eureka math grade 6 module 1 lesson 5 homework answers

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• Grade 5 Eureka - Answer Keys Module 1

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b. 14.204 14.200 14.240 14.210

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Eureka Math Grade 5 Answer Key | Engage NY Math 5th Grade Answer Key Solutions

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• Eureka Math Grade 5 Module 1 Answer Key
• Eureka Math Grade 5 Module 2 Answer Key
• Eureka Math Grade 5 Module 3 Answer Key
• Eureka Math Grade 5 Module 4 Answer Key
• Eureka Math Grade 5 Module 5 Answer Key
• Eureka Math Grade 5 Module 6 Answer Key

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5th Grade Eureka Math Module 1 Lesson 6 PPT

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Eureka Math Grade 6 Module 5 Lesson 5 Answer Key

Engage ny eureka math grade 6 module 5 lesson 5 answer key, eureka math grade 6 module 5 lesson 5 exercise answer key.

Opening Exercise:

Here is an aerial view of a woodlot.

Question 1. If AB = 10 units, FE = 8 units, AF = 6 units, and DE = 7 units, find the lengths of the other two sides. DC = BC = Answer: DC = 2 units BC = 13 units

Question 2. If DC = 10 units, FE = 30 units, AF = 28 units, and BC = 54 units, find the lengths of the other two sides. AB = DE = Answer: AB = 40 units DE = 26 units

Eureka Math Grade 6 Module 5 Lesson 5 Example Answer Key

Example 1: Decomposing Polygons into Rectangles

The Intermediate School is producing a play that needs a special stage built. A diagram of the stage is shown below (not to scale).

Question a. On the first diagram, divide the stage into three rectangles using two horizontal lines. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage. Answer: Dimensions: 2 m by 4m, 2 m by 4 m, and 7 m by 5 m Area: 2 m × 4m = 8 m 2 , 2 m × 4 m = 8 m 2 , 7 m × 5 m = 35 m 2 Total: 8 m 2 + 8 m 2 + 35 m 2 = 51 m 2

Question b. On the second diagram, divide the stage into three rectangles using two vertical lines. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage. Answer: Dimensions: 2 m by 9 m, 2 m by 9 m, and 3 m by 5 m Area: 2 m × 9 m = 18 m × 2 m × 9 m = 18 m 2 , 3 m × 5m = 15 m 2 Total: 51 m 2

i. What are the dimensions of the large rectangle and the small rectangle? Answer: Dimensions: 9 m by 7 m and 3 m by 4 m

ii. What are the areas of the two rectangles? Answer: Area: 9 m × 7 m = 63 m 2 , 3 m × 4 m = 12 m 2

iii. What operation Is needed to find the area of the original figure? Answer: Subtraction

iv. What is the difference in area between the two rectangles? Answer: 63 m 2 – 12 m 2 = 51 m 2

v. What do you notice about your answers to (a), (b), (c), and (d)? 7 m Answer: The area is the same.

vi. Why do you think this is true? Answer: No matter how we decompose the figure, the total area is the sum of its ports. Even if we take the area around the figure and subtract the part that is not included, the area of the figure remains the same, 51 m 2 .

Area of Rectangle 1: b . h 2m . 4m = 8m 2 Area of Rectangle 2: b . h 2 m 4m = 8 m 2 Area of Rectangle 3: b . h 7m 5m 35 m 2 Area of Polygon: 8m 2 + 8m 2 + 35 m 2 = 51 m 2

Area of Rectangle 1: b . h 9m . 2m = 18 m 2 Area of Rectangle 2: b . h 9m . 2m =18 m 2 Area of Rectangle 3: b . h 3m . 5m = 15 m 2 Area of Polygon: 18 m 2 + 18 m 2 + 15 m 2 = 51 m 2

Area of Rectangle: b . h 9m . 7 m = 63 m 2 Area of Missing Rectangle 2: b . h 3m . 4m = 12 m 2 Area of Polygon: 63 m 2 – 12 m 2 = 51 m 2

Example 2: Decomposing Polygons into Rectangles and Triangles

Triangle ABC A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) (6 m) (4 m) A = 12 m 2 .

Triangle ACD A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) (6m)(4m) A = 12 m 2

Triangle ABD A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) (8m)(5m) A = 20 m 2

Triangle BCD A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) (4 m) (5 m) A = 10 m 2

A = 20 m 2 + 10 m 2 = 30 m 2

Area of Rectangle; A = bh A = (8 m) (5 m) A = 40 m 2

Triangle 1 A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) (3 m) (5 m) A = 7.5 m 2

Triangle 2 A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) (1 m) (5 m) A = 2.5 m 2

A = 40 m 2 – 7.5 m 2 – 2.5 m 2 = 30 m 2 (or)

A = 40 m 2 – (7.5 m 2 + 2.5 m 2 ) = 30 m 2

Eureka Math Grade 6 Module 5 Lesson 5 Problem Set Answer Key

Area of Triangle 2 A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) × 3 m × 18 m A = 27 m 2

Area of Trapezoid = Area of Triangle 1 + Area of Triangle 2 Area = 198 m 2 + 27 m 2 = 225 m 2

Area of Triangles 1 and 2 A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) × 7.5 m × 12 m A = 45 m 2

Area of Trapezoid = Area of Rectangle – Area of Triangle 1 – Area of Triangle 2 A = 216 m 2 – 45 m 2 – 45 m 2 = 126 m 2

a. The windows and door will not be painted. Calculate the area of the wall that will be painted. Answer: Whole wall: 12 ft. × 8 ft. = 96 ft 2 Window: 2 ft. × 2 ft. = 4 ft 2 There are two identical windows, 4 ft 2 × 2 = 8 ft 2 Door: 6 ft. × 3 ft. = 18 ft 2 96 ft 2 – 8 ft 2 – 18 ft 2 = 70 ft 2

b. If a quart of Extra-Thick Gooey Sparkle paint covers 30 ft 2 . how many quarts must be purchased for the painting job? Answer: 70 ÷ 30 = 2\(\frac{1}{3}\) Therefore, 3 quarts must be purchased.

Question 6. The figure below shows a floor plan of a new apartment. New carpeting has been ordered, which will cover the living room and bedroom but not the kitchen or bathroom. Determine the carpeted area by composing or decomposing in two different ways, and then explain why they are equivalent.

Alternatively, the whole apartment is 45 ft. × 35 ft. = 1,575 ft 2 Subtracting the kitchen and bath (300 ft 2 and 200 ft 2 ) still gives 1,075 ft 2 . The two areas are equivalent because they both represent the area of the living room and bedroom.

Eureka Math Grade 6 Module 5 Lesson 5 Exit Ticket Answer Key

Area of Triangle 2 A = \(\frac{1}{2}\) bh A = \(\frac{1}{2}\) × 8 mi. × 10 mi. A = 40 mi 2

Area of Parallelogram = Area of Triangle 1 + Area of Triangle 2 A = 40 mi 2 + 40 mi 2 = 80 mi 2 The area of the parallelogram is 80 mi 2 .

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