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Welcome to MM150 Unit 6 Seminar
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Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Dec 27, 2015

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Page 1: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Welcome to MM150 Unit 6

Seminar

Page 2: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

• Line AB AB

• Ray AB AB

• Line segment AB AB

Page 3: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Plane1. Any three points that do not lie on the

same line determine a plane. (Since 2 points determine a line, a line and a point not on the line determine a unique plane).

2. A line in a plane divides the plane into 3 parts: the line and 2 half-planes.

3. The intersection of 2 planes is a line.

Page 4: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

3 Definitions• Parallel planes – 2 planes that do not

intersect

• Parallel lines – 2 lines IN THE SAME PLANE that do not intersect

• Skew lines – 2 lines NOT IN THE SAME PLANE that do not intersect.

Page 5: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Angle

A

D

F

Vertex

Side

Side

Page 6: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Angle MeasuresAcute Angle 0 degrees < acute < 90 degrees

Right Angle 90 degrees

Obtuse Angle 90 degrees < obtuse < 180 degrees

Straight Angle 180 degrees

Page 7: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

More Angle Definitions

B D H

L M

2 angles in the same plane are adjacent angles if they have a common vertex and a common side, but no common interior points. Example: [ang]BDL and [ang]LDM Non-Example: [ang]LDH and [ang]LDM

2 angles are complementary angles if the sum of their measures is 90 degrees.Example: [ang]BDL and [ang]LDM

2 angles are supplementary angles if the sum of their measures is 180 degrees.Example: [ang]BDL and [ang]LDH

Page 8: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

If the measure of [ang]LDM is 33 degrees, find the measures of

the other 2 angles.

B D H

L M

Given information:[ang]BDH is a straight angle[ang]BDM is a right angle

Page 9: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

If the measure of [ang]LDM is 33 degrees, find the measures of

the other 2 angles.

B D H

L M

Given information:[ang]BDH is a straight angle[ang]BDM is a right angle

[ang]BDM=90

[ang]BDL=90-33=57 deg[ang]MDH=90 deg

Page 10: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

If [ang]ABC and [ang]CBD are complementary and [ang]ABC is 10 degrees less than [ang]CBD, find the measure of both angles.

B A

CD

[ang]ABC + [ang]CBD = 90Let x = [ang]CBDThen x – 10 = [ang]ABC

X + (x – 10) = 902x – 10 = 902x = 100X = 50 [ang]CBD = 50 degreesX – 10 = 40 [ang]ABC = 40 degrees

Page 11: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Polygons# of Sides Name

3 Triangle

4 Quadrilateral

5 Pentagon

6 Hexagon

7 Heptagon

8 Octagon

9 Nonagon

10 Decagon

12 Dodecagon

20 Icosagon

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• The sum of the measures of the interior angles of a n-sided polygon is

• (n - 2)*180 degrees

What is the sum of the measures of the interior angles of a nonagon?

n = 9 (9-2) * 180 = 7 * 180 = 1260 degrees

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Sum of Interior Angles

2 * 180 = 360 degrees

3 * 180 = 540 degrees

4 * 180 = 720 degrees

4 - 2 = 2

5 - 2 = 3

6 - 2 = 4

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EVERYONE: How many sides does a polygon have if thesum of the interior angles is 900 degrees?

Formula:

(n - 2)*180 degrees, where n is number of sides of polygon

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EVERYONE: How many sides does a polygon have if thesum of the interior angles is 900 degrees?

• (n - 2) * 180 = 900

• Divide both sides by 180• n - 2 = 5

• Add 2 to both sides• n = 7 The polygon has 7 sides.

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Similar Figures

A

B

C X

Y

Z

80[deg]

80[deg]

50[deg] 50[deg]50[deg]50[deg]

[ang]A has the same measure as [ang]X[ang]B has the same measure as [ang]Y[ang]C has the same measure as [ang]Z

XY = 4 = 2AB 2

22

1 2

4 4

YZ = 4 = 2BC 2

XZ = 2 = 2AC 1

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Page 238 # 73

• Steve is buying a farm and needs to determine the height of a silo. Steve, who is 6 feet tall, notices that when his shadow is 9 feet long, the shadow of the silo is 105 feet long. How tall is the silo?

6 ft

9 ft

105 feet

?

9 = 6105 ?

9 * ? = 105 * 6

9 * ? = 630

? = 70 feet

The silo is 70 feet tall.

Page 18: Welcome to MM150 Unit 6 Seminar. Line AB AB Ray AB AB Line segment AB AB.

Units in measurementLet’s consider a rectangle with length 5

inches and width 3 inches:

Perimeter = 2l + 2w = 10 in + 6 in = 16 inches

Area = l * w = 5 in * 3 in = 15 in*in = 15 in^2 (or 15 sq. in.)

Rectangular box with height 2 inches:

Volume = l * w * h = 5 in * 3 in * 2 in = 30 in*in*in = 30 in^3 (30 cubic inches)

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Area of a Trapezoid

2 m

3 m

4 m

A = (1/2)h(b1 + b2)

A = (1/2)(2)(3 + 4)

A = (1/2)(2)(7)

A = 1(7)

A = 7 square meters

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Circleradius is in greendiameter is in blue

2r = d Twice the radius is the diameter

CircumferenceC = 2∏r or 2r∏

Since 2r = dC = ∏d

AreaA = ∏r2

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Prisms

Pyramids

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Volume• In 3 dimensions, so general rule is that

volume is base (area) times height (length)• For prisms V=Bh• For pyramids V=(1/3)Bh• Similarly with cylinders and cones• Page 255• Spheres• V = (4/3) pi * r^3• SA = 4 pi * r^2

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Examples

Page 263 #8Rectangular prism (box)

V = Bh

V = (6 sq yd)*(6 yard)

V = 36 cubic yards

Page 263 #14cone

V = (1/3)Bh

V = (1/3)(78.5 sq ft)(24 ft)

V = 628 cubic feet

Page 263 #16sphere

V = (4/3)pi*r^3

V = (4/3)(3.14)(7 mi)^3

V = 1436 cubic miles (approx.)

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Surface Area

• Remember surface area is the sum of the areas of the surfaces of a three-dimensional figure.

• Take your time and calculate the area of each side.

• Look for sides that have the same area to lessen the number of calculations you have to perform.

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Examples of surface area

Page 263 #8

Area of the 2 Bases3 yd * 2 yd = 6 sq yd

Area of 2 sides2 yd * 6 yd = 12 sq yd

Area of other 2 sides3 yd * 6 yd = 18 sq yd

Surface area6 + 6 + 12 + 12 + 18 + 18

= 72 sq yd

Page 263 #14

Surface area of a coneSA = [pi]r2 + [pi]r*sqrt[r2 + h2]

SA = 3.14 * (5)2 + 3.14 * 5 * sqrt[52 + 242]

SA = 3.14 * 25 + 3.14 * 5 * sqrt[25 + 576]

SA = 78.5 + 15.7 sqrt[601]

SA = 78.5 + 24.5

SA = 103 sq ft