Copyright 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with...

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 3 Measuring Angles To measure angles more precisely, each degree is subdivided into 60 minutes of arc; and each minute is subdivided into 60 seconds of arc.

Transcript of Copyright 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with...

Page 1: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1

Modeling with

Geometry10

Page 2: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 2

Unit 10B

Problem Solving with Geometry

Page 3: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 3

Measuring Angles

To measure angles more precisely, each degree is subdivided into 60 minutes of arc; and each minute is subdivided into 60 seconds of

arc.

Page 4: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 4

Examplea. Convert 3.6° into degrees, minutes, and seconds of arc.

b. Convert 30°33ʹ31ʺ into decimal form.

Solutiona. Because 3.6° = 3° + 0.6°, we first convert 0.6° into minutes of arc.

So 3.6° = 3°36ʹ.

60'0.6 36'1

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 5

Example (cont)

b. Convert 30°33ʹ31ʺ into decimal form.

SolutionRemember that 1 minutes of arc is 1/60°, and 1 second of an arc is (1/60) of a minute of arc, or

30°33ʹ31ʺ = 30° + 33ʹ + 31ʺ =

1 .60 60

33 3130

60 60 60

30 0.55 0.00861 30.55861

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 6

Latitude and Longitude

We can locate any place on the Earth’s surface by its latitude (north-south position from equator) and longitude (east-west position from prime meridian).

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 7

ExampleAnswer each of the following questions, and explain your answers clearly.a. Suppose you could drill from Miami straight through the center of Earth and continue in a straight line to the other side of Earth. At what latitude and longitude would you emerge?b. Perhaps you’ve heard that if you dug a straight hole from the United States through the center of Earth, you’d come out in China. Is this true?c. Suppose you travel 1° of latitude north or south. How far have you traveled? (Hint: The circumference of Earth is about 25,000 miles.)

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 8

Example (cont)Solutiona. The point on Earth directly opposite Miami must have the opposite latitude (26° S rather than 26°N) and must be 180° away in longitude. Let’s consider going 180° eastward from Miami’s longitude of 80° W. The first 80° eastward would take us to the prime meridian; then we would continue to longitude 100° E to be a full 180° away from Miami. Therefore, the position of the point opposite Miami is latitude 26°S and longitude 100° E (which is in the Indian Ocean).

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 9

Example (cont)Solutionb. It is not true. The United States is in the Northern Hemisphere, so the points on the globe opposite the United States must be in the Southern Hemisphere. China is also in the Northern Hemisphere, so it cannot be opposite the United States.

Page 10: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 10

Example (cont)Solutionc. Every degree of latitude represents the same north or south distance. If you were to traverse the Earth’s 25,000-mile circumference, you would travel through 360° of latitude. Therefore, each degree of latitude represents

Traveling 1° of latitude north or south means traveling a distance of approximately 70 miles.

25,000 mi69.4 mi per degree

360

Page 11: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 11

Angular Size and DistanceThe farther away an object is located from you, the smaller it will appear in angular size (the angle that it covers as seen from your eye).

distancesize physicalsize angular

2360

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 12

Example

a. A quarter is about 1 inch in diameter. Approximately how big will it look in angular size if you hold it 1 yard (36 inches) from your eye?b. The angular diameter of the Moon as seen from Earth is approximately 0.5°, and the Moon is approximately 380,000 kilometers from Earth. What is the real diameter of the Moon?Solutiona.

360angular size = 1 in 1.6

2 36 in

Page 13: Copyright  2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 1 Modeling with Geometry 10.

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 13

Example (cont)b. In this case, we are asked to find the Moon’s physical size (diameter) from its angular size and distance, so we must solve the small-angle formula for physical size. You should confirm that the formula becomes

Now we substitute 380,000 kilometers for the distance and 0.5° for the angular diameter:

The Moon’s real diameter is about 3300 kilometers.

2 distancephysical size = angular size 360

2 380,000physical size = 0.5 3300 km

360

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 14

Pitch, Grade, and SlopeConsider a road that rises uniformly 2 feet in the vertical direction for every 20 feet in the horizontal direction. We say that the road has a pitch of 2 in 20, or 1 in 10. It is also common to describe the rise of the road in terms of its slope. In this case, the slope, or rise over run, is 2/20 = 1/10. As a percentage, it’s called a grade. The grade of the road in this case is 1/10 = 10%.

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 15

Example

a. Suppose a road has a 100% grade. What is its slope? What is its pitch? What angle does it make with the horizontal?

b. Which is steeper: a road with an 8% grade or a road with a pitch of 1 in 9?

c. Which is steeper: a roof with a pitch of 2 in 12 or a roof with a pitch of 3 in 15?

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 16

Example (cont)Solutiona. A 100% grade means a slope of 1 and a pitch of 1 in 1. If you look at a line with a slope of 1 on a graph, you’ll see that it makes an angle of 45° with the horizontal.b. A road with a pitch of 1 in 9 has a slope of 1/9 = 0.11, or 11%. Therefore, this road is steeper than a road with an 8% grade.c. The 2 in 12 roof has a slope of 2/12 = 0.167, while the 3 in 15 roof has a slope of 3/15 = 0.20. The second roof is steeper.

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 17

Pythagorean Theorem

The Pythagorean theorem applies only to right triangles (those with one 90° angle).

For a right triangle with side lengths a, b, and c, in which c is the longest side (or hypotenuse), the Pythagorean theorem states that

a2 + b2 = c2 a

b

c

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 18

Example

Solution1. Start with the Pythagorean theorem: base2 + height2 = hypotenuse2

2. Solve for height by subtracting base2 from both sides, then taking the square root of both sides:

2 2height = hypotenuse - base

Find the area, in acres, of the mountain lot.

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 19

Example (cont)3. Substitute the given values for the base and hypotenuse:

4. Use this height to find the area of the triangle:

5. Convert to acres (1 acre = 43,560 ft2):

2 2height = 1200 ft - 250 ft 1174 ft

2

1area = base height21= 250 ft 1174 ft2

= 146,750 ft

22

1 acre146,750 ft 3.4 acres43,560 ft

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 20

Similar TrianglesTwo triangles are similar if they have the same shape, but not necessarily the same size, meaning that one is a scaled-up or scaled-down version of the other.

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 21

For two similar triangles, corresponding pairs of angles in each triangle are

equal:

the ratios of the side lengths in the two triangles are all equal:

Similar Triangles

CCBBAA angle angle ; angle angle ; angle angle

cc

bb

aa

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 22

Example

' ' '

a b ca b c

Find the lengths of the sideslabeled a and cʹ.Solution

The “unknown” side lengths are a = 6 and cʹ = 8.

9 12

4 6 'a

c

9 9 4 64 6 6a a

9 9 12 6' 84 ' 9

cc

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Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 10, Unit B, Slide 23

You have 132 meters of fence to enclose a corral. What shape should you choose if you want to have the greatest possible area? What is the area of this “optimized” coral?

SolutionSquare corral:

Circular corral:

The circle is the shape with the greatest possible area.

Example

2 233 m 1089 mA

22 2132 m 1387 m

2A r