MAP4C Ch2анаGeometry.notebook. 1. January 23, 2013 ... answer to the lowest
number of decimals in any measurement. e.g. Given the length of a room is ...
MAP4C Ch2 Geometry.notebook
January 23, 2013
Unit 2 Geometry Day 1 Area Applications Learning goals:
Solve area problems involving composite shapes
Due now
NA
AGENDA
Review of Formulas and Conversions Investigate Minigolf hole (p. 67) Example 1 Use Trigonometry
MSIP / Home Learning: Read pp. 60 66 pp. 7174 #56, (7 or 8), 10, 12, 13
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Recall... p. 60 Imperial Metric Conversions
Distance
p. 63
Volume
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p. 61 Area
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Significant digits
When dealing with decimal measurements, you must round your final answer to the lowest number of decimals in any measurement.
e.g. Given the length of a room is 10.8m and the width is 7.78m, when calculating the area, your answer is only accurate when rounded to ONE decimal place. 7.78 m
A = 84.0 m2 10.8 m
e.g. Given a triangular table has base 6.30 ft and height 4.125 feet, when calculating the area, your answer is only accurate when rounded to TWO decimal places. 4.125 ft A = 12.99 ft2
6.30 ft
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p. 67
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Solution
The minigolf hole consists of: • a trapezoid with height 2.3 m, top 1.2 m and bottom 2.1 m • a trapezoid with top 1.8 m and bottom 4.6 m • a rectangle with length 4.6 m and width 2.5 m • a circle with radius 1.25 m
NOTE: trapezoids should be oriented so that the top and bottom are parallel
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2.5m
Acircle = πr2 ÷ 2
1.2m
= 3.14m x 1.252m ÷ 2 = 2.45 m2
Atrap1 = (1.2 + 2.1) x 2.3 ÷ 2 = 3.795 = 3.8 m2 Arect = L x W = 4.6m x 2.5m = 11.5 m2
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y
h x
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b = (4.6 1.8) ÷ 2 = 1.4m
h2 = y2 x2 = 2.52 1.42 = 4.29
h = √4.29 = 2.1
Atrap1 = (top + bottom) x h ÷ 2 = (1.8 + 4.6) x 2.1 ÷ 2 = 6.7 m2
Atotal = 2.5 + 11.5 + 6.7 + 3.8 = 24.5 m2
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3
Atotal = 24.5 m2
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Example 1
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Solution
h x
1) Find the hroof 2) Aroof 3) Awall 4) Adoor 5) Calculate Aroof + Awall Adoor .
1. The height of the roof is the length of the side opposite the 22° angle.
Use tan θ = h ÷ x
Atriangle = b x h ÷2
tan 22° = h ÷ 4.265 4.265 x tan 22° = h h = 1.723 m
= 8.53 x 1.723 ÷2 = 7.35 m2
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Awall = L x W = 8.53 x 5.75 ≈ 49.05 m2 Adoor = L x W = 33.5 in x 81.5 in = 2730.3 in2
We need to convert this to m2. 1 in = 2.54 cm = 0.0254 m 1 in2 = (0.0254)2 m2 So 2730.3 in2 = 2730.3 x (0.0254)2 m2 = 1.8 m2
Atotal = Aroof + Awall Adoor = 7.35 + 49.05 1.8 = 54.6 m2 The carpenters will need 54.6 m2 of pressboard for the wall of the house.
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Ch2 Geometry
January 23, 2013
Day 2 2.2 Working With Composite Objects
Learning goals: Find the volume and Surface Area of composite objects Due now pp. 7174 #56, (7 or 8), 10, 12, 13 AGENDA
Example 1 Example 2
MSIP / Home Learning:
Finding V and SA of a cylinder Finding V and SA of a composite object Read Examples 23 on pp. 7881 Follow this format when solving problems Complete pp. 8185 #35, 12, 15, 1718 Also see examples on pp. 6465
Youtube 4Layer Pentagonal Prism
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Example 1)
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Net for a cylinder
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Solution
a) The formula for volume of a cylinder is V = πr2h = π x 2.52 x 5 = 98.17... ≈ 98 ft3
The shrink wrap covering it will equal the surface area. SA = (area of ends) + (area of lateral surface) = 2πr2 + 2πrh = 2π(2.5)2 + 2π(2.5)(5) = 39.25 + 78.5 = 117.75 ≈ 118 ft2 ∴ 118 ft2 of shrink wrap are required.
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b) SA = (area of lateral surface / side) = 2πrh = 2π(2.5)(5) = 78.5
If the ends do not need to be covered, only 79 ft2 of shrink wrap is required.
c) 1 ft = 0.3048 m So 1 ft2 = (0.3048 m)2 = 0.09290304 m2 To convert from ft2 to m2 we multiply by 0.09290304
79 x 0.09290304 = 7.339... So 8 m2 of shrink wrap are required.
NOTE: The answer in the book is 7 m2. What is the problem with this?
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Example 2)
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Method 1 The shed is a triangular prism mounted on top of a rectangular prism. Find the volume of each one and add them. V = (volume of triangular prism) + (volume of rectangular prism) = (area of triangular front) x (height) + (area of base) x (height) = (1/2 x L x ht) x w + (L x w) x hr
ht
hr
w
w
ht L
L
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Solution (p. 79)
width
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To convert from cm3 to m3, divide by 1 000 000
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Method 2 The volume of any prism with identical top and bottom, and parallel edges is A = (area of base) x (height) The shed can be viewed as a pentagonal prism. Find the area of the 'base' the front wall, and multiply it by the width. so A = (Arect + Atri) x w = (L x hr + 1/2 x L x ht) x w
ht
hr
w
L
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Example 2 cont'd
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Faces: • 2 roof panels (rectangles) • front + back (triangle + rectangle) • 2 sides (identical rectangles) • floor (rectangle)
Full solution p. 80
slant height
174.0 cm
s
s2
s2
s slant height
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Geometry Day 34
January 23, 2013
2.3 Optimizing Areas and Perimeters
Due now: pp. 8185 #35, 12, 15, 1718
Learning goals:
AGENDA
Solve 2D optimization problems
Investigation Finding Maximum Areas and Minimum Perimeters Example 1 Finding Dimensions of Optimal Rectangles Optimizing With Constraints Example 2
Toothpicks pp. 87 top 88 Graphing calculator emulator pp. 8889
MSIP / Home Learning: pp. 94 96 #1, (25)ac, 9, 13, 1516, 19
A gardener wants to determine the greatest rectangular area that can be enclosed by a given length of edging. A dog breeder wants to find the rectangle with a given area that has the least perimeter. Both are optimization problems.
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First 20 mins:
With a partner, complete one of the following Investigations:
pp. 8788 using manipulatives (toothpicks) pp. 8889 using Graphing calculator emulator START>All Programs>Math>TI Virtual Calculator OR Desktop>Math>TI Virtual Calculator
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Part A Maximum Area
Part B Minimum Perimeter
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Recall from Grade 9... For a 2 or 4sided enclosure, a square is the optimal* shape.
However, for a 3sided enclosure, a rectangle with L = 2W is optimal*!
W
W
L = 2W
* minimizes perimeter for a given area, OR maximizes area for a given perimeter.
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Example 1
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Finding Dimensions of Optimal Rectangles
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Example 2)
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Optimizing With Constraints
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Solution a) Create a table to show all dimensions that will give an area of 40m2. Length (m)
Width (m)
Perimeter (m) (L + 2W)
Area (m2)
1
40
81
40
2
20
42
40
4
10
24
40
5
8
21
40
8
5
18
40
10
4
18
40
20
2
24
40
40
1
42
40
The minimum perimeter, if the dimensions must be whole numbers, is 18, which occurs when the dimensions are L=8m and W=5m or L=10m and W= 4m.
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b) If the dimensions can be decimals, find the dimensions of a rectangle with L = 2W.
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Example 3) Enclosing NonRectangular Areas
HINT: The perimeter will be 900m. Let P = 900 and solve for the unknown dimension.
Square Circle P = 4s C = 2πr
A = s2 A = πr2
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Unit 2 Geometry
Learning goals:
January 23, 2013
Day 5 2.4 Optimizing Area and Perimeter Using a Spreadsheet
Students will calculate Area and Perimeter Students will create and analyze a scatter plot Students will convert measurements
AGENDA Investigation Part A Maximum Area for a Given Perimeter Part B Minimum Perimeter for a Given Area Reflect
MSIP / Home Learning: pp. 101 – 102 #1, 7, 8
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Debrief 2.4 Spreadsheets
Part A
What are the dimensions of the pen with maximum area? Is it possible to build a pen with a larger area?
8.0m x 8.0m = 26.2 ft x 26.2 ft No a square maximizes area.
Part B
What are the dimensions of the pen with minimal perimeter? 5.0m x 4.8m s = √24 = 4.8989 Is it possible to build a pen with a smaller perimeter?
P = 4(4.8989) = 19.60
p. 102
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Minds On
Think about the different shapes and sizes of cans.
What is the purpose behind...
a tuna can? a soup can? an energy drink? others?
Dropping the Maple Leaf ham 'puck' at centre ice, Maple Lieff Gardens
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Unit 2 Geometry
January 23, 2013
Day 6 2.5 Dynamic Investigations of Optimal Measurements
Due now:
Investigate Handout / p. 110 pp. 101 – 102 #1, 7, 8
Due Wed:
Rewrite Prep p. 57#16 (by Wed.)
Learning goals:
Students will identify figures with minimum surface area Students will identify figures with maximum volume
AGENDA
Option Sheets Food Drive Inquiry (Geometer's Sketchpad Challenge) 1, 2 Minimizing Surface Area 3, 4 Maximizing Volume
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Geometer's Sketchpad Challenge
1. Copy and complete the following table 2. Complete the Inquiry on pp. 103104 and record your findings
#
Shape
Smallest Largest Possible Possible Surface Volume Area (cm^3) (cm^2)
Rectangular
1 prism
NA
2 Cylinder
NA
Rectangular
3 Prism
Dim 1 (cm)
Dim 2 (cm)
Dim 3 (cm)
NA
NA
NA 4 Cylinder 2.5 DynamicVolumeInvestigations.gsp
NA
2.5 DynamicVolumeInvestigations_done.gsp
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Geometer's Sketchpad Challenge
1. Copy and complete the following table 2. Complete the Inquiry on pp. 103104 and record your findings
#
Shape
Smallest Largest Possible Possible Surface Volume Area (cm^3) (cm^2)
Rectangular
Dim 1 (cm)
Dim 2 (cm)
1 prism
NA
L =
W =
2 Cylinder
NA
r =
h =
Rectangular
3 Prism
NA
L =
W =
4 Cylinder
NA
r =
h =
Dim 3 (cm)
H = NA
H = NA
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MAP4C Ch2 Geometry.notebook
Unit 3 Geometry
Learning goals:
AGENDA
January 23, 2013
Day 7 2.6 Optimizing Volume and Surface Area
Students will... compute dimensions that minimize SA compute dimensions that maximize V
Investigate Example 1 Example 2
Minimize the SA of a Rectangular Prism Optimizing Rectangular Prisms Optimizing Other Objects
MSIP / Home Learning: pp. 110 112 #1, 3, 7, 11, 12
1) Download file to M: or USB drive 2) Open GSP (Desktop>Math) 3) Click File > Open
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Unit 2 Geometry Day 7 2.6 Optimizing V and SA
Due now: pp. 105 Investigate Due Wed: Rewrite Prep p. 57#16 (by Wed.)
Learning goals:
AGENDA
Determine the dimensions of optimal figures (with restrictions)
Warm up: 1000cm3 cuboid Optimizing Rectangular Prisms Optimizing with Constraints Optimizing a Cylinder
MSIP / Home Learning: pp. 110112 #15, 1113
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Investigate Minimize the Surface Area of a Rectangular Prism
Work with a partner. Find the dimensions of 3 boxes with volume 1000cm3. All the boxes must be rectangular prisms. Calculate the surface area of each box.
Which of your boxes box has the least surface area? What are its dimensions?
Class' least SA: L: W: H:
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Among all rectangular prisms (cuboids) with a given surface area, a cube has the maximum volume.
Among all rectangular prisms with a given volume, a cube has the minimum surface area
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To find the dimensions of an object (3D) with restrictions:
1. Start with the formula corresponding to the given information (SA or V). 2. Sub in the given value, plus any other restrictions, and solve for one dimension. 3. Find all dimensions. 4. Perform any necessary calculations.
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Example 1 Optimizing Rectangular Prisms
Solution
a) The formula for SA of a cube is SA = 6s2
V = s3
b) The formula for volume of a cube is V = s3
SA = 6s2
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Example 2 Optimizing Other Objects
Solution V = πr2h so 350 = πr2h
But in the optimal cylinder, h = 2r
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1. Enter equations:
y = 350 πx2
z = 2πx2 + 2πxy
2. Set up graph.
3. Plot graph and determine minimum point.
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1. Enter equations:
y = 350 πx2
z = 2πx2 + 2πxy
2. Set up graph.
3. Plot graph and determine minimum point.
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Unit 3 Geometry Day 8 2.7 Optimizing Surface Area Using a Spreadsheet
Learning goals:
AGENDA
Students will calculate Surface Area Students will create and analyze a scatter plot
Complete pp. 115118 Answer Q. #112 NOTE: For #6 & 8 you will need to fill past row A23. Refer to the §2.4 Handout
MSIP / Home Learning: Complete pp. 115118 pp. 110112 #15, 1113
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MAP4C Ch2 Geometry.notebook Due now:
January 23, 2013
2.6 pp. 110112 #15, 1113
Questions?
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Due now:
January 23, 2013
2.6 pp. 110112 #15, 1113
Questions?
Review pp. 120 122 #1, 2, 4ab, 7, 8, (910)ac, 11, 12, 15, 17, 18
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Key Concepts: Chapter 2 Geometry
2.1 Area Applications Find the area of composite shapes (2D)
A mailbox consists of a semicylinder on top of a squarebased rectangular prism. The width is 15 cm and the length is 45 cm. a) Find the area of the door at one end. b) Find the volume and surface area.
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2.2 Composite Objects Find the SA and V of composite objects (3D)
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2.3/2.4 Optimizing Area and Perimeter (Rectangles)
For 2 or 4sided, optimal is a square For 3sided, optimal has L = 2W (long side) = 2 x (short side)
Ex) The CPHS athletic department is going to build a skating rink out of some old 2x4s they found in the wood shop. What are the optimal dimensions if...
a) They want a 144m2 ice surface? What length of 2x4s is required for the border? b) They have 80m of 2x4s to create a border, but will be building the rink along one of the exterior walls (3 sides)? What is the area of the rink?
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2.52.7 Optimal Measurements For fixed SA, use SA = 6s2 Cuboid: For fixed V, use V = s3 Cylinder:
For fixed V, h = 2r then use SA = 2πrh + 2πr2
You will not be tested on spreadsheets. Ex 1a) What are the optimal dimensions of a cuboid tetra pack that will hold 512ml of Crimson Cow energy drink? How much waxed cardboard is required to construct one tetrapack?
Crimson Cow
Ex 1b) What are the optimal dimensions of a cylindrical can that will hold the same amount? How much aluminum is required to construct one can?
Crimson Cow
Ex 1c) Which can costs less to manufacture if waxed cardboard (tetra pack) costs $0.04/cm2 and aluminum costs $0.05/cm2?
Review pp. 120 122 #1, 2, 4ab, 7, 8, (910)ac, 11, 12, 15, 17, 18
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Attachments
2.5 DynamicVolumeInvestigations.gsp 2.5 DynamicVolumeInvestigations_done.gsp
