chapter 1 sections 2 and 3

16
CHAPTER 1 SECTIONS 2 AND 3 Points, Lines, and Planes Perimeter and Area

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Chapter 1 Sections 2 and 3. Points, Lines, and Planes Perimeter and Area. Warm-Up. 1) In a coordinate system, which quadrant is in the lower left-hand portion of the plane?  2) Graph three points that lie on the graph of y= 4x – 5. - PowerPoint PPT Presentation

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Page 1: Chapter 1 Sections 2 and 3

CHAPTER 1SECTIONS 2

AND 3Points, Lines, and Planes

Perimeter and Area

Page 2: Chapter 1 Sections 2 and 3

WARM-UP1) In a coordinate system, which quadrant is in the lower left-hand portion of the plane?  2) Graph three points that lie on the graph of y= 4x – 5.    3) Points N(5, -2) and M(2, -4) lie on the graph of 2x – 3y = 16. Determine whether P(8, 0) is collinear to N and M.    

Page 3: Chapter 1 Sections 2 and 3

1) In a coordinate system, which quadrant is in the lower left-hand portion of the plane?III2) Graph three points that lie on the graph of y= 4x – 5.  

y

x

III

III IVX Y (x,y)

0

1

2

y= 4x - 5y= 4(0) – 5y= 0 – 5y = -5

y= 4x - 5y= 4(1) – 5y= 4 – 5y = -1

y= 4x - 5y= 4(2) – 5y= 8 – 5y = 3

(0,-5)

(1,-1)

(2,3)

y

x

Page 4: Chapter 1 Sections 2 and 3

 3) Points N(5, -2) and M(2, -4) lie on the graph of 2x – 3y = 16. Determine whether P(8, 0) is collinear to N and M.  

2x – 3y = 162(8) – 3(0) = 1616 – 0 = 1616 = 16

Yes point P is collinear to points N and M.

Page 5: Chapter 1 Sections 2 and 3

VOCABULARYPlanes- A basic undefined term of geometry. Flat surfaces that extend indefinitely in all directions and have no thickness.

Lines- A basic undefined term of geometry. Lines extend indefinitely and have no thickness or width.

Points- A basic undefined term of geometry. Points have no size.

Page 6: Chapter 1 Sections 2 and 3

VOCABULARY CONT.Space- A boundless three-dimensional figure consisting of all of its surface points and all of its interior points.

Coplanar- points on the same plane

Area- The number of square units contained in the interior of a figure.Area = Length * Width

Perimeter- The distance around a two-dimensional figure. Perimeter of a rectangle = 2(Length) + 2(Width)

l

w

l

w

Page 7: Chapter 1 Sections 2 and 3

Example 1: Jessie is choosing her lunch in the cafeteria. She may choose one vegetable from peas, carrots, and green beans; one fruit from watermelon, strawberries, or an apple; and the type of bread for her sandwich from white, wheat, rye. How many different combinations of lunch items can Jessie choose?Peas, Watermelon, WhitePeas, Watermelon, WheatPeas, Watermelon, RyePeas, Strawberries, WhitePeas, Strawberries, WheatPeas, Strawberries, RyePeas, Apple, WhitePeas, Apple, WheatPeas, Apple, Rye

Carrots, Watermelon, WhiteCarrots, Watermelon, WheatCarrots, Watermelon, RyeCarrots, Strawberries, WhiteCarrots, Strawberries, WheatCarrots, Strawberries, RyeCarrots, Apple, WhiteCarrots, Apple, WheatCarrots, Apple, Rye

Green Beans, Watermelon, WhiteGreen Beans, Watermelon, WheatGreen Beans, Watermelon, RyeGreen Beans, Strawberries, WhiteGreen Beans, Strawberries, WheatGreen Beans, Strawberries, RyeGreen Beans, Apple, WhiteGreen Beans, Apple, WheatGreen Beans, Apple, Rye

There are nine options if she chooses peas. Since there are 3 different options for vegetables. We want to multiply 9 by 3.

So there are 21 different combinations to choose.

Page 8: Chapter 1 Sections 2 and 3

Example 2: Using the figure to answer the following questions.a) Name a point that is not collinear to F and C.Points A, B, or D

b) Identify a point that is not in the plane N.Points E or F

c) What is the intersection of plane ADE and plane N ?Line DA

E

F

DC

BAN

Page 9: Chapter 1 Sections 2 and 3

Example 3: Draw and label a figure showing lines NP and QR intersecting at point S for points N(3, -1), P(5, 2), Q(-3, 1), and R(0, -4). y

x

N

R

Q

S

P

Page 10: Chapter 1 Sections 2 and 3

Example 4: State whether each is best modeled by a point, line, or plane.

A) A star in the sky.A point.

B) An ice skating rink.A plane.

C) A telephone wire strung between two poles.A line.

Page 11: Chapter 1 Sections 2 and 3

Example 4: Mr. and Mrs. Pinel are planning a rectangular vegetable garden. They want the garden to have an area of at least 15 square yards, but they have only 18 yards of wire fence to surround it. What are the possible dimensions of the garden if all the fence is used and the sides have whole-number lengths?

a) Draw all possibilities.

1

8

8

3

77

2

2

88

111

3

6 6

6

4

4

5 5 44

5

5

3 3

6

2 2

7

7

Page 12: Chapter 1 Sections 2 and 3

Example 4: Mr. and Mrs. Pinel are planning a rectangular vegetable garden. They want the garden to have an area of at least 15 square yards, but they have only 18 yards of wire fence to surround it. What are the possible dimensions of the garden if all the fence is used and the sides have whole-number lengths?

b)Fill in the table. Width Length Area = Length * Width1

2

7

6

3

4

8

5

8

7

65

4

3

2

1

8 * 1 = 8

1 * 8 = 8

2 * 7 = 14

7 * 2 = 146 * 3 = 18

3 * 6 = 18

4 * 5 = 20

5 * 4 = 20

Page 13: Chapter 1 Sections 2 and 3

Example 4: Mr. and Mrs. Pinel are planning a rectangular vegetable garden. They want the garden to have an area of at least 15 square yards, but they have only 18 yards of wire fence to surround it.

c) What are the possible dimensions of the garden if all the fence is used and the sides have whole-number lengths?

6 by 3 or 5 by 4.

Page 14: Chapter 1 Sections 2 and 3

Example 5: Find the are and perimeter.

10 cm

4 cm

Area = Length * WidthA = 10 * 4A = 40 cm^2

Perimeter = 2(Length) + 2(Width)P = 2(10) + 2(4)P = 20 + 8P = 28 cm

Page 15: Chapter 1 Sections 2 and 3

Example 6: Find the missing measure in each formula.

a) P = 2l + 2w; l = 3, P = 16, w = ?P = 2l + 2w16 = 2(3) + 2w16 = 6 + 2w10 = 2w5 = w

b) I = prt; p = 350, r = 6%, I = 42, t = ?I = prt42 = 350(0.06)(t)42 = 21(t)2 = t

Page 16: Chapter 1 Sections 2 and 3

Example 7: Find the maximum area for the given perimeter of a rectangle. State the length and width of the rectangle.

Perimeter = 24 millimetersLength Width Area

1 mm2 mm

11 mm

3 mm4 mm5 mm

9 mm

10 mm

7 mm

8 mm

6 mm

6 mm

1 * 11 = 11 mm^2

3 * 9 = 27 mm^24 * 8 = 32 mm^25 * 7 = 35 mm^26 * 6 = 36 mm^2

2 * 10 = 20 mm^2

The maximum area for the given perimeter is 36 mm^2 and the length and width are 6 mm by 6 mm.

Short cut- If the perimeter is divisible by 4, then divide by 4 and square that number. EX: 24/4 = 6, Area is 6 * 6 = 36