Lesson 20: Real-World Area Problems AWS
Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Lesson 20: Real-World Area Problems
175
Student Outcomes ο§
Students determine the area of composite figures in real-life contextual situations using composition and decomposition of polygons and circular regions.
Lesson Notes Students apply their understanding of the area of polygons and circular regions to real-world contexts.
Classwork Opening Exercise (8 minutes) Opening Exercise Find the area of each shape based on the provided measurements. Explain how you found each area.
Triangular Region, half base times height. ππππ =
π β ππ πππππ β π. π πππππ π
= ππ. π ππππππ
Regular Hexagon, area of the shown triangle times six for the six triangles that fit into the hexagon. π ππππ = π ( β π πππππ β π. π πππππ) π = ππ. π ππππππ
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Parallelogram, base times height. ππππ = ππ πππππ β π πππππ = πππ ππππππ
Semi-circle, half the area of a circle with the same radius. π ππππ = (π. π)π ππππππ π = ππ. ππππ ππππππ β ππ. ππ ππππππ
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Example 1 (10 minutes) Students should first attempt this question without assistance. Once students have had time to work on their own, lead the class through a discussion using the following prompts according to student need. Example 1 A landscape company wants to plant lawn seed. A ππ lb. bag of lawn seed will cover up to πππ sq. ft. of grass and costs $ππ. ππ plus the π% sales tax. A scale drawing of a rectangular yard is given. The length of the longest side is πππ ft. The house, driveway, sidewalk, garden areas, and utility pad are shaded. The unshaded area has been prepared for planting grass. How many ππ lb. bags of lawn seed should be ordered, and what is the cost?
π΄2
π΄3
π΄1 π΄4
π΄7 π΄6
π΄5
ο§
The following calculations demonstrate how to find the area of the lawn by subtracting the area of the home from the area of the entire yard.
ο§
Non-grassy sections in the map of the yard and their areas.
ο§
οΊ
π΄1 = 4 π’πππ‘π β 4 π’πππ‘π = 16 π’πππ‘π 2
οΊ
π΄2 = 1 π’πππ‘π β 13 π’πππ‘π = 13 π’πππ‘π 2
οΊ
π΄3 = 7 π’πππ‘π β 13 π’πππ‘π = 91 π’πππ‘π 2
οΊ
π΄4 = 1 π’πππ‘π β 6 π’πππ‘π = 6 π’πππ‘π 2
οΊ
π΄5 = 6 π’πππ‘π β 1 π’πππ‘π = 6 π’πππ‘π 2
οΊ
π΄6 = 1 π’πππ‘π β 6 π’πππ‘π = 6 π’πππ‘π 2
οΊ
π΄7 = 2 π’πππ‘π β 1 π’πππ‘π = 2 π’πππ‘π 2
MP.2
The scale of the map is 5 ft. for every one unit and 25 ft2 for every 1 unit2.
What is the grassy area in square feet? οΊ
ο§
Subtract the area of the non-grassy sections from the area of the yard. π΄ = (20 π’πππ‘π β 18 π’πππ‘π ) β 140 π’πππ‘π 2 = 220 π’πππ‘π 2
What is the scale factor of the map of the yard? οΊ
ο§
π΄1 + π΄2 + π΄3 + π΄4 + π΄5 + π΄6 + π΄7 = 16 π’πππ‘π 2 + 13 π’πππ‘π 2 + 91 π’πππ‘π 2 + 6 π’πππ‘π 2 + 6 π’πππ‘π 2 + 6 π’πππ‘π 2 + 2 π’πππ‘π 2 = 140 π’πππ‘π 2
What is the area of the grassy section of the yard? οΊ
ο§
ο§ An alternative image with fewer regions is provided after the initial solution.
What is the total area of the non-grassy sections? οΊ
ο§
Scaffolding:
220 β 25 = 5,500
If one 20 lb. bag covers 420 square feet, write a numerical expression for the number of bags needed to cover the grass in the yard. Explain your expression. οΊ
Grassy area Γ· area that one bag of seed covers
οΊ
5,500 Γ· 420
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
ο§ MP.2 ο§
ο§
7β’6
How many bags are needed to cover the grass in the yard? οΊ
5,500 Γ· 420 β 13.1
οΊ
It will take 14 bags to seed the yard.
What is the final cost of seeding the yard? οΊ
1.08 β 14 β $49.98 β $755.70
οΊ
Final cost with sales tax is $755.70
Encourage students to write or state an explanation for how they solved the problem.
Alternative image of property:
100 ft.
ο§
ο§
Non-grassy sections in the map of the yard and their areas. οΊ
π΄1 = 14 π’πππ‘π β 14 π’πππ‘π = 196 π’πππ‘π 2
οΊ
π΄2 = 4 π’πππ‘π β 8 π’πππ‘π = 32 π’πππ‘π 2
What is the total area of the non-grassy sections? οΊ
ο§
ο§
π΄1 + π΄2 = 196 π’πππ‘π 2 + 32 π’πππ‘π 2 = 228 π’πππ‘π 2
What is the area of the grassy section of the yard? οΊ
Subtract the area of the non-grassy sections from the area of the yard.
οΊ
π΄ = (20 π’πππ‘π β 15 π’πππ‘π ) β 228 π’πππ‘π 2 = 72 π’πππ‘π 2
What is the scale factor of the map of the yard? οΊ
The scale of the map is 5 ft. for every one unit and 25 ft2 for every 1 unit2.
ο§
What is the grassy area in square feet?
ο§
If one 20 lb. bag covers 420 square feet, write a numerical expression for the number of bags needed to cover the grass in the yard. Explain your expression.
οΊ
MP.2 ο§
ο§
72 β 25 = 1,800
οΊ
Grassy area Γ· area that one bag of seed covers
οΊ
1,800 Γ· 420
How many bags are needed to cover the grass in the yard? οΊ
1,800 Γ· 420 β 4.3
οΊ
It will take 5 bags to seed the yard.
What is the final cost of seeding the yard? οΊ
1.08 β 5 β $49.98 β $269.89
οΊ
Final cost with sales tax is $269.89.
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exercise 1 (6 minutes) Exercise 1 A landscape contractor looks at a scale drawing of a yard and estimates that the area of the home and garage is the same as the area of a rectangle that is πππ ft. Γ ππ ft. The contractor comes up with π, πππ ft2. How close is this estimate? The entire yard (home and garage) has an area of πππ ft. Γ ππ ft. = π, πππ ft2. The contractorβs estimate is π, πππ ft2. He is π, πππ ft2 over the actual area, which is quite a bit more (π, πππ ft2 is roughly ππ% of the actual area).
Example 2 (10 minutes) Example 2 Ten dartboard targets are being painted as shown in the following figure. The radius of the smallest circle is π in. and each successive, larger circle is π in. more in radius than the circle before it. A βtesterβ can of red and of white paint is purchased to paint the target. Each π oz. can of paint covers ππ ft2. Is there enough paint of each color to create all ten targets?
C4
Let each circle be labeled as in the diagram.
C3 C 2 C 1
Radius of πͺπ is π in.; area of πͺπ is ππ in2. Radius of πͺπ is π in.; area of πͺπ is πππ in2. Radius of πͺπ is π in.; area of πͺπ is πππ in2. Radius of πͺπ is ππ in.; area of πͺπ is ππππ in2.
ο§ MP.2 & MP.7
Write a numerical expression that represents the area painted red. Explain how your expression represents the situation. οΊ
The area of red and white paint in square inches is found by finding the area between circles of the target board:
οΊ
Red paint: (144π in2 β 81π in2) + (36π in2 β 9π in2)
οΊ
White paint: (81π in2 β 36π in2) + 9π in2
The following calculations demonstrate how to find the area of red and white paint in the target. Target area painted red The area between πͺπ and πͺπ :
ππππ in2 β πππ in2 = πππ in2
The area between πͺπ and πͺπ :
πππ in2 β ππ in2 = πππ in2
Area painted red in one target:
πππ in2 +πππ in2 = πππ in2; approximately πππ. π in2
Area of red paint for one target in sq. ft.:
πππ. π πππ (
Area to be painted red for ten targets in sq. ft.:
π. ππ ft2 Γ ππ = ππ. π ft2
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
π
π ππ
πππ πππ
) β π. ππ πππ
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Target area painted white The area between πͺπ and πͺπ :
πππ in2 βπππ in2 = πππ in2
The area of πͺπ :
ππ in2
Area painted white in one target:
πππ in2 +ππ in2 = πππ in2; approximately πππ. π in2
Area of white paint for one target sq. ft.:
πππ. π πππ (
Area of white paint needed for ten targets in sq. ft.:
π. ππ ft2 Γ ππ = ππ. π ft2
π
π ππ
πππ ππ
π)
β π. ππ πππ
There is not enough red paint in one tester can of paint to complete all ten targets; however, there is enough white paint in one tester can of paint for all ten targets.
Closing (2 minutes) ο§
What is a useful strategy when tackling area problems with real-world context? οΊ
Decompose drawings into familiar polygons and circular regions, and identify all relevant measurements.
οΊ
Pay attention to the unit needed in a response to each question.
Exit Ticket (9 minutes)
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
Name
7β’6
Date
Lesson 20: Real-World Area Problems Exit Ticket A homeowner called in a painter to paint bedroom walls and ceiling. The bedroom is 18 ft. long, 12 ft. wide, and 8 ft. high. The room has two doors each 3 ft. by 7 ft. and three windows each 3 ft. by 5 ft. The doors and windows do not have to be painted. A gallon of paint can cover 300 ft2. A hired painter claims he will need 4 gal. Show that the estimate is too high.
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exit Ticket Sample Solutions A homeowner called in a painter to paint bedroom walls and ceiling. The bedroom is ππ ft. long, ππ ft. wide, and π ft. high. The room has two doors each π ft. by π ft. and three windows each π ft. by π ft. The doors and windows do not have to be painted. A gallon of paint can cover πππ ft2. A hired painter claims he will need π gal. Show that the estimate is too high. Area of 2 walls:
π(ππ ft. β π ft.) = πππ ft2
Area of remaining 2 walls:
π(ππ ft. β π ft.) = πππ ft2
Area of ceiling:
ππ ft. β ππ ft. = πππ ft2
Area of 2 doors:
π(π ft. β π ft. ) = ππ ft2
Area of 3 windows
π(π ft. β π ft. ) = ππ ft2
Area to be painted:
(πππ ft2 + πππ ft2 + πππ ft2 ) β (ππ ft2 + ππ ft2 ) = πππ ft2
Gallons of paint needed:
πππ Γ· πππ = π. ππ The painter will need a little more than π gal.
The painterβs estimate for how much paint is necessary was too high.
Problem Set Sample Solutions 1.
A farmer has four pieces of unfenced land as shown below in the scale drawing where the dimensions of one side are given. The farmer trades all of the land and $ππ, πππ for π acres of similar land that is fenced. If one acre is equal to ππ, πππ ft2, how much per square foot for the extra land did the farmer pay rounded to the nearest cent?
π π
π΄1
π΄3
π¨π = (π πππππ β π πππππ) = ππ ππππππ
π΄2
π΄4
π π
π¨π = (π πππππ + π πππππ)(π πππππ) = ππ ππππππ π¨π = (π πππππ β π πππππ) + (π πππππ β π πππππ) = ππ ππππππ π π
π¨π = (π πππππ β π πππππ) + (π πππππ β π πππππ) + (π πππππ β π πππππ) = ππ ππππππ The sum of the farmerβs four pieces of land: π¨π + π¨π + π¨π + π¨π = ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ = πππ ππππππ The sum of the farmerβs four pieces of land in sq. ft.: π units = πππ ft. divide each side by π π unit = ππ ft. and π unit2 = π, πππ ft2 πππ β π, πππ = πππ, πππ The total area of the farmer's four pieces of land: πππ, πππ ft2.
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Real-World Area Problems 7/12/15
217 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
The sum of the farmerβs four pieces of land in acres: πππ, πππ Γ· ππ, πππ β π. ππ The farmerβs four pieces of land total about π. ππ acres. Extra land purchased with $ππ, πππ: π acres β π. ππ acres = π. ππ acres Extra land in square feet: ππ, πππ πππ π. ππ πππππ ( ) β ππ, πππ. π πππ π ππππ Price per square foot for extra land: $ππ, πππ ( ) β $π. ππ ππ, πππ. π πππ
2.
An ordinance was passed that required farmers to put a fence around their property. The least expensive fences cost $ππ for each foot. Did the farmer save money by moving the farm? At $ππ for each foot, $ππ, πππ would purchase π, πππ feet of fencing. The perimeter of the third piece of land (labeled π¨π ) has perimeter π, πππ ft. So it would have cost over $ππ, πππ just to fence that piece of property. The farmer did save money by moving the farm.
3.
A stop sign is an octagon (i.e., a polygon with eight sides) with eight equal sides and eight equal angles. The dimensions of the octagon are given. One side of the stop sign is to be painted red. If Timmy has enough paint to paint πππ ft2, can he paint πππ stop signs? Explain your answer. π π
area of top trapezoid= (ππ in. + ππ in.)(π. π in.) = πππ. ππ in2 area of middle rectangle = ππ in. β ππ in. = πππ in2 π π
area of bottom trapezoid = (ππ in.+ ππ in.)(π. π in.) = πππ. ππ in2 Total area of stop sign in square inches: π¨π + π¨π + π¨π = πππ. ππ in2 + πππ in2 + πππ. ππ in2 = πππ. π in2 Total area of stop sign in square feet: πππ. π πππ (
π
π ππ
πππ πππ
) β π. ππ πππ
Yes, the area of one stop sign is less than π ft2 (π. ππ ft2). Therefore, πππ stop signs would be less than πππ ft2.
π΄1
π΄2
π΄3
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
4.
7β’6
The Smith family is renovating a few aspects of their home. The following diagram is of a new kitchen countertop. Approximately how many square feet of counter space is there?
π΄1
π΄2
π¨π = (ππ ππ. +ππ ππ. )(ππ ππ. +ππ ππ. ) = π, πππ πππ π π¨π = (ππ ππ.β π ππ. ) + (πππ πππ ) π β (πππ πππ + ππ πππ ) = πππ πππ π΄3) = πππ πππ π¨π = (ππ ππ.β ππ ππ. ) β (ππ ππ.β ππ ππ.
Total area of counter space in square inches: π¨π + π¨π + π¨π β π, πππ πππ + πππ πππ + πππ πππ π¨π + π¨π + π¨π β π, πππ πππ
Total area of counter-space in square feet: π πππ π, πππ πππ ( ) β ππ. π πππ πππ πππ There is approximately ππ. π ft2 of counter space.
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
7β’6
In addition to the kitchen renovation, the Smithsβ are laying down new carpet. Everything but closets, bathrooms, and the kitchen will have new carpet. How much carpeting must be purchased for the home?
π΄5
π΄2
π΄1
π΄4 π΄6
π΄3
π¨π = (π πππππ β π πππππ) + π ππππππ = ππ ππππππ π¨π = (π πππππ β π πππππ) β π ππππππ = ππ ππππππ π¨π = (π πππππ β π πππππ) β π ππππππ = ππ ππππππ π¨π = π πππππ β ππ πππππ = ππ ππππππ π¨π = (π πππππ β π πππππ) + (π πππππ β π πππππ) = ππ ππππππ π¨π = π πππππ β π πππππ = ππ ππππππ
Total area that needs carpeting: π¨π + π¨π + π¨π + π¨π + π¨π + π¨π = ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ = πππ ππππππ
Scale factor: π ππππ = π ππ.; π πππππ = π πππ. Total area that needs carpeting in square feet: π πππ πππ ππππππ ( ) = πππ πππ π πππππ
6.
π π
Jamie wants to wrap a rectangular sheet of paper completely around cans that are π in. high and π in. in diameter. π π
She can buy a roll of paper that is π in. wide and ππ ft. long. How many cans will this much paper wrap? π inch diameter cans have a circumference of ππ in., approximately ππ. ππ in. ππ ft. is the same as πππ in.; πππ in. Γ· ππ. ππ in. is approximately ππ. π in., so this paper will cover ππ cans.
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Real-World Area Problems 7/12/15
220 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Lesson 20: Real-World Area Problems
175
Student Outcomes ο§
Students determine the area of composite figures in real-life contextual situations using composition and decomposition of polygons and circular regions.
Lesson Notes Students apply their understanding of the area of polygons and circular regions to real-world contexts.
Classwork Opening Exercise (8 minutes) Opening Exercise Find the area of each shape based on the provided measurements. Explain how you found each area.
Triangular Region, half base times height. ππππ =
π β ππ πππππ β π. π πππππ π
= ππ. π ππππππ
Regular Hexagon, area of the shown triangle times six for the six triangles that fit into the hexagon. π ππππ = π ( β π πππππ β π. π πππππ) π = ππ. π ππππππ
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Parallelogram, base times height. ππππ = ππ πππππ β π πππππ = πππ ππππππ
Semi-circle, half the area of a circle with the same radius. π ππππ = (π. π)π ππππππ π = ππ. ππππ ππππππ β ππ. ππ ππππππ
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Example 1 (10 minutes) Students should first attempt this question without assistance. Once students have had time to work on their own, lead the class through a discussion using the following prompts according to student need. Example 1 A landscape company wants to plant lawn seed. A ππ lb. bag of lawn seed will cover up to πππ sq. ft. of grass and costs $ππ. ππ plus the π% sales tax. A scale drawing of a rectangular yard is given. The length of the longest side is πππ ft. The house, driveway, sidewalk, garden areas, and utility pad are shaded. The unshaded area has been prepared for planting grass. How many ππ lb. bags of lawn seed should be ordered, and what is the cost?
π΄2
π΄3
π΄1 π΄4
π΄7 π΄6
π΄5
ο§
The following calculations demonstrate how to find the area of the lawn by subtracting the area of the home from the area of the entire yard.
ο§
Non-grassy sections in the map of the yard and their areas.
ο§
οΊ
π΄1 = 4 π’πππ‘π β 4 π’πππ‘π = 16 π’πππ‘π 2
οΊ
π΄2 = 1 π’πππ‘π β 13 π’πππ‘π = 13 π’πππ‘π 2
οΊ
π΄3 = 7 π’πππ‘π β 13 π’πππ‘π = 91 π’πππ‘π 2
οΊ
π΄4 = 1 π’πππ‘π β 6 π’πππ‘π = 6 π’πππ‘π 2
οΊ
π΄5 = 6 π’πππ‘π β 1 π’πππ‘π = 6 π’πππ‘π 2
οΊ
π΄6 = 1 π’πππ‘π β 6 π’πππ‘π = 6 π’πππ‘π 2
οΊ
π΄7 = 2 π’πππ‘π β 1 π’πππ‘π = 2 π’πππ‘π 2
MP.2
The scale of the map is 5 ft. for every one unit and 25 ft2 for every 1 unit2.
What is the grassy area in square feet? οΊ
ο§
Subtract the area of the non-grassy sections from the area of the yard. π΄ = (20 π’πππ‘π β 18 π’πππ‘π ) β 140 π’πππ‘π 2 = 220 π’πππ‘π 2
What is the scale factor of the map of the yard? οΊ
ο§
π΄1 + π΄2 + π΄3 + π΄4 + π΄5 + π΄6 + π΄7 = 16 π’πππ‘π 2 + 13 π’πππ‘π 2 + 91 π’πππ‘π 2 + 6 π’πππ‘π 2 + 6 π’πππ‘π 2 + 6 π’πππ‘π 2 + 2 π’πππ‘π 2 = 140 π’πππ‘π 2
What is the area of the grassy section of the yard? οΊ
ο§
ο§ An alternative image with fewer regions is provided after the initial solution.
What is the total area of the non-grassy sections? οΊ
ο§
Scaffolding:
220 β 25 = 5,500
If one 20 lb. bag covers 420 square feet, write a numerical expression for the number of bags needed to cover the grass in the yard. Explain your expression. οΊ
Grassy area Γ· area that one bag of seed covers
οΊ
5,500 Γ· 420
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Real-World Area Problems 7/12/15
212 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
ο§ MP.2 ο§
ο§
7β’6
How many bags are needed to cover the grass in the yard? οΊ
5,500 Γ· 420 β 13.1
οΊ
It will take 14 bags to seed the yard.
What is the final cost of seeding the yard? οΊ
1.08 β 14 β $49.98 β $755.70
οΊ
Final cost with sales tax is $755.70
Encourage students to write or state an explanation for how they solved the problem.
Alternative image of property:
100 ft.
ο§
ο§
Non-grassy sections in the map of the yard and their areas. οΊ
π΄1 = 14 π’πππ‘π β 14 π’πππ‘π = 196 π’πππ‘π 2
οΊ
π΄2 = 4 π’πππ‘π β 8 π’πππ‘π = 32 π’πππ‘π 2
What is the total area of the non-grassy sections? οΊ
ο§
ο§
π΄1 + π΄2 = 196 π’πππ‘π 2 + 32 π’πππ‘π 2 = 228 π’πππ‘π 2
What is the area of the grassy section of the yard? οΊ
Subtract the area of the non-grassy sections from the area of the yard.
οΊ
π΄ = (20 π’πππ‘π β 15 π’πππ‘π ) β 228 π’πππ‘π 2 = 72 π’πππ‘π 2
What is the scale factor of the map of the yard? οΊ
The scale of the map is 5 ft. for every one unit and 25 ft2 for every 1 unit2.
ο§
What is the grassy area in square feet?
ο§
If one 20 lb. bag covers 420 square feet, write a numerical expression for the number of bags needed to cover the grass in the yard. Explain your expression.
οΊ
MP.2 ο§
ο§
72 β 25 = 1,800
οΊ
Grassy area Γ· area that one bag of seed covers
οΊ
1,800 Γ· 420
How many bags are needed to cover the grass in the yard? οΊ
1,800 Γ· 420 β 4.3
οΊ
It will take 5 bags to seed the yard.
What is the final cost of seeding the yard? οΊ
1.08 β 5 β $49.98 β $269.89
οΊ
Final cost with sales tax is $269.89.
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exercise 1 (6 minutes) Exercise 1 A landscape contractor looks at a scale drawing of a yard and estimates that the area of the home and garage is the same as the area of a rectangle that is πππ ft. Γ ππ ft. The contractor comes up with π, πππ ft2. How close is this estimate? The entire yard (home and garage) has an area of πππ ft. Γ ππ ft. = π, πππ ft2. The contractorβs estimate is π, πππ ft2. He is π, πππ ft2 over the actual area, which is quite a bit more (π, πππ ft2 is roughly ππ% of the actual area).
Example 2 (10 minutes) Example 2 Ten dartboard targets are being painted as shown in the following figure. The radius of the smallest circle is π in. and each successive, larger circle is π in. more in radius than the circle before it. A βtesterβ can of red and of white paint is purchased to paint the target. Each π oz. can of paint covers ππ ft2. Is there enough paint of each color to create all ten targets?
C4
Let each circle be labeled as in the diagram.
C3 C 2 C 1
Radius of πͺπ is π in.; area of πͺπ is ππ in2. Radius of πͺπ is π in.; area of πͺπ is πππ in2. Radius of πͺπ is π in.; area of πͺπ is πππ in2. Radius of πͺπ is ππ in.; area of πͺπ is ππππ in2.
ο§ MP.2 & MP.7
Write a numerical expression that represents the area painted red. Explain how your expression represents the situation. οΊ
The area of red and white paint in square inches is found by finding the area between circles of the target board:
οΊ
Red paint: (144π in2 β 81π in2) + (36π in2 β 9π in2)
οΊ
White paint: (81π in2 β 36π in2) + 9π in2
The following calculations demonstrate how to find the area of red and white paint in the target. Target area painted red The area between πͺπ and πͺπ :
ππππ in2 β πππ in2 = πππ in2
The area between πͺπ and πͺπ :
πππ in2 β ππ in2 = πππ in2
Area painted red in one target:
πππ in2 +πππ in2 = πππ in2; approximately πππ. π in2
Area of red paint for one target in sq. ft.:
πππ. π πππ (
Area to be painted red for ten targets in sq. ft.:
π. ππ ft2 Γ ππ = ππ. π ft2
Lesson 20: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
π
π ππ
πππ πππ
) β π. ππ πππ
Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Target area painted white The area between πͺπ and πͺπ :
πππ in2 βπππ in2 = πππ in2
The area of πͺπ :
ππ in2
Area painted white in one target:
πππ in2 +ππ in2 = πππ in2; approximately πππ. π in2
Area of white paint for one target sq. ft.:
πππ. π πππ (
Area of white paint needed for ten targets in sq. ft.:
π. ππ ft2 Γ ππ = ππ. π ft2
π
π ππ
πππ ππ
π)
β π. ππ πππ
There is not enough red paint in one tester can of paint to complete all ten targets; however, there is enough white paint in one tester can of paint for all ten targets.
Closing (2 minutes) ο§
What is a useful strategy when tackling area problems with real-world context? οΊ
Decompose drawings into familiar polygons and circular regions, and identify all relevant measurements.
οΊ
Pay attention to the unit needed in a response to each question.
Exit Ticket (9 minutes)
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
Name
7β’6
Date
Lesson 20: Real-World Area Problems Exit Ticket A homeowner called in a painter to paint bedroom walls and ceiling. The bedroom is 18 ft. long, 12 ft. wide, and 8 ft. high. The room has two doors each 3 ft. by 7 ft. and three windows each 3 ft. by 5 ft. The doors and windows do not have to be painted. A gallon of paint can cover 300 ft2. A hired painter claims he will need 4 gal. Show that the estimate is too high.
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exit Ticket Sample Solutions A homeowner called in a painter to paint bedroom walls and ceiling. The bedroom is ππ ft. long, ππ ft. wide, and π ft. high. The room has two doors each π ft. by π ft. and three windows each π ft. by π ft. The doors and windows do not have to be painted. A gallon of paint can cover πππ ft2. A hired painter claims he will need π gal. Show that the estimate is too high. Area of 2 walls:
π(ππ ft. β π ft.) = πππ ft2
Area of remaining 2 walls:
π(ππ ft. β π ft.) = πππ ft2
Area of ceiling:
ππ ft. β ππ ft. = πππ ft2
Area of 2 doors:
π(π ft. β π ft. ) = ππ ft2
Area of 3 windows
π(π ft. β π ft. ) = ππ ft2
Area to be painted:
(πππ ft2 + πππ ft2 + πππ ft2 ) β (ππ ft2 + ππ ft2 ) = πππ ft2
Gallons of paint needed:
πππ Γ· πππ = π. ππ The painter will need a little more than π gal.
The painterβs estimate for how much paint is necessary was too high.
Problem Set Sample Solutions 1.
A farmer has four pieces of unfenced land as shown below in the scale drawing where the dimensions of one side are given. The farmer trades all of the land and $ππ, πππ for π acres of similar land that is fenced. If one acre is equal to ππ, πππ ft2, how much per square foot for the extra land did the farmer pay rounded to the nearest cent?
π π
π΄1
π΄3
π¨π = (π πππππ β π πππππ) = ππ ππππππ
π΄2
π΄4
π π
π¨π = (π πππππ + π πππππ)(π πππππ) = ππ ππππππ π¨π = (π πππππ β π πππππ) + (π πππππ β π πππππ) = ππ ππππππ π π
π¨π = (π πππππ β π πππππ) + (π πππππ β π πππππ) + (π πππππ β π πππππ) = ππ ππππππ The sum of the farmerβs four pieces of land: π¨π + π¨π + π¨π + π¨π = ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ = πππ ππππππ The sum of the farmerβs four pieces of land in sq. ft.: π units = πππ ft. divide each side by π π unit = ππ ft. and π unit2 = π, πππ ft2 πππ β π, πππ = πππ, πππ The total area of the farmer's four pieces of land: πππ, πππ ft2.
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
The sum of the farmerβs four pieces of land in acres: πππ, πππ Γ· ππ, πππ β π. ππ The farmerβs four pieces of land total about π. ππ acres. Extra land purchased with $ππ, πππ: π acres β π. ππ acres = π. ππ acres Extra land in square feet: ππ, πππ πππ π. ππ πππππ ( ) β ππ, πππ. π πππ π ππππ Price per square foot for extra land: $ππ, πππ ( ) β $π. ππ ππ, πππ. π πππ
2.
An ordinance was passed that required farmers to put a fence around their property. The least expensive fences cost $ππ for each foot. Did the farmer save money by moving the farm? At $ππ for each foot, $ππ, πππ would purchase π, πππ feet of fencing. The perimeter of the third piece of land (labeled π¨π ) has perimeter π, πππ ft. So it would have cost over $ππ, πππ just to fence that piece of property. The farmer did save money by moving the farm.
3.
A stop sign is an octagon (i.e., a polygon with eight sides) with eight equal sides and eight equal angles. The dimensions of the octagon are given. One side of the stop sign is to be painted red. If Timmy has enough paint to paint πππ ft2, can he paint πππ stop signs? Explain your answer. π π
area of top trapezoid= (ππ in. + ππ in.)(π. π in.) = πππ. ππ in2 area of middle rectangle = ππ in. β ππ in. = πππ in2 π π
area of bottom trapezoid = (ππ in.+ ππ in.)(π. π in.) = πππ. ππ in2 Total area of stop sign in square inches: π¨π + π¨π + π¨π = πππ. ππ in2 + πππ in2 + πππ. ππ in2 = πππ. π in2 Total area of stop sign in square feet: πππ. π πππ (
π
π ππ
πππ πππ
) β π. ππ πππ
Yes, the area of one stop sign is less than π ft2 (π. ππ ft2). Therefore, πππ stop signs would be less than πππ ft2.
π΄1
π΄2
π΄3
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
4.
7β’6
The Smith family is renovating a few aspects of their home. The following diagram is of a new kitchen countertop. Approximately how many square feet of counter space is there?
π΄1
π΄2
π¨π = (ππ ππ. +ππ ππ. )(ππ ππ. +ππ ππ. ) = π, πππ πππ π π¨π = (ππ ππ.β π ππ. ) + (πππ πππ ) π β (πππ πππ + ππ πππ ) = πππ πππ π΄3) = πππ πππ π¨π = (ππ ππ.β ππ ππ. ) β (ππ ππ.β ππ ππ.
Total area of counter space in square inches: π¨π + π¨π + π¨π β π, πππ πππ + πππ πππ + πππ πππ π¨π + π¨π + π¨π β π, πππ πππ
Total area of counter-space in square feet: π πππ π, πππ πππ ( ) β ππ. π πππ πππ πππ There is approximately ππ. π ft2 of counter space.
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Real-World Area Problems 7/12/15
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Lesson 20
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
7β’6
In addition to the kitchen renovation, the Smithsβ are laying down new carpet. Everything but closets, bathrooms, and the kitchen will have new carpet. How much carpeting must be purchased for the home?
π΄5
π΄2
π΄1
π΄4 π΄6
π΄3
π¨π = (π πππππ β π πππππ) + π ππππππ = ππ ππππππ π¨π = (π πππππ β π πππππ) β π ππππππ = ππ ππππππ π¨π = (π πππππ β π πππππ) β π ππππππ = ππ ππππππ π¨π = π πππππ β ππ πππππ = ππ ππππππ π¨π = (π πππππ β π πππππ) + (π πππππ β π πππππ) = ππ ππππππ π¨π = π πππππ β π πππππ = ππ ππππππ
Total area that needs carpeting: π¨π + π¨π + π¨π + π¨π + π¨π + π¨π = ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ + ππ ππππππ = πππ ππππππ
Scale factor: π ππππ = π ππ.; π πππππ = π πππ. Total area that needs carpeting in square feet: π πππ πππ ππππππ ( ) = πππ πππ π πππππ
6.
π π
Jamie wants to wrap a rectangular sheet of paper completely around cans that are π in. high and π in. in diameter. π π
She can buy a roll of paper that is π in. wide and ππ ft. long. How many cans will this much paper wrap? π inch diameter cans have a circumference of ππ in., approximately ππ. ππ in. ππ ft. is the same as πππ in.; πππ in. Γ· ππ. ππ in. is approximately ππ. π in., so this paper will cover ππ cans.
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