Engage NY Eureka Math 3rd Grade Module 7 End of Module Assessment Answer Key
Eureka Math Grade 3 Module 7 End of Module Assessment Task Answer Key
Question 1.
Katy and Jane construct a four-sided wall to surround their castle. The wall has a perimeter of 100 feet. One side measures 16 feet. A different side measures 16 feet. A third side measures 34 feet.
a. Draw and label a diagram of the wall. Use a letter to represent the unknown side length.
b. What is the unknown side length? Show your work, or explain how you know.
c. Katy and Jane build a square fence around the castle’s pool. It has a perimeter of 36 feet. What is the area that the fence encloses? Use a letter to represent the unknown. Show your work.
Answer:
a.
b. Length of the unknown side = 34 feet.
c.
Area of the square wall = 81 square feet.
Explanation:
a.
ABCD is the wall constructed to surround Katy and Jane castle.
b. Perimeter of the wall constructed to surround Katy and Jane castle = 100 feet
Length of the AB side of the wall constructed = 34 feet
Length of the BC side of the wall constructed = 16feet
Length of the DA side of the wall constructed = 16 feet
Let the unknown side be as X.
=> Length of the CD side of the wall constructed = X
Perimeter of the wall constructed to surround Katy and Jane castle = Length of the AB side of the wall constructed + Length of the BC side of the wall constructed + Length of the DA side of the wall constructed + Length of the CD side of the wall constructed
100 feet = 34 feet + 16 feet + 16 feet + X
=> 100 feet = 50 feet + 16 feet + X
=> 100 feet = 66 feet + X
=> 100 feet – 66 feet = X
=> 34 feet = X.
c.
Perimeter of the square Wall = 36 feet.
Length of the FG side in the square wall = Y
Perimeter of the square Wall = 4 × Side
=> 36 feet = 4 × Y
=> 36 feet ÷ 4 = Y
=> 9 feet = Y.
Area of the square wall = Side × Side
= 9 feet × 9 feet
= 81 Square Feet.
Question 2.
Each shape has a missing side length labeled with a letter. The perimeter of the shape is labeled inside. Find the unknown side length for each shape.
Answer:
Length of the unknown BC side of 1 ABCD Square Figure = 6 cm.
Length of the HE unknown side of 2 EFGH Rectangle figure = 6 cm.
Length of the unknown KL side of 3 IJKL Rhombus figure = 6 cm.
Length of the unknown MN side of 4 MNOP Trapezium figure = 6 cm.
Length of the unknown RS side of 5 QRST Square figure = 6 cm.
Explanation:
Perimeter of 1 ABCD Square Figure = 24 cm
Length of the AB side of 1 ABCD Square Figure = 6 cm
Length of the unknown BC side of 1 ABCD Square Figure = a cm
Length of the CD side of 1 ABCD Square Figure = 6 cm
Length of the DA side of 1 ABCD Square Figure = 6 cm
Perimeter of 1 ABCD Square Figure = AB + BC + CD + DA
=> 24 cm = 6 cm + a cm + 6 cm + 6 cm
=> 24 cm = 12cm + 6 cm + a cm +
=> 24 cm = 18 cm + a cm
=> 24 cm – 18 cm = a cm
=> 6 cm = a cm.
Perimeter of 2 EFGH Rectangle figure = 30 cm.
Length of the EF side of 2 EFGH Rectangle figure = 9 cm.
Length of the FG side of 2 EFGH Rectangle figure = 6 cm.
Length of the GH side of 2 EFGH Rectangle figure = 9 cm.
Length of the HE unknown side of 2 EFGH Rectangle figure = b cm.
Perimeter of 2 EFGH Rectangle figure = EF + FG + GH + HE
=> 30 cm = 9 cm + 6 cm + 9 cm + b cm
=> 30 cm = 15 cm + 9 cm + b cm
=> 30 cm = 24 cm + b cm
=> 30 cm – 24 cm = b cm
=> 6 cm = b cm.
Perimeter of 3 IJKL Rhombus figure = 20 cm.
Length of the IJ side of 3 IJKL Rhombus figure = 6 cm.
Length of the JK side of 3 IJKL Rhombus figure = 4 cm.
Length of the unknown KL side of 3 IJKL Rhombus figure = e cm.
Length of the LI side 3 IJKL Rhombus figure = 4 cm.
Perimeter of 3 IJKL Rhombus figure = IJ+ JK+ KL+ LI
=> 20 cm = 6 cm + 4 cm + e cm + 4 cm
=> 20 cm = 10 cm + e cm + 4 cm
=> 20 cm = 14 cm + e cm
=> 20 cm – 14 cm = e cm
=> 6 cm = e cm.
Perimeter of 4 MNOP Trapezium figure = 22 cm.
Length of the unknown MN side of 4 MNOP Trapezium figure = c cm.
Length of the NO side of 4 MNOP Trapezium figure = 7 cm.
Length of the OP side 4 MNOP Trapezium figure = 2 cm.
Length of the PM side 4 MNOP Trapezium figure = 7 cm.
Perimeter of 4 MNOP Trapezium figure = MN + NO + OP + PM
=> 22 cm = c cm + 7 cm + 2 cm + 7 cm
=> 22 cm = c cm + 9 cm + 7 cm
=> 22 cm = c cm + 16 cm
=> 22 cm – 16 cm = c cm
=> 6 cm = c cm.
Perimeter of 5 QRST Square figure = 24 cm.
Length of the QR side of 5 QRST Square figure = 6 cm.
Length of the unknown RS side of 5 QRST Square figure = d cm.
Length of the ST side 5 QRST Square figure = 6 cm.
Length of the TQ side 5 QRST Square figure = 6 cm.
Perimeter of 5 QRST Square figure = QR + RS + ST + TQ
=> 24 cm = 6 cm + d cm + 6 cm + 6 cm
=> 24 cm = 12 cm + 6 cm + d cm
=> 24 cm = 18 cm + d cm
=> 24 cm – 18 cm = d cm
=> 6 cm = d cm.
Question 3.
Suppose each is 1 square centimeter.
a. Find the area and perimeter of each shape.
b. John says, “If two shapes have the same area, they must also have the same perimeter.” Is John correct? Use your answer from part (a) above to explain why or why not.
Answer:
a. Area of figure 1 = 14 square centimeter.
Perimeter of figure 1 = 16 square centimeter.
Area of figure 2 = 14 square centimeter.
Perimeter of figure 2 = 18 square centimeter.
b. No, John is incorrect because area and perimeter of any shape or figures are not inner linked. They both are two completely different concepts of any figures or shapes.
Explanation:
a.
Each is 1 square centimeter.
Figure 1 has 14 squares boxes.
Area of aeft rectangle in figure 1 = Length × Width
= 4 square centimeter × 1 square centimeter
= 4 square centimeter.
Area of rsfg rectangle in figure 1 = Length × Width
= 3 square centimeter × 1 square centimeter
= 3 square centimeter.
Area of pqhi rectangle in figure 1 = Length × Width
= 3 square centimeter × 1 square centimeter
= 3 square centimeter.
Area of opjk rectangle in figure 1 = Side × Side
= 4 square centimeter × 1 square centimeter
= 4 square centimeter.
Area of figure 1 = Area of aeft rectangle in figure 1 + Area of rsfg rectangle in figure 1 + Area of pqhi rectangle in figure 1 + Area of opjk rectangle in figure 1
= 4 square centimeter + 3 square centimeter + 3 square centimeter + 4 square centimeter
= 7 square centimeter + 3 square centimeter + 4 square centimeter
= 10 square centimeter + 4 square centimeter
= 14 square centimeter.
Perimeter of figure 1 = ae + ek + ko + oa
= 4 square centimeter+ 6 square centimeter + 4 square centimeter+ 6 square centimeter
= 10 square centimeter + 4 square centimeter+ 6 square centimeter
= 14 square centimeter+ 6 square centimeter
= 20 square centimeter.
Figure 2 has 14 squares boxes.
Area of admp Figure = Side × Side
= 3 square centimeter × 3 square centimeter
= 9 square centimeter.
Area of ldfj Figure = Side × Side
= 2 square centimeter × 2 square centimeter
= 4 square centimeter.
Area of fghi Figure = Side × Side
= 1 square centimeter × 1 square centimeter
= 1 square centimeter.
Area of figure 2 = Area of admp Figure + Area of ldfj Figure + Area of fghi Figure
= 9 square centimeter + 4 square centimeter + 1 square centimeter.
= 14 square centimeter.
Perimeter of figure 2 = ap + ag + gm + mp
= 3 square centimeter + 6 square centimeter + 6 square centimeter + 3 square centimeter
= 9 square centimeter + 6 square centimeter + 3 square centimeter
= 15 square centimeter + 3 square centimeter
= 18 square centimeter.
b. Area is a measure of how much space there is inside a shape. Perimeter Perimeter is the total length of the sides of a two dimensional shape. John is incorrect as they both are two different concepts of any figure or shape. Here, figure 1 and figure 2 are having same areas yet different in perimeter.
Question 4.
Mr. Jackson’s class finds all possible perimeters for a rectangle composed of 36 centimeter tiles. The chart below shows how many students found each rectangle.
Perimeter |
Number of Students |
24 cm | 6 |
26 cm | 9 |
30 cm | 5 |
40 cm | 7 |
74 cm | 4 |
a. Check the students’ work. Did they find all the possible perimeters? How do you know?
b. Use the chart. Estimate to construct a line plot of how many students found each perimeter.
Number of Students Who Found Each Perimeter
Answer:
a.
b.
Explanation:
a.
b.
Question 5.
The square to the right has an area of 16 square centimeters.
a. What is the length of each side? Explain how you know.
b. Draw copies of the square above to make a figure with a perimeter of 32 centimeters.
c. Write a number sentence to show that your figure has the correct perimeter of 32 centimeters.
Answer:
a. Length of the each Side of ABCD Square figure = 4 cm.
b.
c. Perimeter of EFGH Square figure = EF + FG + GH + HE
= 8 cm + 8 cm + 8 cm + 8 cm
= 16 cm + 8 cm + 8 cm
= 24 cm + 8 cm
= 32 cm.
Explanation:
a.
Let the Length of the Side of ABCD Square figure = X
Perimeter of ABCD Square figure = 16 square cm
Perimeter of ABCD Square figure = 4 × Side
=> 16 square cm = 4 × X
=> 16 square cm ÷ 4 = X
=> 4 cm = X.
b.
Perimeter of EFGH Square figure = 32 cm.
c. Perimeter of EFGH Square figure = 32 cm.
Length of the side of EFGH Square figure = X cm
Perimeter of EFGH Square figure = 4 × Side
=> 32 cm = 4 × Side
=> 32 cm ÷ 4 = Side
=> 8 cm = side.