{"id":72267,"date":"2021-03-24T19:33:30","date_gmt":"2021-03-24T14:03:30","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=72267"},"modified":"2022-04-01T10:27:40","modified_gmt":"2022-04-01T04:57:40","slug":"eureka-math-grade-7-module-6-lesson-20","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/eureka-math-grade-7-module-6-lesson-20\/","title":{"rendered":"Eureka Math Grade 7 Module 6 Lesson 20 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 6 Lesson 20 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 6 Lesson 20 Example Answer Key<\/h3>\n

Example 1.
\nA landscape company wants to plant lawn seed. A 20 lb. bag of lawn seed will cover up to 420 sq.ft. of grass and costs $49.98 plus the 8% sales tax. A scale drawing of a rectangular yard is given. The length of the longest side is 100 ft. The house, driveway, sidewalk, garden areas, and utility pad are shaded. The unshaded area has been prepared for planting grass. How many 20 lb. bags of lawn seed should be ordered, and what is the cost?
\n\"Engage
\nAnswer:
\nThe 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.
\n\u2192 Find the non – grassy sections in the map of the yard and their areas.
\nA1<\/sub> = 4 units \u2022 4 units = 16 units2<\/sup>
\nA2<\/sub> = 1 units \u2022 13 units = 13 units2<\/sup>
\nA3<\/sub> = 7 units \u2022 13 units = 91 units2<\/sup>
\nA4<\/sub> = 1 units \u2022 6 units = 6 units2<\/sup>
\nA5<\/sub> = 6 units \u2022 1 units = 6 units2<\/sup>
\nA6<\/sub> = 1 units \u2022 6 units = 6 units2<\/sup>
\nA7<\/sub> = 2 units \u2022 1 units = 2 units2<\/sup>
\n\"Engage
\n\u2192 What is the total area of the non – grassy sections?
\nA1<\/sub> + A2<\/sub> + A3<\/sub> + A4<\/sub> + A5<\/sub> + A6<\/sub> + A7<\/sub> = 16 units2<\/sup> + 13 units2<\/sup> + 91 units2<\/sup> + 6 units2<\/sup> + 6 units2<\/sup> + 6 units2<\/sup> + 2 units2<\/sup> = 140 units2<\/sup><\/p>\n

\u2192 What is the area of the grassy section of the yard?
\nSubtract the area of the non – grassy sections from the area of the yard.
\nA = (20 units \u2022 18 units) – 140 units2<\/sup> = 220 units2<\/sup><\/p>\n

\u2192 What is the scale of the map of the yard?
\nThe scale of the map is 5 ft.
\nWhat is the grassy area in square feet?
\n220 units2<\/sup> \u2022 25 (ft2<\/sup>)\/(units2<\/sup> ) = 5,500 ft2<\/sup>
\nThe area of the grassy space is 5,500 ft2<\/sup>.<\/p>\n

\u2192 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.
\n\u2192 Grassy area \u00f7 area that one bag of seed covers
\n5,500 \u00f7 420
\n\u2192 How many bags are needed to cover the grass in the yard?
\n5,500 \u00f7 420\u224813.1
\nIt will take 14 bags to seed the yard.
\n\u2192 What is the final cost of seeding the yard?
\n1.08 \u2022 14 \u2022 $49.98\u2248$755.70
\nThe final cost with sales tax is $755.70.<\/p>\n

Example 2.
\nTen dartboard targets are being painted as shown in the following figure. The radius of the smallest circle is 3 in., and each successive larger circle is 3 in. more in radius than the circle before it. A can of red paint and a can of white paint is purchased to paint the target. Each 8 oz. can of paint covers 16 ft2<\/sup>. Is there enough paint of each color to create all ten targets?
\n\"Engage
\nAnswer:
\nLet each circle be labeled as in the diagram.
\n\"Engage
\nRadius of C1<\/sub> is 3 in.; area of C1<\/sub> is 9\u03c0 in2<\/sup>.
\nRadius of C2<\/sub> is 6 in.; area of C2<\/sub> is 36\u03c0 in2<\/sup>.
\nRadius of C3<\/sub> is 9 in.; area of C3<\/sub> is 81\u03c0 in2<\/sup>.
\nRadius of C4<\/sub> is 12 in.; area of C4<\/sub> is 144\u03c0 in2<\/sup>.<\/p>\n

\u2192 Write a numerical expression that represents the area painted red. Explain how your expression represents the situation.
\n\u2192 The area of red and white paint in square inches is found by finding the area between circles of the target board.
\nRed paint: (144\u03c0 in2<\/sup> – 81\u03c0 in2<\/sup>) + (36\u03c0 in2<\/sup> – 9\u03c0 in2<\/sup>)
\nWhite paint: (81\u03c0 in2<\/sup> – 36\u03c0 in2<\/sup>) + 9\u03c0 in2<\/sup>
\nThe following calculations demonstrate how to find the area of red and white paint in the target.
\nTarget area painted red
\nThe area between C4<\/sub> and C3<\/sub>: 144\u03c0 in2<\/sup> – 81\u03c0 in2<\/sup> = 63\u03c0 in2<\/sup>
\nThe area between C2<\/sub> and C1<\/sub>: 36\u03c0 in2<\/sup> – 9\u03c0 in2<\/sup> = 27\u03c0 in2<\/sup>
\nArea painted red in one target: 63\u03c0 in2<\/sup> + 27\u03c0 in2<\/sup> = 90\u03c0 in2<\/sup>; approximately 282.7 in2<\/sup>
\nArea of red paint for one target in square feet: 282.7 in2<\/sup> (\\(\\frac{1 \\mathrm{ft}^{2}}{144 \\mathrm{in}^{2}}\\))\u22481.96 ft2<\/sup>
\nArea to be painted red for ten targets in square feet: 1.96 ft2<\/sup>\u00d710 = 19.6 ft2<\/sup><\/p>\n

Target area painted white
\nThe area between C3<\/sub> and C2<\/sub>: 81\u03c0 in2<\/sup> – 36\u03c0 in2<\/sup> = 45\u03c0 in2<\/sup>
\nThe area of C1<\/sub>: 9\u03c0 in2<\/sup>
\nArea painted white in one target: 45\u03c0 in2<\/sup> + 9\u03c0 in2<\/sup> = 54\u03c0 in2<\/sup>; approximately 169.6 in2<\/sup>
\nArea of white paint for one target in square feet: 169.6 in2<\/sup> (\\(\\frac{1 \\mathrm{ft}^{2}}{144 \\mathrm{in}^{2}}\\))\u22481.18 ft2<\/sup>
\nArea of white paint for ten targets in square feet: 1.18 ft2<\/sup>\u00d710 = 11.8 ft2<\/sup>
\nThere is not enough red paint in one 8 oz. can of paint to complete all ten targets; however, there is enough white paint in one 8 oz. can of paint for all ten targets.<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 20 Exercise Answer Key<\/h3>\n

Opening Exercise
\nFind the area of each shape based on the provided measurements. Explain how you found each area.
\n\"Engage
\nAnswer:
\n\"Engage
\nTriangular region: half base times height
\narea = \\(\\frac{1}{2}\\) \u2022 10 units \u2022 7.5 units
\n= 37.5 units2<\/sup><\/p>\n

\"Engage
\nParallelogram: base times height
\narea = 12 units \u2022 9 units
\n= 108 units2<\/sup><\/p>\n

\"Engage
\nRegular hexagon: area of the shown triangle times six for the six triangles that fit into the hexagon
\narea = 6(\\(\\frac{1}{2}\\)6 units \u2022 5.2 units)
\n= 93.6 units2<\/sup><\/p>\n

\"Engage
\nSemicircle: half the area of a circle with the same radius
\narea = \\(\\frac{\\pi}{2}\\) (4.5 unit)2<\/sup>
\n= 10.125\u03c0 units2<\/sup>
\n\u224831.81 units2<\/sup><\/p>\n

Exercise 1.
\nA 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 100 ft. \u00d7 35 ft. The contractor comes up with 5,500 ft2<\/sup>. How close is this estimate?
\nAnswer:
\nThe entire yard (home and garage) has an area of 100 ft.\u00d735 ft. = 3,500 ft2<\/sup>. The contractor\u2019s estimate is 5,500 ft2<\/sup>. He is 2,000 ft2<\/sup> over the actual area, which is quite a bit more (2,000 ft2<\/sup> is roughly 57% of the actual area).<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 20 Problem Set Answer Key<\/h3>\n

Question 1.
\nA farmer has four pieces of unfenced land as shown to the right in the scale drawing where the dimensions of one side are given. The farmer trades all of the land and $10,000 for 8 acres of similar land that is fenced. If one acre is equal to 43,560 ft2<\/sup>, how much per square foot for the extra land did the farmer pay rounded to the nearest cent?
\n\"Eureka
\nAnswer:
\n\"Eureka
\nA1<\/sub> = \\(\\frac{1}{2}\\) (6 units\u22194 units) = 12 units2<\/sup>
\nA2<\/sub> = \\(\\frac{1}{2}\\) (6 units + 7 units)(4 units) = 26 units2<\/sup>
\nA3<\/sub> = (3 units \u2022 6 units) + (3 units \u2022 5 units) = 33 units2<\/sup>
\nA4<\/sub> = (4 units \u2022 7 units) + (3 units \u2022 3 units) + \\(\\frac{1}{2}\\)(3 units\u22194 units) = 43 units2<\/sup><\/p>\n

The sum of the farmer\u2019s four pieces of land:
\nA1<\/sub> + A2<\/sub> + A3<\/sub> + A4<\/sub> = 12 units2<\/sup> + 26 units2<\/sup> + 33 units2<\/sup> + 43 units2<\/sup> = 114 units2<\/sup><\/p>\n

The sum of the farmer\u2019s four pieces of land in square feet:
\n6 units = 300 ft.; divide each side by 6.
\n1 unit = 50 ft. and 1 unit2<\/sup> = 2,500 ft2<\/sup>
\n114 \u2022 2,500 = 285,000
\nThe total area of the farmer\u2019s four pieces of land: 285,000 ft2<\/sup>.<\/p>\n

The sum of the farmer\u2019s four pieces of land in acres:
\n285,000 \u00f7 43,560\u22486.54
\nThe farmer\u2019s four pieces of land total about 6.54 acres.
\nExtra land purchased with $10,000: 8 acres – 6.54 acres = 1.46 acres
\nExtra land in square feet:
\n1.46 acres(\\(\\frac{43,560 \\mathrm{ft}^{2}}{1 \\text { acre }}\\))\u224863597.6 ft2<\/sup>
\nPrice per square foot for extra land:
\n(\\(\\frac{\\$ 10,000}{63,597.6 \\mathrm{ft}^{2}}\\))\u2248$0.16<\/p>\n

Question 2.
\nAn ordinance was passed that required farmers to put a fence around their property. The least expensive fences cost $10 for each foot. Did the farmer save money by moving the farm?
\nAnswer:
\nAt $10 for each foot, $10,000 would purchase 1,000 feet of fencing. The perimeter of the third piece of land (labeled A3<\/sub>) has a perimeter of 1,200 ft. So, it would have cost over $10,000 just to fence that piece of property. The farmer did save money by moving the farm.<\/p>\n

Question 3.
\nA 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 cover 500 ft2<\/sup>, can he paint 100 stop signs? Explain your answer.
\n\"Eureka
\nAnswer:
\nArea of top trapezoid = \\(\\frac{1}{2}\\)(12 in. + 29 in.)(8.5 in.) = 174.25 in2<\/sup>
\nArea of middle rectangle = 12 in. \u2022 29 in. = 348 in2<\/sup>
\nArea of bottom trapezoid = \\(\\frac{1}{2}\\)(12 in. + 29 in.)(8.5 in.) = 174.25 in2<\/sup>
\nTotal area of stop sign in square inches:
\nA1<\/sub> + A2<\/sub> + A3<\/sub> = 174.25 in2<\/sup> + 348 in2<\/sup> + 174.25 in2<\/sup> = 696.5 in2<\/sup>
\nTotal area of stop sign in square feet:
\n696.5 in 2<\/sup> (\\(\\frac{1 \\mathrm{ft}^{2}}{144 \\mathrm{in}^{2}}\\))\u22484.84 ft 2<\/sup>
\nYes, the area of one stop sign is less than 5 ft 2<\/sup> (approximately 4.84 ft 2<\/sup>). Therefore, 100 stop signs would be less than 500 ft 2<\/sup>.
\n\"Eureka<\/p>\n

Question 4.
\nThe 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?
\n\"Eureka
\nAnswer:
\n\"Eureka
\nA1<\/sub> = (20 in. + 16 in.)(18 in. + 14 in.) = 1,152 in2<\/sup>
\nA2<\/sub> = (18 in. \u2022 7 in.) + 1\/2 (49\u03c0 in2<\/sup> )
\n\u2248(126 in2<\/sup> + 77 in2<\/sup> )
\n\u2248203 in2<\/sup>
\nA3<\/sub> = (50 in. \u2022 16 in.) – (17 in. \u2022 16 in.) = 528 in2<\/sup><\/p>\n

Total area of counter space in square inches:
\nA1<\/sub> + A2<\/sub> + A3 <\/sub>\u2248 1,152 in2<\/sup> + 203 in2<\/sup> + 528 in2<\/sup>
\nA1<\/sub> + A2<\/sub> + A3 <\/sub>\u2248 1,883 in2<\/sup><\/p>\n

Total area of counter space in square feet:
\n1,883 in2<\/sup> (\\(\\frac{1 \\mathrm{ft}^{2}}{144 \\mathrm{in}^{2}}\\))\u224813.1 ft2<\/sup>
\nThere is approximately 13.1 ft2<\/sup> of counter space.<\/p>\n

Question 5.
\nIn addition to the kitchen renovation, the Smiths 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?
\n\"Eureka
\nAnswer:
\n\"Eureka
\nA1<\/sub> = (9 units \u2022 7 units) + 3 units2<\/sup> = 66 units2<\/sup>
\nA2<\/sub> = (6 units \u2022 7 units) – 4 units2<\/sup> = 38 units2<\/sup>
\nA3<\/sub> = (6 units \u2022 7 units) – 4 units2<\/sup> = 38 units2<\/sup>
\nA4<\/sub> = 2 units \u2022 11 units = 22 units2<\/sup>
\nA5<\/sub> = (5 units \u2022 3 units) + (4 units\u22c56 units) = 39 \u3016units2<\/sup>
\nA6<\/sub> = 5 units \u2022 8 units = 40 \u3016units2<\/sup><\/p>\n

Total area that needs carpeting:
\nA1<\/sub> + A2<\/sub> + A3<\/sub> + A4<\/sub> + A5<\/sub> + A6<\/sub> = 66 units2<\/sup> + 38 units2<\/sup> + 38 units2<\/sup> + 22 units2<\/sup> + 39 units2<\/sup> + 40 units2<\/sup>
\n= 243 units2<\/sup><\/p>\n

Scale: 1 unit = 2 ft.; 1 unit2<\/sup> = 4 ft2<\/sup>
\nTotal area that needs carpeting in square feet:
\n243 units2<\/sup> (\\(\\frac{4 \\mathrm{ft}^{2}}{1 \\mathrm{unit}^{2}}\\)) = 972 ft2<\/sup><\/p>\n

Question 6.
\nJamie wants to wrap a rectangular sheet of paper completely around cans that are 8\\(\\frac{1}{2}\\) in. high and 4 in. in diameter. She can buy a roll of paper that is 8\\(\\frac{1}{2}\\) in. wide and 60 ft. long. How many cans will this much paper wrap?
\nAnswer:
\nA can with a 4 – inch diameter has a circumference of 4\u03c0 in. (i.e., approximately 12.57 in.). 60 ft. is the same as 720 in.; 720 in. \u00f7 12.57 in. is approximately 57.3 in., so this paper will cover 57 cans.<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 20 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nA homeowner called in a painter to paint the 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<\/sup>. A hired painter claims he will need 4 gallons. Show that the estimate is too high.
\nAnswer:
\nArea of 2 walls: 2(18 ft. \u2022 8 ft.) = 288 ft2<\/sup>
\nArea of remaining 2 walls: 2(12 ft. \u2022 8 ft.) = 192 ft2<\/sup>
\nArea of ceiling: 18 ft. \u2022 12 ft. = 216 ft2<\/sup>
\nArea of 2 doors: 2(3 ft. \u2022 7 ft.) = 42 ft2<\/sup>
\nArea of 3 windows: 3(3 ft. \u2022 5 ft.) = 45 ft2<\/sup>
\nArea to be painted: (288 ft2<\/sup> + 192 ft2<\/sup> + 216 ft2<\/sup>) – (42 ft2<\/sup> + 45 ft2<\/sup>) = 609 ft2<\/sup>
\nGallons of paint needed: 609 \u00f7 300 = 2.03; The painter will need a little more than 2 gal.
\nThe painter\u2019s estimate for how much paint is necessary is too high.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 6 Lesson 20 Answer Key Eureka Math Grade 7 Module 6 Lesson 20 Example Answer Key Example 1. A landscape company wants to plant lawn seed. A 20 lb. bag of lawn seed will cover up to 420 sq.ft. of grass and costs $49.98 plus the 8% sales … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72267"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=72267"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72267\/revisions"}],"predecessor-version":[{"id":92188,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72267\/revisions\/92188"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=72267"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=72267"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=72267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}