = 165.6 n
180 (n-2) n =
165.6 1
180n – 360 = 165.6 n
– 360 = – 14.4 n n =25 sides
180 – 165.6 =
360˚ 14.4
First, find the measure of an exterior angle.
= 25 sides
14.4˚
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9. Find x.
3x + 90 + 70 + 80 + 60 = 360°x = 20
?
?90
70
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What’s the measure of each interior angle of a regular pentagon?
What’s the measure of each interior angle of a regular hexagon?
180(3) 5 = 108°
180(4) 6 = 120°
A soccer ball is made up of 20 hexagons and 12 pentagons.
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Lesson 11.2
Perimeter & Areas of Regular Polygons
Today, we are going to…> find the perimeter and area of regular polygons
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10
10
10
1. Find the area of this equilateral triangle
5
5 3
60˚10
(5 3 )(10)
2
25 3
≈ 43.3
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24
24
24
2. Find the area of this equilateral triangle
12 3
(12 3 )(24)
2
144 3
≈ 249.4
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s
s
s
Equilateral Triangles
Area = s
2 3
2
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A regular polygon’s area can be covered with isosceles triangles.
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side
Area =
Perimeter =
½ side · apothem
side · # sides
· # sides
side
apothem
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Area of Regular Polygons
A = ½(side)(apothem)(# sides)
area of each isosceles triangle
number of isosceles triangles
A = ½ sa n
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Perimeter of
Regular Polygons
P = (side)(# sides)
P = s n
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Find the area of the polygon.
3. a pentagon with an apothem of 0.8 cm and side length of 1.2 cm
A = ½ (1.2) (0.8) (5)
A = 2.4 cm2
A = ½ (s) (a) (n)
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Find the area of the polygon.
4. a polygon with perimeter 120 m and apothem 1.7 m
A = ½ (1.7) (120)
(s)(n)
A = 102 m2
A = ½ (s) (a) (n)
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72° 60° 45°
360° 5
360° 6
360° 8
36° 30° 22.5°360°2(5)
360°2(6)
360°2(8)
5. Find the central angle of the polygon.6. Find the measure of this angle
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The measure of the angle formed by the apothem and a radius of a polygon is…
360
2(n)
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5 cm10 cm
7. Find the length of the apothem if the side length is 10 cm.
apothem = 5 cm3
30°
360° 2(6)
60˚
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A =
10 cm
8. Find the area of the polygon.
259.8 cm2
5 3
½ (10)(5 )(6) = 3
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tan A =opposite adjacent
cos A =adjacent
hypotenuse
sin A =opposite
hypotenuse
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10
5
a30°
360°12
s=a=n=
105 36 A = ½ san = 259.8
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12
6
a36°
360°10
s=a=n=
12
5 A = ½ san
8.3
tan 36° = 6a
= 249
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12
x
a
22.5° 360°16
s=a=n=
9.2
8A = ½ san
sin 22.5° = x12
cos 22.5° = a12
11.1
= 408
12
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A =( )( )=
14. Find the area of the square.
72 units2
3 2 6
x
s = 6 2
3 2
6 2 6 2
45°
a
72 units2A = ½ ( )( )(4)= 6 2 3 2
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Worksheet
Practice Problems
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Lesson 11.3Perimeters and Areas of
Similar Figures
Today, we are going to…> explore the perimeters and areas of
similar figures
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1. Find the ratio of their …Sides Perimeters Areas3 : 2
8
6
12
9
1015
3 : 2 9 : 436 : 54 :24 24
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68
28
2112
9
2. Find the ratio of their …Sides Perimeters Areas3 : 4 3 : 4 9 : 1660 : 126 :80 224
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Theorem 11.5
In similar polygons …
Sides Perimeters Areas 3 : 2 3 : 2 9 : 43 : 4 3 : 4 9 : 16
Do you notice a pattern?
a : b a2 : b2a : b
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9 25
3 5
=PV = 6
3. The ratio of the area of Δ PVQ to the area of Δ RVT is 9:25. If the length of RV is 10 and the two
triangles are similar, find PV.
What is the ratio of their sides?
x 10
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4. The ratio of the sides of two similar polygons is 4 : 7. If the area of the smaller polygon is
36 cm2, find the area of the larger polygon.
4 7
110.25 cm2 16 49
=What is the ratio of their areas?
36 x
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Lesson 12.7Similar Solids
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1. Find the ratio of sides.
2:3
Are they similar solids?
4
82
6
123
3 6 122 4 8
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2. Find the ratio of their surface areas.
6
123
4
82
112 : 252 reduces to 4:9
112 units2 252 units2
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3. Find the ratio of their volumes.
6
123
4
82
64 : 216 reduces to 8 : 27
64 units3 216 units3
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Theorem 12.13
a2 : b2 a3 : b3
Ratio ofsides
Ratio ofSurface Area
Ratio ofVolume
a : b
2 : 3 4 : 9 8 : 27
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4.
1 : 16 1 : 64
Scale Factor SA V1 : 42 : 5 4 : 25 8 : 125
9 : 1003 : 10 27 : 1000
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5. A right cylinder with a surface area of 48π square centimeters and a volume of 45π cubic centimeters is similar to another larger cylinder. Their scale factor is 2:3. Find the surface area and volume of the larger solid.
ratio of surface areas? 4:9
49
48πx
=4x = 9(48)
x = 108π 339.3
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5. A right cylinder with a surface area of 48π square centimeters and a volume of 45π cubic centimeters is similar to another larger cylinder. Their scale factor is 2:3. Find the surface area and volume of the larger solid.
ratio of volumes? 8:27
827
45πx
=8x = 27(45)
x = 151.875π 477.1
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Lesson 12.2, 12.3, 12.6Surface Area
Today, we will learn to…> find the Surface Area of prisms,
cylinders, pyramids, cones, and spheres
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Why would we need to find surface area?
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Surface Area of a Right Prism
S = 2B + PH
Area of the Base
Perimeter of the Base
Height of the Solid
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S = 749 282( ) + ( )( ) =
294cm2
Shape of the base? square
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S = 2( ) + ( )( ) = 91615 174m2
Shape of the base?
rectangle
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2( )+( )( ) =S = 54 36 3 216 cm2
Shape of the base?triangle
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Surface Area of a Cylinder
S = 2B + PH
S = 2π r 2 + 2π r H
area of base
circumference of base
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S = 2π ( ) 2 + 2π ( )( )
6 cm
4 cm
4
251.3cm2
Shape of the base?circle
r = 4
4 6
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8
10
8
Slant Height
2= H2 + x2
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Surface Area of a Regular Pyramid
S = B + ½ P
Area of the Base
Perimeter of the Base
Slant Height of the Solid
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10.8
64 32( ) + ½ ( )( ) = 236.8m2
S =
2 = 102 + 42 =
10.8
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Surface Area of a Cone
S = B + ½ P
area of base
½ the circumference
of base
S = π r 2 + π r
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17 =
2 = 152 + 82
628.3in2
r = 8
S = π ( ) 2 + π ( )( )8 8 17
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Surface Area of a Sphere
S = 4 π r2
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8 inr =
S=
8
4π(64)
804.2in2
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Each layer of this cake is 3 inches high. One can of frosting will cover 130 square inches
of cake, how many cans do we need?
6 in
10 in
S =641.34 in2
B = ½ san
3
360˚123 3
½ (6)(3 3 )(6) + 36 (3)½ (6)(5 3 )(10) + 60 (3)+
5
5
We need 5 cans
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Find the surface area.
72 cm22( )+( )( ) =S = 4 8 8
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Find the surface area.
48 in2
x2 = 32 + 42x
x = 5
3 + 4 + 5
2( )+( )( ) =S = 6 12 3
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Find the surface area.
S = 2π( )2 + 2π( )( )
967.61 in2
7 7 15
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Find the surface area.
S = ( ) + ½( )( )
144 in2
4
l l 2 = 32 + 42
l = 5
64 32 5
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Find the surface area.
S =
63.33 m2
( 5/2)2( 3 )
( ) + ½( )( )10.8 15 7
10.8
B?
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Find the surface area.
103.67 in2
S = π ( ) 2 + π ( )( )3 3 8
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Find the surface area.
282.74 cm2
x 2 = 52 + 122
x = 13
S = π ( ) 2 + π ( )( )5 5 13
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Find the surface area.
S = 4π( )2
452.39 cm2
6
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Lesson 12.4,12.5,12.6Volume
Today, we will learn to…> find the Volume of prisms,
cylinders, pyramids, cones, and spheres
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Volume?
5 times 3 is 15 4 layers of 15
60 cubes
5(3)(4) ?
Why do we need to find volume?
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Theorem 12.6
Cavaleiri’s PrincipleIf 2 solids have the same height and the same cross-sectional area at every level,
then they have the same volume.
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Rectangular Prism
V = L•W•HVolume of a
Triangular Prism
V = ½ b•h•H
Cube V = s3
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V = ( )3=
343 cm3
7
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V = ( )( )( )=
135 m3
3 5 9
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V = ½ ( )( )( )= 162 cm312
9 3
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Volume of any Prism
V = B•H
Height of the Solid
Area of the Base
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Volume of a Cylinder
V = B•H
V = π r 2 H
area of base
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V =
4 cm
6 cm
(π)( 2)( )= 301.6 cm34 6
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Volume of
Pyramids and Cones
V = B ▪ H 3
Experiment
V = π r 2 H
3
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V =
213.3 m3
( )( ) 364 10
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V =
1005.3in3
(π)( 2)( ) 3
8 15
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Volume of a sphere
3V = 4π r3
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5 mr =
V =
5
523.6 m3
34 π ( )35
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Since a hemisphere is ½ of a sphere its volume
is ½ the volume of the sphere.
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Find the volume.
V = ( )( )( )
32 cm3
2 2 8
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Find the volume.
V = ½ ( )( )( ) =18 in34 3 3
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V = (π)( 2)( ) = 2309.07 in3
Find the volume.
7 15
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V =( )( ) 3 = 64 in3
Find the volume.
64 3
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V =π( 2)( ) 3
69.74 in3
Find the volume.
8 2 = x2 + 32
x =
x3 7.4
7.4
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V =
314.16 cm3
Find the volume.
π( 2)( ) 35 12
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V = 4(π)( 3) 3
904.78 cm3
Find the volume.
6
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V =( )( ) 336 6
V = ( )( )( )6 6 6
288 ft3
+
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V =
97.9 m3
π( 2)( ) 32.55 5.1
V = ( )( )( )5.1 5.1 5.1
–
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V =
173.4 in3
V =(π)( 2)( ) =2.5 7.5
π( 2)( ) 32.5 4
+
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Find the volume of the sand.
V =π( 2)( ) 33.9 3.9
62.12 in3
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1. Figure ABCDE has interior angle measures of 110˚, 90˚, 125˚, 130˚, and x˚. Find x.2. A regular polygon has 13 sides. Find the sum of the measures of the interior angles.3. Find the measures of one interior angle of a regular 22-gon.4. What is the sum of the measures of one set of exterior angles of a
25-gon?
85°
1980°
163.64°
360°
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5. What is the measure of each exterior angle of a regular octagon?
6. The measure of each exterior angle of a regular polygon is 36˚. How many
sides does the polygon have?
7. Find the number of sides in a regular polygon if its interior angles are each 162°.
45°
10 sides
20 sides
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10
5
a 30°
360°12
s=a=n=
105 36
A = ½ san
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6. A polygon has interior angle measures of 120˚, 80˚, 135˚, 120˚, 100˚, and x˚. Find x.7. A regular polygon has 15 sides. Find the sum of the measures of the interior angles.8. Find the measure of one interior angle of a regular 24-gon.9. What is the sum of the measures of one set of exterior angles of a
50-gon?
165°
2340°
165°
360°
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10. What is the measure of each exterior angle of a regular 40-gon?
11. The measure of each exterior angle of a regular polygon is 7.2˚. How many
sides does the polygon have?
12. Find the number of sides in a regular polygon if its interior angles are each 174°.
9°
50 sides
60 sides
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12
6
a30°
360°12
s=a=n=
126 36 A = ½ san = 374
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12
6
a36°
360°10
s=a=n=
12
5 A = ½ san
8.3
tan 36° = 6a
= 249
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12
x
a
22.5° 360°16
s=a=n=
9.2
8A = ½ san
sin 22.5° = x12
cos 22.5° = a12
11.1
= 408.5
12
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