## 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 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?

Answer:

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.

→ Find the non – grassy sections in the map of the yard and their areas.

A_{1} = 4 units • 4 units = 16 units^{2}

A_{2} = 1 units • 13 units = 13 units^{2}

A_{3} = 7 units • 13 units = 91 units^{2}

A_{4} = 1 units • 6 units = 6 units^{2}

A_{5} = 6 units • 1 units = 6 units^{2}

A_{6} = 1 units • 6 units = 6 units^{2}

A_{7} = 2 units • 1 units = 2 units^{2}

→ What is the total area of the non – grassy sections?

A_{1} + A_{2} + A_{3} + A_{4} + A_{5} + A_{6} + A_{7} = 16 units^{2} + 13 units^{2} + 91 units^{2} + 6 units^{2} + 6 units^{2} + 6 units^{2} + 2 units^{2} = 140 units^{2}

→ What is the area of the grassy section of the yard?

Subtract the area of the non – grassy sections from the area of the yard.

A = (20 units • 18 units) – 140 units^{2} = 220 units^{2}

→ What is the scale of the map of the yard?

The scale of the map is 5 ft.

What is the grassy area in square feet?

220 units^{2} • 25 (ft^{2})/(units^{2} ) = 5,500 ft^{2}

The area of the grassy space is 5,500 ft^{2}.

→ 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.

→ Grassy area ÷ area that one bag of seed covers

5,500 ÷ 420

→ How many bags are needed to cover the grass in the yard?

5,500 ÷ 420≈13.1

It will take 14 bags to seed the yard.

→ What is the final cost of seeding the yard?

1.08 • 14 • $49.98≈$755.70

The final cost with sales tax is $755.70.

Example 2.

Ten 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 ft^{2}. Is there enough paint of each color to create all ten targets?

Answer:

Let each circle be labeled as in the diagram.

Radius of C_{1} is 3 in.; area of C_{1} is 9π in^{2}.

Radius of C_{2} is 6 in.; area of C_{2} is 36π in^{2}.

Radius of C_{3} is 9 in.; area of C_{3} is 81π in^{2}.

Radius of C_{4} is 12 in.; area of C_{4} is 144π in^{2}.

→ Write a numerical expression that represents the area painted red. Explain how your expression represents the situation.

→ The area of red and white paint in square inches is found by finding the area between circles of the target board.

Red paint: (144π in^{2} – 81π in^{2}) + (36π in^{2} – 9π in^{2})

White paint: (81π in^{2} – 36π in^{2}) + 9π in^{2}

The following calculations demonstrate how to find the area of red and white paint in the target.

Target area painted red

The area between C_{4} and C_{3}: 144π in^{2} – 81π in^{2} = 63π in^{2}

The area between C_{2} and C_{1}: 36π in^{2} – 9π in^{2} = 27π in^{2}

Area painted red in one target: 63π in^{2} + 27π in^{2} = 90π in^{2}; approximately 282.7 in^{2}

Area of red paint for one target in square feet: 282.7 in^{2} (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈1.96 ft^{2}

Area to be painted red for ten targets in square feet: 1.96 ft^{2}×10 = 19.6 ft^{2}

Target area painted white

The area between C_{3} and C_{2}: 81π in^{2} – 36π in^{2} = 45π in^{2}

The area of C_{1}: 9π in^{2}

Area painted white in one target: 45π in^{2} + 9π in^{2} = 54π in^{2}; approximately 169.6 in^{2}

Area of white paint for one target in square feet: 169.6 in^{2} (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈1.18 ft^{2}

Area of white paint for ten targets in square feet: 1.18 ft^{2}×10 = 11.8 ft^{2}

There 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.

### Eureka Math Grade 7 Module 6 Lesson 20 Exercise Answer Key

Opening Exercise

Find the area of each shape based on the provided measurements. Explain how you found each area.

Answer:

Triangular region: half base times height

area = \(\frac{1}{2}\) • 10 units • 7.5 units

= 37.5 units^{2}

Parallelogram: base times height

area = 12 units • 9 units

= 108 units^{2}

Regular hexagon: area of the shown triangle times six for the six triangles that fit into the hexagon

area = 6(\(\frac{1}{2}\)6 units • 5.2 units)

= 93.6 units^{2}

Semicircle: half the area of a circle with the same radius

area = \(\frac{\pi}{2}\) (4.5 unit)^{2}

= 10.125π units^{2}

≈31.81 units^{2}

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 100 ft. × 35 ft. The contractor comes up with 5,500 ft^{2}. How close is this estimate?

Answer:

The entire yard (home and garage) has an area of 100 ft.×35 ft. = 3,500 ft^{2}. The contractor’s estimate is 5,500 ft^{2}. He is 2,000 ft^{2} over the actual area, which is quite a bit more (2,000 ft^{2} is roughly 57% of the actual area).

### Eureka Math Grade 7 Module 6 Lesson 20 Problem Set Answer Key

Question 1.

A 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 ft^{2}, how much per square foot for the extra land did the farmer pay rounded to the nearest cent?

Answer:

A_{1} = \(\frac{1}{2}\) (6 units∙4 units) = 12 units^{2}

A_{2} = \(\frac{1}{2}\) (6 units + 7 units)(4 units) = 26 units^{2}

A_{3} = (3 units • 6 units) + (3 units • 5 units) = 33 units^{2}

A_{4} = (4 units • 7 units) + (3 units • 3 units) + \(\frac{1}{2}\)(3 units∙4 units) = 43 units^{2}

The sum of the farmer’s four pieces of land:

A_{1} + A_{2} + A_{3} + A_{4} = 12 units^{2} + 26 units^{2} + 33 units^{2} + 43 units^{2} = 114 units^{2}

The sum of the farmer’s four pieces of land in square feet:

6 units = 300 ft.; divide each side by 6.

1 unit = 50 ft. and 1 unit^{2} = 2,500 ft^{2}

114 • 2,500 = 285,000

The total area of the farmer’s four pieces of land: 285,000 ft^{2}.

The sum of the farmer’s four pieces of land in acres:

285,000 ÷ 43,560≈6.54

The farmer’s four pieces of land total about 6.54 acres.

Extra land purchased with $10,000: 8 acres – 6.54 acres = 1.46 acres

Extra land in square feet:

1.46 acres(\(\frac{43,560 \mathrm{ft}^{2}}{1 \text { acre }}\))≈63597.6 ft^{2}

Price per square foot for extra land:

(\(\frac{\$ 10,000}{63,597.6 \mathrm{ft}^{2}}\))≈$0.16

Question 2.

An 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?

Answer:

At $10 for each foot, $10,000 would purchase 1,000 feet of fencing. The perimeter of the third piece of land (labeled A_{3}) 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.

Question 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 cover 500 ft^{2}, can he paint 100 stop signs? Explain your answer.

Answer:

Area of top trapezoid = \(\frac{1}{2}\)(12 in. + 29 in.)(8.5 in.) = 174.25 in^{2}

Area of middle rectangle = 12 in. • 29 in. = 348 in^{2}

Area of bottom trapezoid = \(\frac{1}{2}\)(12 in. + 29 in.)(8.5 in.) = 174.25 in^{2}

Total area of stop sign in square inches:

A_{1} + A_{2} + A_{3} = 174.25 in^{2} + 348 in^{2} + 174.25 in^{2} = 696.5 in^{2}

Total area of stop sign in square feet:

696.5 in ^{2} (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈4.84 ft ^{2}

Yes, the area of one stop sign is less than 5 ft ^{2} (approximately 4.84 ft ^{2}). Therefore, 100 stop signs would be less than 500 ft ^{2}.

Question 4.

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?

Answer:

A_{1} = (20 in. + 16 in.)(18 in. + 14 in.) = 1,152 in^{2}

A_{2} = (18 in. • 7 in.) + 1/2 (49π in^{2} )

≈(126 in^{2} + 77 in^{2} )

≈203 in^{2}

A_{3} = (50 in. • 16 in.) – (17 in. • 16 in.) = 528 in^{2}

Total area of counter space in square inches:

A_{1} + A_{2} + A_{3 }≈ 1,152 in^{2} + 203 in^{2} + 528 in^{2}

A_{1} + A_{2} + A_{3 }≈ 1,883 in^{2}

Total area of counter space in square feet:

1,883 in^{2} (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈13.1 ft^{2}

There is approximately 13.1 ft^{2} of counter space.

Question 5.

In 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?

Answer:

A_{1} = (9 units • 7 units) + 3 units^{2} = 66 units^{2}

A_{2} = (6 units • 7 units) – 4 units^{2} = 38 units^{2}

A_{3} = (6 units • 7 units) – 4 units^{2} = 38 units^{2}

A_{4} = 2 units • 11 units = 22 units^{2}

A_{5} = (5 units • 3 units) + (4 units⋅6 units) = 39 〖units^{2}

A_{6} = 5 units • 8 units = 40 〖units^{2}

Total area that needs carpeting:

A_{1} + A_{2} + A_{3} + A_{4} + A_{5} + A_{6} = 66 units^{2} + 38 units^{2} + 38 units^{2} + 22 units^{2} + 39 units^{2} + 40 units^{2}

= 243 units^{2}

Scale: 1 unit = 2 ft.; 1 unit^{2} = 4 ft^{2}

Total area that needs carpeting in square feet:

243 units^{2} (\(\frac{4 \mathrm{ft}^{2}}{1 \mathrm{unit}^{2}}\)) = 972 ft^{2}

Question 6.

Jamie 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?

Answer:

A can with a 4 – inch diameter has a circumference of 4π in. (i.e., approximately 12.57 in.). 60 ft. is the same as 720 in.; 720 in. ÷ 12.57 in. is approximately 57.3 in., so this paper will cover 57 cans.

### Eureka Math Grade 7 Module 6 Lesson 20 Exit Ticket Answer Key

Question 1.

A 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 ft^{2}. A hired painter claims he will need 4 gallons. Show that the estimate is too high.

Answer:

Area of 2 walls: 2(18 ft. • 8 ft.) = 288 ft^{2}

Area of remaining 2 walls: 2(12 ft. • 8 ft.) = 192 ft^{2}

Area of ceiling: 18 ft. • 12 ft. = 216 ft^{2}

Area of 2 doors: 2(3 ft. • 7 ft.) = 42 ft^{2}

Area of 3 windows: 3(3 ft. • 5 ft.) = 45 ft^{2}

Area to be painted: (288 ft^{2} + 192 ft^{2} + 216 ft^{2}) – (42 ft^{2} + 45 ft^{2}) = 609 ft^{2}

Gallons of paint needed: 609 ÷ 300 = 2.03; The painter will need a little more than 2 gal.

The painter’s estimate for how much paint is necessary is too high.