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Eureka Math Grade 3 Module 7 Lesson 21 Answer Key

Engage ny eureka math 3rd grade module 7 lesson 21 answer key, eureka math grade 3 module 7 lesson 21 sprint answer key.

Question 1. 2 × 3 = Answer: 2 × 3 = 6.

Question 2. 3 × 3 = Answer: 3 × 3 = 9.

Question 3. 4 × 3 = Answer: 4 × 3 = 12.

Question 4. 5 × 3 = Answer: 5 × 3 = 15.

Question 5. 1 × 3 = Answer: 1 × 3 = 3.

Question 6. 6 ÷ 3 = Answer: 6 ÷ 3 = 2.

Question 7. 9 ÷ 3 = Answer: 9 ÷ 3 = 3.

Question 8. 15 ÷ 3 = Answer: 15 ÷ 3 = 5.

Question 9. 3 ÷ 3 = Answer: 3 ÷ 3 = 1.

Question 10. 12 ÷ 3 = Answer: 12 ÷ 3 = 4.

Question 11. 6 × 3 = Answer: 6 × 3 = 18.

Question 12. 7 × 3 = Answer: 7 × 3 = 21.

Question 13. 8 × 3 = Answer: 8 × 3 = 24.

Question 14. 9 × 3 = Answer: 9 × 3 = 27.

Question 15. 10 × 3 = Answer: 10 × 3 = 30.

Question 16. 24 ÷ 3 = Answer: 24 ÷ 3 = 8.

Question 17. 21 ÷ 3 = Answer: 21 ÷ 3 = 7.

Question 18. 27 ÷ 3 = Answer: 27 ÷ 3 = 9.

Question 19. 18 ÷ 3 = Answer: 18 ÷ 3 = 6.

Question 20. 30 ÷ 3 = Answer: 30 ÷ 3 = 10.

Question 21. ___ × 3 = 15 Answer: 5 × 3 = 15.

Question 22. ___ × 3 = 3 Answer: 1 × 3 = 3.

Question 23. ___ × 3 = 30 Answer: 10 × 3 = 30.

Question 24. ___ × 3 = 6 Answer: 2 × 3 = 6.

Question 25. ___ × 3 = 9 Answer: 3 × 3 = 9.

Question 26. 30 ÷ 3 = Answer: 30 ÷ 3 = 10.

Question 27. 15 ÷ 3 = Answer: 15 ÷ 3 = 5.

Question 28. 3 ÷ 3 = Answer: 3 ÷ 3 = 1.

Question 29. 6 ÷ 3 = Answer: 6 ÷ 3 = 2.

Question 30. 9 ÷ 3 = Answer: 9 ÷ 3 = 3.

Question 31. ___ × 3 = 18 Answer: 6 × 3 = 18.

Question 32. ___ × 3 = 21 Answer: 7 × 3 = 21.

Question 33. ___ × 3 = 27 Answer: 9 × 3 = 27.

Question 34. ___ × 3 = 24 Answer: 8 × 3 = 24.

Question 35. 21 ÷ 3 = Answer: 21 ÷ 3 = 7.

Question 36. 27 ÷ 3 = Answer: 27 ÷ 3 = 9.

Question 37. 18 ÷ 3 = Answer: 18 ÷ 3 = 6.

Question 38. 24 ÷ 3 = Answer: 24 ÷ 3 = 8.

Question 39. 11 × 3 = Answer: 11 × 3 = 33.

Question 40. 33 ÷ 3 = Answer: 33 ÷ 3 = 11.

Question 41. 12 × 3 = Answer: 12 × 3 = 36.

Question 42. 36 ÷ 3 = Answer: 36 ÷ 3 = 12.

Question 43. 13 × 3 = Answer: 13 × 3 = 39.

Question 44. 39 ÷ 3 = Answer: 39 ÷ 3 = 13.

Question 1. 1 × 3 = Answer: 1 × 3 = 3.

Question 2. 2 × 3 = Answer: 2 × 3 = 6.

Question 3. 3 × 3 = Answer: 3 × 3 = 9.

Question 4. 4 × 3 = Answer: 4 × 3 = 12.

Question 5. 5 × 3 = Answer: 5 × 3 = 15.

Question 6. 9 ÷ 3 = Answer: 9 ÷ 3 = 3.

Question 7. 6 ÷ 3 = Answer: 6 ÷ 3 = 2.

Question 8. 12 ÷ 3 = Answer: 12 ÷ 3 = 4.

Question 10. 15 ÷ 3 = Answer: 15 ÷ 3 = 5.

Question 11. 10 × 3 = Answer: 10 × 3 = 30.

Question 12. 6 × 3 = Answer: 6 × 3 = 18.

Question 13. 7 × 3 = Answer: 7 × 3 = 21.

Question 14. 8 × 3 = Answer: 8 × 3 = 24.

Question 15. 9 × 3 = Answer: 9 × 3 = 27.

Question 16. 21 ÷ 3 = Answer: 21 ÷ 3 = 7.

Question 17. 18 ÷ 3 = Answer: 18 ÷ 3 = 6.

Question 18. 24 ÷ 3 = Answer: 24 ÷ 3 = 8.

Question 19. 30 ÷ 3 = Answer: 30 ÷ 3 = 10.

Question 20. 27 ÷ 3 = Answer: 27 ÷ 3 = 9.

Question 21. ___ × 3 = 3 Answer: 1 × 3 = 3.

Question 22. ___ × 3 = 15 Answer: 5 × 3 = 15.

Question 23. ___ × 3 = 6 Answer: 2 × 3 = 6.

Question 24. ___ × 3 = 30 Answer: 10 × 3 = 30.

Question 26. 6 ÷ 3 = Answer: 6 ÷ 3 = 2.

Question 27. 3 ÷ 3 = Answer: 3 ÷ 3 = 1.

Question 28. 30 ÷ 3 = Answer: 30 ÷ 3 = 10.

Question 29. 15 ÷ 3 = Answer: 15 ÷ 3 = 5.

Question 32. ___ × 3 = 24 Answer: 8 × 3 = 24.

Question 34. ___ × 3 = 21 Answer: 7 × 3 = 21.

Question 35. 24 ÷ 3 = Answer: 24 ÷ 3 = 8.

Question 38. 21 ÷ 3 = Answer: 21 ÷ 3 = 7.

Eureka Math Grade 3 Module 7 Lesson 21 Problem Set Answer Key

Figure 2 : EFGH Rectangle Length of the EFGH Rectangle = 6cm Width of the EFGH Rectangle = 2cm Area of the EFGH Rectangle = Length  × Width = 6cm × 2cm = 12 square cm.

Figure 3 : IJKL Rectangle Length of the IJKL Rectangle = 5cm Width of the IJKL Rectangle = 3cm Area of the IJKL Rectangle = Length  × Width = 5cm × 3cm = 15 square cm.

Figure 1 : ABCD Rectangle Length of the ABCD Rectangle = 8cm Width of the ABCD Rectangle = 1cm Area of the ABCD Rectangle = Length  × Width = 8cm × 1cm = 8 square cm.

Figure 2 : EFGH Rectangle Length of the EFGH Rectangle = 7cm Width of the EFGH Rectangle = 2cm Area of the EFGH Rectangle = Length  × Width = 7cm × 2cm = 14 square cm.

Figure 3 : IJKL Rectangle Length of the IJKL Rectangle = 6cm Width of the IJKL Rectangle = 3cm Area of the IJKL Rectangle = Length  × Width = 6cm × 3cm = 18 square cm.

Figure 4 : MNOP Rectangle Length of the MNOP Rectangle = 5cm Width of the MNOP Rectangle = 4cm Area of the MNOP Rectangle = Length  × Width = 5cm × 4cm = 20 square cm.

Question 3. Use centimeter grid paper to shade in as many rectangles as you can with the given perimeters. a. Use the charts below to show how many rectangles you shaded for each given perimeter. You might not use all the spaces in the charts.

b. Did you make a square with either of the given perimeters? How do you know? Answer: Yes, I  have made a square with lengths of the given perimeters of 20cm. I know it  because a rectangle with the side length all  equal to 5cm and a rectangle with all equal sides is to be known as Square.

Explanation: Yes, because a rectangle with side length all  equal to 5cm and a rectangle with all equal sides is to be known as Square.

Figure2: Gavin’s rectangle: Length of the EFGH Gavin’s rectangle = 7cm Width of the EFGH Gavin’s rectangle = 1cm Area of the EFGH Gavin’s rectangle = Length × Width = 6cm ×  2cm = 12 Square cm. Perimeter of the EFGH Gavin’s rectangle =  2 ( Length + Width ) = 2 ( 6cm + 2cm) = 2 ×  8cm = 16 cms .

Eureka Math Grade 3 Module 7 Lesson 21 Exit Ticket Answer Key

Explanation: I have drawn three rectangles having lengths as ABCD Rectangle ( 8cm,2cm) , EFGH Rectangle  ( 6cm,4cm)  and IJKL Rectangle ( 7cm,3cm).

Eureka Math Grade 3 Module 7 Lesson 21 Homework Answer Key

b. Area of the ABCD Rectangle = 12 square cm. Area of the EFGH Rectangle = 10 square cm. Area of the IJKL Rectangle = 6 square cm.

c. Perimeters of the three rectangles are the same having 14 cms, where as their areas are completely different from each other.

b. Figure 1 : ABCD Rectangle Length of the ABCD Rectangle = 4cm Width of the ABCD Rectangle = 3cm Area of the ABCD Rectangle = Length  × Width = 4cm × 3cm = 12 square cm.

Figure 2 : EFGH Rectangle Length of the EFGH Rectangle = 5cm Width of the EFGH Rectangle = 2cm Area of the EFGH Rectangle = Length  × Width = 5cm × 2cm = 10 square cm.

Figure 3 : IJKL Rectangle Length of the IJKL Rectangle = 6cm Width of the IJKL Rectangle = 1cm Area of the IJKL Rectangle = Length  × Width = 6cm × 1cm = 6 square cm.

c. Perimeter of the ABCD  rectangle =  2 ( Length + Width ) = 2 ( 4cm + 3cm) = 2 ×  7cm = 14 cms.

Perimeter of the EFGH rectangle =  2 ( Length + Width ) = 2 ( 5cm + 2cm) = 2 ×  7cm = 14 cms.

Perimeter of the IJKL rectangle =  2 ( Length + Width ) = 2 ( 6cm + 1cm) = 2 ×  7cm = 14 cms. Perimeters of the three rectangles are the same having 14 cms, where as their areas are completely different from each other. Perimeter requires only the lengths of the four sides of rectangle to be added where as for area we require length and width of the rectangle.

Question 2. Tanner uses unit squares to build rectangles that have a perimeter of 18 units. He creates the chart below to record his findings. a. Complete Tanner’s chart. You might not use all the spaces in the chart.

b. Explain how you found the widths and lengths in the chart above. Answer: The chart in itself has the labelling heading of the columns as length and width and their areas. If not always length value is going to be more than the value of width in the measurements of rectangle.

Explanation: The chart in itself has the labelling heading of the columns as length and width and their areas. If not always length value is going to be more than the value of width in the measurements of rectangle. This is how I found the widths and lengths in the chart above.

Explanation: Jason’s ABCD Rectangle: Length of the Jason’s ABCD Rectangle = 4cm Width of the Jason’s ABCD Rectangle = 2cm Area of the Jason’s ABCD Rectangle = Length × Width = 4cm ×  2cm = 8 square cm. Perimeter of the Jason’s ABCD Rectangle = 2 ( Length + Width ) = 2 ( 4cm + 2cm ) = 2 × 6cm = 12 cm.

Dina’s EFGH Rectangle: Length of the Dina’s EFGH Rectangle = 5cm Width of the Dina’s EFGH Rectangle = 1cm Area of the Dina’s EFGH Rectangle = Length × Width = 5cm ×  1cm = 5 square cm. Perimeter of the Dina’s EFGH Rectangle = 2 ( Length + Width ) = 2 ( 5cm + 1cm ) = 2 × 6cm = 12 cm. Jason’s ABCD rectangle and Dina’s EFGH rectangle are having same perimeter yet different areas.

Eureka Math Grade 3 Module 7 Lesson 21 Data Sheet Answer Key

Use the data you gathered from Problem Sets 20 and 21 to complete the charts to show how many rectangles you can create with a given perimeter. You might not use all the spaces in the charts.

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