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Eureka Math Grade 4 Module 3 Lesson 2 Answer Key

Engage ny eureka math 4th grade module 3 lesson 2 answer key, eureka math grade 4 module 3 lesson 2 problem set answer key.

b. Find the perimeter of the porch. Answer: Perimeter of the porch is 32 feet,

Explanation: Given width = 4 feet and long is 12 feet, so perimeter of rectangular porch is 2 × ( 12 feet + 4 feet) = 2 × (16 feet) = 32 feet.

b. Find the perimeter and area of the banner. Answer: Perimeter of the banner : 70 inches, Area of the banner : 150 inches 2 ,

Explanation: Given a narrow rectangular banner is 5 inches wide. It is 6 times as long as it is wide. So wide = 5 inches, long = 6 X 5 inches = 30 inches, Perimeter of the banner = 2 X (l + w) = 2 X (30 inches + 5 inches) = 2 X ( 35 inches) = 70 inches and area of the banner = l X w = 30 inches X 5 inches = 150 square inches.

Question 3. The area of a rectangle is 42 square centimeters. Its length is 7 centimeters. a. What is the width of the rectangle? Answer: The width of the rectangle is 6 centimeters,

Explanation: Given The area of a rectangle is 42 square centimeters. Its length is 7 centimeters. So let us take width as x cm, so 42 sq cm = 7 cm X x cm, x cm = 42 sq cm X cm ÷ 7 cm = 6 cm, therefore, the width of the rectangle is 6 centimeters.

Explanation: Given Charlie wants to draw a second rectangle that is the same length but is 3 times as wide means length is 7 cm and wide is 3 × 7 cm = 21 cm, Drawn and labeled Charlie’s second rectangle as shown above.

c. What is the perimeter of Charlie’s second rectangle? Answer: Perimeter of Charlie’s second rectangle is 56 centimeters,

Explanation: Given Charlie’s second rectangle has length 7 cm and width as 21 cm so perimeter is 2 X ( 7 cm + 21 cm) = 2 × (28 cm) = 56 cm, therefore perimeter of Charlie’s second rectangle is 56 centimeters.

c. What is the relationship between the two perimeters? Answer: The perimeter of sandbox at the park is twice the perimeter of Betsy’s sandbox,

Explanation: As both length and width of sandbox at park is twice of Betsy’s sandbox, so their perimeter of sandbox at park is twice the perimeter of Betsy’s sandbox.

d. Find the area of the park’s sandbox using the formula A = l × w. Answer: Area of the park’s sandbox is 80 square feet,

Explanation: As sandbox at park has 10 feet long and 8 feet wide so area of sandbox at park is 10 feet X 8 feet = 80 square feet.

e. The sandbox at the park has an area that is how many times that of Betsy’s sandbox? Answer: The sandbox at the park has an area that is four times that of Betsy’s sandbox,

Explanation: Given area of Betsy’s rectangular sandbox is 20 square feet and area of the park’s sandbox is 80 square feet, so number of times more is 80 square feet ÷ 20 square feet = 4, therefore the sandbox at the park has an area that is four times that of Betsy’s sandbox.

f. Compare how the perimeter changed with how the area changed between the two sandboxes. Explain what you notice using words, pictures, or numbers. Answer: Sand box at park has twice perimeter and area has became four times that of Betsy’s sandbox,

Explanation: As we know Perimeter of Betsy’s sandbox is 18 feet and perimeter of the sand box at park is 36 feet, Area of Betsy’s rectangular sandbox is 20 square feet and area of the park’s sandbox is 80 square feet, Now on comparing Sand box at park has twice perimeter and area has became four times to that of Betsy’s sandbox.

Eureka Math Grade 4 Module 3 Lesson 2 Exit Ticket Answer Key

b. Find the perimeter of the table. Answer: Perimeter of the table is 28 feet,

Explanation: Given wide is 2 feet and long is 12 feet, So Perimeter of the table is 2 × ( 12 feet + 2 feet) = 2 × 14 feet = 28 feet.

b. Find the perimeter and area of the blanket. Answer: Perimeter is 32 feet and area of the blanket is 48 square feet,

Explanation: Given wide as 4 feet and long is 12 feet, So perimeter of the blanket is 2 × (12 feet + 4 feet) = 2 × (16 feet) = 32 feet and area is 12 feet × 4 feet = 48 square feet.

Eureka Math Grade 4 Module 3 Lesson 2 Homework Answer Key

b. Find the perimeter of the pool. Answer: Perimeter of the rectangular pool is 56 feet,

Explanation: Given rectangular pool is 7 feet wide and 21 feet long, So perimeter is 2 X (21 feet + 7 feet) = 2 X 28 feet = 56 feet.

b. Find the perimeter and area of the poster. Answer: Perimeter of the poster is 30 inches and area of the poster is 36 square inches,

Explanation: Given poster as 3 inches long and 4 X 3 inches = 12 inches wide, So perimeter of the poster is 2 X ( 3 inches + 12 inches) = 2 × 15 inches = 30 inches and area of the poster is 3 inches × 12 inches = 36 square inches.

Question 3. The area of a rectangle is 36 square centimeters, and its length is 9 centimeters. a. What is the width of the rectangle? Answer: The width of the rectangle is 4 centimeters,

Explanation: Given the area of a rectangle is 36 square centimeters and its length is 9 centimeters so width of the rectangle is 36 sq cm = 9 cm X width , width = 36 cm X cm ÷ 9 cm = 4 cm, therefore the width of the rectangle is 4 centimeters.

c. What is the perimeter of Elsa’s second rectangle? Answer: Perimeter of Elsa’s second rectangle is 72 centimeters,

Explanation: Given Elsa’s second rectangle is with length 9 cm and wide is 27 cm, So perimeter is 2 X (9 cm + 27 cm) = 2 × (36 cm) = 72 centimeters. therefore, Perimeter of Elsa’s second rectangle is 72 centimeters.

Explanation: Given the area of Nathan’s bedroom rug is 15 square feet. The longer side measures 5 feet so width is 15 feet × feet ÷ 5 feet = 3 feet, now preimeter of Nathan’s bedroom rug is 2 X (5 feet + 3 feet) = 2 × 8 feet = 16 feet, Drawn and labeled a diagram of Nathan’s bedroom rug as shown above.

Explanation: Drawn and labeled a diagram of Nathan’s living room rug as shown above with long as 2 × 5 feet = 10 feet and width as 2 × 3 feet = 6 feet. Perimeter of Nathan’s living room rug is 2 × (10 feet + 6 feet) = 2 × 16 feet = 32 feet.

c. What is the relationship between the two perimeters? Answer: The perimeter of Nathan’s living room rug is twice the perimeter of Nathan’s bedroom rug,

Explanation: As perimeter of Nathan’s bedroom rug is 16 feet and perimeter of Nathan’s living room rug is 32 feet and as both length and width of Nathan’s living room rug is twice of Nathan’s bedroom rug.

d. Find the area of the living room rug using the formula A = l × w. Answer: The area of the living room rug is 60 square feet,

Explanation: Given living room rug has long 10 feet and width 6 feet, So area of the living room rug is 10 feet X 6 feet = 60 square feet.

e. The living room rug has an area that is how many times that of the bedroom rug? Answer: The living room rug has an area that is four times that of the bedroom rug,

Explanation: The area of the living room rug is 60 square feet and the area of Nathan’s bedroom rug is 15 square feet. So number of times the area of the living room rug to the area of Nathan’s bedroom rug is 60 sq feet ÷ 15 square feet = 4, Therefore, the living room rug has an area that is four times that of the bedroom rug.

f. Compare how the perimeter changed with how the area changed between the two rugs. Explain what you notice using words, pictures, or numbers. Answer: Nathan’s living room has twice perimeter and area has became four times that of Nathan’s bedroom rug,

Explanation: As we know Perimeter of  is Nathan’s bedroom rug 16 feet and perimeter of Nathan’s living room is 32 feet, Area of Nathan’s bedroom rug is 15 square feet and area of Nathan’s living room is 60 square feet, Now on comparing Nathan’s living room has twice perimeter and area has became four times to that of Nathan’s bedroom rug.

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Home > CC3 > Chapter 3 > Lesson 3.2.4

Lesson 3.1.1, lesson 3.1.2, lesson 3.1.3, lesson 3.1.4, lesson 3.1.5, lesson 3.1.6, lesson 3.1.7, lesson 3.2.1, lesson 3.2.2, lesson 3.2.3, lesson 3.2.4, lesson 3.2.5.

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