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Common Core Grade 3 Math (Worksheets, Homework, Lesson Plans)

Looking for video lessons that will help you in your Common Core Grade 3 Math classwork or homework? Looking for Common Core Math Worksheets and Lesson Plans that will help you prepare lessons for Grade 3 students?

The following lesson plans and worksheets are from the New York State Education Department Common Core-aligned educational resources. The Lesson Plans and Worksheets are divided into seven modules.

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Grade 3 Homework, Lesson Plans And Worksheets

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Math Expressions 3, Volume 2, Grade: 3 Publisher: Houghton Mifflin Harcourt

Math expressions 3, volume 2, title : math expressions 3, volume 2, publisher : houghton mifflin harcourt, isbn : 547060726, isbn-13 : 9780547060729, use the table below to find videos, mobile apps, worksheets and lessons that supplement math expressions 3, volume 2., textbook resources.

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Math Expressions Grade 3 Unit 2 Lesson 2 Answer Key Solve Area Word Problems

Solve the questions in Math Expressions Grade 3 Homework and Remembering Answer Key Unit 2 Lesson 2 Answer Key Solve Area Word Problems to attempt the exam with higher confidence. https://mathexpressionsanswerkey.com/math-expressions-grade-3-unit-2-lesson-2-answer-key/

Math Expressions Common Core Grade 3 Unit 2 Lesson 2 Answer Key Solve Area Word Problems

Math Expressions Grade 3 Unit 2 Lesson 2 Homework

Complete each Unknown Number puzzle.

Explanation: 9 × ?? = 36 => ?? = 36 ÷ 9 => ?? = 4. 2 × 4 = 8. ?? × 4 = 12 => ?? = 12 ÷ 4 => ?? = 3. 3 × ?? = 9 => ?? = 9 ÷ 3 => ?? = 3. 9 × 3 = 27. 2 × 3 = 6. 9 × 6 = 54. 2 × 6 = 12. 3 × 6 = 18.

Explanation: ?? × 7 = 28. => ?? = 28 ÷ 7 => ?? = 4. ?? × 7 = 42. => ?? = 42 ÷ 7 => ?? = 6. ?? × 7 = 56 => ?? = 56 ÷ 7 => ?? = 8. 6 × ?? = 30. => ?? = 30 ÷ 6 => ?? = 5. 4 × 5 = 20. 8 × 5 = 40. 4 × 6 = 24. 6 × 6 = 36. 8 × 6 = 48.

Explanation: 7 × ?? = 56 => ?? = 56 ÷ 7 => ?? = 8. 5 × 8 = 40. 5 × ?? = 30. => ?? = 30 ÷ 5 => ?? = 6. 7 × 6 = 42. 5 × 4 = 20. 7 × 4 = 28. ?? × 4 = 12. => ?? = 12 ÷ 4 => ?? = 3. 8 × 3 = 24. 8 × 6 = 18.

Solve each problem. Label your answers with the correct units.

Question 4. Raul built a rectangular tabletop with a length of 3 feet and a width of 6 feet. What is the area of the tabletop? Answer: Area of the tabletop = 18 square feet.

Explanation: Length of the rectangular tabletop Raul built = 3 feet. Width of the rectangular tabletop Raul built = 6 feet. Area of the tabletop = Length of the rectangular tabletop Raul built  × Width of the rectangular tabletop Raul built = 3 × 6 = 18 square feet.

Question 5. Li Fong covered the rectangular floor of his tree house with 48 square feet of carpeting. If one side of the floor has a length of 6 feet, what is the length of the adjacent side? Answer: Length of the adjacent side = 8 feet.

Explanation: Li Fong covered the rectangular floor of his tree house with 48 square feet of carpeting. => Area of the rectangular floor Li Fong covered = 48 square feet. Length of the one side of the floor = 6 feet. Length of the adjacent side = Area of the rectangular floor Li Fong covered  ÷ Length of the one side of the floor = 48 ÷ 6 = 8 feet.

Question 6. Frances wants to paint a rectangular wall that has a width of 8 feet and a length of 9 feet. She has a quart of paint that will cover 85 square feet. What is the area of the wall? Does Frances have enough paint? Answer: Area of the wall = 72 square feet. Yes, she has (a quart of paint that will cover 85 square feet) the enough paint to paint the wall (area of the wall is 72 square feet) .

Explanation: Width of the rectangular wall Frances wants to paint = 8 feet. Length of the rectangular wall Frances wants to paint = 9 feet. Area of the quart of paint she has can cover the wall = 85 square feet. Area of the wall = Width of the rectangular wall Frances wants to paint × Length of the rectangular wall Frances wants to paint = 8 × 9 = 72 square feet.

Question 7. Willis cut out a paper rectangle with an area of 42 square centimeters. If one side has a length of 6 centimeters, what is the length of the adjacent side? Answer: Length of the adjacent side = 7 centimeters.

Explanation: Area of the paper rectangle Willis cut out = 42 square centimeters. Length of one side = 6 centimeters. Length of the adjacent side = Area of the paper rectangle Willis cut out ÷ Length of one side = 42 ÷ 6 = 7 centimeters.

Math Expressions Grade 3 Unit 2 Lesson 2 Remembering

Question 1. 3 × (5 × 1) = ___ Answer: 3 × (5 × 1) = 3 × 5 = 15.

Question 2. (2 × 5) × 3 = ___ Answer: (2 × 5) × 3 = 10 × 3 = 30.

Question 3. (0 × 4) × 9 = ___ Answer: (0 × 4) × 9 = 0 × 9 = 0.

Question 4. 22 × 1 = ___ Answer: 22 × 1 = 22.

Question 5. 4 × 7 = 7 × __ = ___ Answer: 4 × 7 = 7 × __4__ = 28.

Explanation: 4 × 7 = 7 × ?? => (4 × 7) = 7 × ?? => 28 ÷ 7 = ?? => 4 = ?? => 4 × 7 = 28. => 7 × 4 = 28. Hence, 4 × 7 = 7 × 4 = 28.

Question 6. (3 × 3) × 6 = ___ Answer: (3 × 3) × 6 = 9 × 6 = 54.

Read the problem and decide what type of problem it is. Write the letter from the list below. Then write an equation and solve the problem.

a. Array Multiplication b. Array Division c. Equal Groups of Multiplication d. Equal Division with Unknown Group Size e. Equal Division with an Unknown Multiplier (number of groups)

Question 7. Andrew has 18 invitations to write. If he writes 3 invitations a day, how many days will it take him to finish? Answer: d. Equal Division with Unknown Group Size. Number of days it takes him to finish = 6.

Explanation: Number of invitations Andrew has to write = 18. Number of invitations Andrew writes a day = 3. Number of days it takes him to finish = Number of invitations Andrew has to write ÷ Number of invitations Andrew writes a day = 18 ÷ 3 = 6.

Solve each problem.

Question 8. Brian buys 6 video games. They cost $10 each. How much does he spend on the video games? Answer: c. Equal Groups of Multiplication. Total amount of money he spends on the video games = $60.

Explanation: Number of video games Brian buys = 6. Cost of the each video games Brian buys = $10. Total amount of money he spends on the video games = Number of video games Brian buys × Cost of the each video games Brian buys = 6 × 10 = $60.

Question 9. Sharon plants 48 rose bushes. Each row has 6 rose bushes. How many rows of rose bushes does Sharon plant? Answer: b. Array Division. Number of rows of rose bushes Sharon plant = 8.

Explanation: Number of rose bushes Sharon plants = 48. Number of rose bushes in each row = 6. Number of rows of rose bushes Sharon plant = Number of rose bushes Sharon plants ÷ Number of rose bushes in each row = 48 ÷ 6 = 8.

Question 10. Stretch Your Thinking Ming’s rug has a length that is 2 times its width. The area of the rug is 8 square feet. What is the length and width of Ming’s rug? Answer: e. Equal Division with an Unknown Multiplier (number of groups) Length of the Ming’s rug  = 4 feet. Width of the Ming’s rug  = 2 feet.

Explanation: Let the width of the Ming’s rug be w feet. Ming’s rug has a length that is 2 times its width. => Length of the Ming’s rug = 2 × Width of the Ming’s rug => 2 × w => 2w. Given: Area of the rug = 8 square feet. Area of the Ming’s rug = Length of the Ming’s rug  × Width of the Ming’s rug 8 = 2w × w 8 ÷ 2= w × w 4 = 2w 2 feet = w. Length of the Ming’s rug  = 2 × w = 2 ×2 = 4 feet.

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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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