## Engage NY Eureka Math 3rd Grade Module 7 Lesson 18 Answer Key

### Eureka Math Grade 3 Module 7 Lesson 18 Problem Set Answer Key

Question 1.
Use unit squares to build as many rectangles as you can with an area of 24 square units. Shade in squares on your grid paper to represent each rectangle that you made with an area of 24 square units.
a. Estimate to draw and label the side lengths of each rectangle you built in Problem 1. Then, find the perimeter of each rectangle. One rectangle is done for you.

P = 24 units + 1 unit + 24 units + 1 unit = 50 units
b. The areas of the rectangles in part (a) above are all the same. What do you notice about the perimeters?
a. Perimeter of ABCD rectangle = 22units.
Perimeter of EFGH rectangle = 28units.
Perimeter of IJKL rectangle = 20units.

b. All rectangles drawn in the above1.a are not the same sided figures. They are not having same perimeters because their lengths are different compared to one another.

Explanation:
a.

Perimeter of ABCD rectangle =  Side + Side + Side + Side
= AB + BC + CD + DA
= 8units + 3units + 8units + 3units
= 11units + 8units + 3units
= 19units + 3units
= 22units.
Perimeter of EFGH rectangle = Side + Side + Side + Side
= EF + FG + GH + HE
= 12units + 2units + 12units + 2units
= 14units + 12units + 2units
= 26units + 2units
= 28units.
Perimeter of IJKL rectangle = Side + Side + Side + Side
= IJ + JK + KL + KI
= 6units + 4 units + 6units + 4 units
= 10units + 6units + 4units
= 16units + 4units
= 20units.

b. All rectangles drawn in the above 1a are not the same figures. They are not having same perimeters.

Question 2.
Use unit square tiles to build as many rectangles as you can with an area of 16 square units. Estimate to draw each rectangle below. Label the side lengths.
a. Find the perimeters of the rectangles you built.
b. What is the perimeter of the square? Explain how you found your answer.
a. Perimeter of OPQR Rectangle = 14units.
Perimeter of EFGH Rectangle = 34units.

b. Perimeter of ABCD Square= 16units. As, the sides in a square are equal, we can multiple the number of sides into the side value to get the Perimeter of square instead of adding them separately.

Explanation:
a.

Perimeter of OPQR Rectangle = Side + Side + Side + Side
= OP + PQ + QR + RO
= 5units + 2units + 5units + 2units
= 7units + 5units + 2units
= 12units + 2units
= 14units.
Perimeter of EFGH Rectangle = Side + Side + Side + Side
= EF + FG+ GH + HE
= 16units + 1unit + 16units + 1unit
= 17units + 16units + 1unit
= 33units + 1unit
= 34unit.

b.

Perimeter of ABCD Square= 4 × Side
= 4 × 4units
= 16units.

Question 3.
Doug uses square unit tiles to build rectangles with an area of 15 square units. He draws the rectangles as shown below but forgets to label the side lengths. Doug says that Rectangle A has a greater perimeter than Rectangle B. Do you agree? Why or why not?

Yes, Doug is correct, the perimeter of rectangle A is greater than the perimeter of the rectangle B because in the appearance itself we notice that rectangle A is bigger in size than that of rectangle B.

Explanation:

Well, comparing the rectangles drawn by Doug its easy to say rectangle A is going to have the greater perimeter that compared to the perimeter of the rectangle B because the size of the rectangle A is bigger than that of rectangle B.

### Eureka Math Grade 3 Module 7 Lesson 18 Exit Ticket Answer Key

Tessa uses square-centimeter tiles to build rectangles with an area of 12 square centimeters. She draws the rectangles as shown below. Label the unknown side lengths of each rectangle. Then, find the perimeter of each rectangle.

P = _____

P = _____

P = _____
Perimeter of the ABCD Rectangle = 26cm.
Perimeter of the EFGH Rectangle = 12cm.
Perimeter of the IJKL Rectangle = 16cm.

Explanation:

The unknown side are measured by using a ruler by me.
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 12cm + 1cm + 12cm + 1cm
= 13cm + 12cm + 1cm
= 25cm + 1cm
= 26cm.
Perimeter of the EFGH Rectangle = Side + Side + Side + Side
= EF + FG + GH+ HE
= 3cm + 3cm + 3cm + 3cm
= 6cm + 3cm + 3cm
= 9cm + 3cm
= 12cm.
Perimeter of the IJKL Rectangle = Side + Side + Side + Side
IJ + JK + KL + KI
= 6cm + 2cm + 6cm + 2cm
= 8cm + 6cm + 2cm
= 14cm + 2cm
= 16cm.

Question 1.
Shade in squares on the grid below to create as many rectangles as you can with an area of 18 square centimeters.

ABCD rectangle
EFGH rectangle
IJKL rectangle.

Explanation:

Question 2.
Find the perimeter of each rectangle in Problem 1 above.
Perimeter of the ABCD rectangle = 20units.
Perimeter of the EFGH rectangle = 12units.
Perimeter of the IJKL rectangle = 18units.

Explanation:

Perimeter of the ABCD rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 8units + 2units + 8units + 2units
= 10units + 8units + 2units
= 18units + 2units
= 20units.
Perimeter of the EFGH rectangle = Side + Side + Side + Side
= EF + FG + GH+ HE
= 4units + 2units + 4units + 2units
= 6units + 4units + 2units
= 10units + 2units
= 12units.
Perimeter of the IJKL rectangle = Side + Side + Side + Side
= IJ + JK + KL + LI
=  5units + 4units + 5units + 4units
= 9units + 5units + 4units
= 14units + 4units
= 18units.

Question 3.
Estimate to draw as many rectangles as you can with an area of 20 square centimeters. Label the side lengths of each rectangle.
a. Which rectangle above has the greatest perimeter? How do you know just by looking at its shape?
b. Which rectangle above has the smallest perimeter? How do you know just by looking at its shape?
a. Among all the three rectangles drawn, through looks  IJKL Rectangles is having greater perimeter compared to other rectangles. We can say that by seeing the length of the rectangles, which length is high that going to have greater perimeter.

b. Among all the three rectangles drawn, through looks  EFGH Rectangles is having smallest perimeter compared to other rectangles. We can say that by seeing the length of the rectangles.

Explanation:
a.

Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 7units + 4units + 7units + 4units
= 11units + 7units + 4units
= 18units + 4units
= 22units.
Perimeter of the EFGH Rectangle = Side + Side + Side + Side
= EF + FG + GH + HE
= 5units + 3units + 5units + 3units
= 8units + 5units + 3units
= 13units + 3units
= 16units.
Perimeter of the IJKL Rectangle = Side + Side + Side + Side
= IJ + JK + KL + LI
= 11units + 2units + 11units + 2units
= 13units + 11units + 2units
= 24units + 2units
= 26 units.

b. Among all the three rectangles drawn, through looks  EFGH Rectangles is having the smallest perimeter compared to other rectangles. We can say that by seeing the length of the rectangles, which length is small that going to have the smallest perimeter.