Lesson 12.1 Exploring Solids

Lesson 12.1Exploring Solids

Today, we will learn to…> use vocabulary associated with solids> use Euler’s Formula

Polyhedron

~ a solid formed by polygons

Prismsare polyhedra named by their basesname the base

Prisms have 2 parallel bases, Pyramids have 1 base.

Cylinders, Cones, & Spheres

with curved surfaces are

NOT polyhedra

Parts of Solids FacesEdgesVertex

Bases vs Lateral FacesHeight

F + V = E =

+ = 6 8 1412

F + V = E =

+ = 5 5 108

Euler’s Theorem

# Faces + # Vertices = # Edges + 2

F + V = E + 2

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

2. In a polyhedron,F = 5 V = ? E = 8find the number of vertices.

F + V = E + 25 + V = 8 + 25 + V = 10

___ vertices5

3. In a polyhedron,F = ? V = 6 E = 9find the number of faces.

F + V = E + 2F + 6 = 9 + 2F + 6 = 11

___ faces5

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

A polyhedron can be convex or concave.

4. Describe the cross section shown.

square


pentagon


triangle

A polyhedron is regular if all of its faces are congruent

regular polygons.

Platonic Solids

http://matti.usu.edu/nlvm/nav/frames_asid_128_g_3_t_3.html?open=instructions

Lesson 11.1Angle Measures

in Polygons

Today, we are going to…> find angle measures in polygons

# 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

Theorem 11.1Polygon Interior Angles

Theorem

The sum of the measures of the interior angles is

____________(n - 2)180°

The measure of one interior angle in a regular

polygon is…

(n - 2) 180°n

Corollary to Theorem 11.1

1. Find the sum of the measures of the interior angles of a 30-gon.

(30 - 2) 180

(28) 180

5040˚

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

3. What is the measure of each interior angle of a regular

nonagon?

1260˚ 9 =

(9 - 2)180 9

(7)180 9

140

4. Find x.

180(7 - 2)180(5)900˚

How many angles? 7

900° – given angles = x

900 – 783 = 117

Sum?

5. Find x.

180(5 - 2)180(3)540˚

How many angles? 5

sum of all angles = 54043x – 19 = 540

13

Sum?

x =

1 2

34

5

Exploring Exterior Angles

GSP

Theorem 11.2Polygon Exterior Angles

Theorem

The sum of the measures of one set of exterior

angles in any polygon is

_________360°

The measure of one exterior angle in any

regular polygon is

360°n

The number of sides in any regular polygon is…

360°ext. # sides =

One exterior angle and its interior angle are always

________________.supplementary

6. What is the measure of each exterior angle in a regular

decagon?

360˚ 10

36˚

7. How many sides does a regular polygon have if each exterior angle measures 40˚?

360˚ 40° = 9 sides

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˚

9. Find x.

3x + 90 + 70 + 80 + 60 = 360°x = 20

?

?90

70

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.

Lesson 11.2

Perimeter & Areas of Regular Polygons

Today, we are going to…> find the perimeter and area of regular polygons

10

10

10

1. Find the area of this equilateral triangle

5

5 3

60˚10

(5 3 )(10)

2

25 3

≈ 43.3

24

24

24

2. Find the area of this equilateral triangle

12 3

(12 3 )(24)

2

144 3

≈ 249.4

s

s

s

Equilateral Triangles

Area = s

2 3

2

A regular polygon’s area can be covered with isosceles triangles.

side

Area =

Perimeter =

½ side · apothem

side · # sides

· # sides

side

apothem

Area of Regular Polygons

A = ½(side)(apothem)(# sides)

area of each isosceles triangle

number of isosceles triangles

A = ½ sa n

Perimeter of

Regular Polygons

P = (side)(# sides)

P = s n

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)

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)

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

The measure of the angle formed by the apothem and a radius of a polygon is…

360

2(n)

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˚

A =

10 cm

8. Find the area of the polygon.

259.8 cm2

5 3

½ (10)(5 )(6) = 3

tan A =opposite adjacent

cos A =adjacent

hypotenuse

sin A =opposite

hypotenuse

10

5

a30°

360°12

s=a=n=

105 36 A = ½ san = 259.8

12

6

a36°

360°10

s=a=n=

12

5 A = ½ san

8.3

tan 36° = 6a

= 249

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

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

Worksheet

Practice Problems

Lesson 11.3Perimeters and Areas of

Similar Figures

Today, we are going to…> explore the perimeters and areas of

similar figures

1. Find the ratio of their …Sides Perimeters Areas3 : 2

8

6

12

9

1015

3 : 2 9 : 436 : 54 :24 24

68

28

2112

9

2. Find the ratio of their …Sides Perimeters Areas3 : 4 3 : 4 9 : 1660 : 126 :80 224

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

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

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

Lesson 12.7Similar Solids

1. Find the ratio of sides.

2:3

Are they similar solids?

4

82

6

123

3 6 122 4 8

2. Find the ratio of their surface areas.

6

123

4

82

112 : 252 reduces to 4:9

112 units2 252 units2

3. Find the ratio of their volumes.

6

123

4

82

64 : 216 reduces to 8 : 27

64 units3 216 units3

Theorem 12.13

a2 : b2 a3 : b3

Ratio ofsides

Ratio ofSurface Area

Ratio ofVolume

a : b

2 : 3 4 : 9 8 : 27

4.

1 : 16 1 : 64

Scale Factor SA V1 : 42 : 5 4 : 25 8 : 125

9 : 1003 : 10 27 : 1000

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

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

Lesson 12.2, 12.3, 12.6Surface Area

Today, we will learn to…> find the Surface Area of prisms,

cylinders, pyramids, cones, and spheres

Why would we need to find surface area?

Surface Area of a Right Prism

S = 2B + PH

Area of the Base

Perimeter of the Base

Height of the Solid

S = 749 282( ) + ( )( ) =

294cm2

Shape of the base? square

S = 2( ) + ( )( ) = 91615 174m2

Shape of the base?

rectangle

2( )+( )( ) =S = 54 36 3 216 cm2

Shape of the base?triangle

Surface Area of a Cylinder

S = 2B + PH

S = 2π r 2 + 2π r H

area of base

circumference of base

S = 2π ( ) 2 + 2π ( )( )

6 cm

4 cm

4

251.3cm2

Shape of the base?circle

r = 4

4 6

8

10

8

Slant Height

2= H2 + x2

Surface Area of a Regular Pyramid

S = B + ½ P

Area of the Base

Perimeter of the Base

Slant Height of the Solid

10.8

64 32( ) + ½ ( )( ) = 236.8m2

S =

2 = 102 + 42 =

10.8

Surface Area of a Cone

S = B + ½ P

area of base

½ the circumference

of base

S = π r 2 + π r

17 =

2 = 152 + 82

628.3in2

r = 8

S = π ( ) 2 + π ( )( )8 8 17

Surface Area of a Sphere

S = 4 π r2

8 inr =

S=

8

4π(64)

804.2in2

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

Find the surface area.

72 cm22( )+( )( ) =S = 4 8 8


48 in2

x2 = 32 + 42x

x = 5

3 + 4 + 5

2( )+( )( ) =S = 6 12 3


S = 2π( )2 + 2π( )( )

967.61 in2

7 7 15


S = ( ) + ½( )( )

144 in2

4

l l 2 = 32 + 42

l = 5

64 32 5


S =

63.33 m2

( 5/2)2( 3 )

( ) + ½( )( )10.8 15 7

10.8

B?


103.67 in2

S = π ( ) 2 + π ( )( )3 3 8


282.74 cm2

x 2 = 52 + 122

x = 13

S = π ( ) 2 + π ( )( )5 5 13


S = 4π( )2

452.39 cm2

6

Lesson 12.4,12.5,12.6Volume

Today, we will learn to…> find the Volume of prisms,

cylinders, pyramids, cones, and spheres

Volume?

5 times 3 is 15 4 layers of 15

60 cubes

5(3)(4) ?

Why do we need to find volume?

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.

Rectangular Prism

V = L•W•HVolume of a

Triangular Prism

V = ½ b•h•H

Cube V = s3

V = ( )3=

343 cm3

7

V = ( )( )( )=

135 m3

3 5 9

V = ½ ( )( )( )= 162 cm312

9 3

Volume of any Prism

V = B•H

Height of the Solid

Area of the Base

Volume of a Cylinder

V = B•H

V = π r 2 H

area of base

V =

4 cm

6 cm

(π)( 2)( )= 301.6 cm34 6

Volume of

Pyramids and Cones

V = B ▪ H 3

Experiment

V = π r 2 H

3

V =

213.3 m3

( )( ) 364 10

V =

1005.3in3

(π)( 2)( ) 3

8 15

Volume of a sphere

3V = 4π r3

5 mr =

V =

5

523.6 m3

34 π ( )35

Since a hemisphere is ½ of a sphere its volume

is ½ the volume of the sphere.

Find the volume.

V = ( )( )( )

32 cm3

2 2 8

Find the volume.

V = ½ ( )( )( ) =18 in34 3 3

V = (π)( 2)( ) = 2309.07 in3

Find the volume.

7 15

V =( )( ) 3 = 64 in3

Find the volume.

64 3

V =π( 2)( ) 3

69.74 in3

Find the volume.

8 2 = x2 + 32

x =

x3 7.4

7.4

V =

314.16 cm3

Find the volume.

π( 2)( ) 35 12

V = 4(π)( 3) 3

904.78 cm3

Find the volume.

6

V =( )( ) 336 6

V = ( )( )( )6 6 6

288 ft3

+

V =

97.9 m3

π( 2)( ) 32.55 5.1

V = ( )( )( )5.1 5.1 5.1

–

V =

173.4 in3

V =(π)( 2)( ) =2.5 7.5

π( 2)( ) 32.5 4

+

Find the volume of the sand.

V =π( 2)( ) 33.9 3.9

62.12 in3

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°

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

10

5

a 30°

360°12

s=a=n=

105 36

A = ½ san

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°

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

12

6

a30°

360°12

s=a=n=

126 36 A = ½ san = 374

12

6

a36°

360°10

s=a=n=

12

5 A = ½ san

8.3

tan 36° = 6a

= 249

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

Lesson 12.1 Exploring Solids

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