## Engage NY Eureka Math 7th Grade Module 6 End of Module Assessment Answer Key

### Eureka Math Grade 7 Module 6 End of Module Assessment Task Answer Key

Question 1.

In the following two questions, lines AB and CD intersect at point O. When necessary, assume that seemingly straight lines are indeed straight lines. Determine the measures of the indicated angles.

a. Find the measure of ∠XOC.

b. Find the measures of ∠AOX, ∠YOD, and ∠DOB.

Answer:

a. x + 10 = 25 + 45

x + 10 – 10 = 70 – 10

x = 60

∠XOC = 60˚

b. 2x + 90 + x + (60 – x) = 180

2x + 150 – 150 = 180 – 150

2x = 30

\(\frac{1}{2}\) (2x) = \(\frac{1}{2}\)(30)

x = 15

∠AOX = 2(15)° = 30˚

∠YOD = 15˚

∠DOB = (60 – 15)° = 45˚

Question 2.

Is it possible to draw two different triangles that both have angle measurements of 40° and 50° and a side length of 5 cm? If it is possible, draw examples of these conditions, and label all vertices and angle and side measurements. If it is not possible, explain why.

Answer:

One possible solution:

Question 3.

In each of the following problems, two triangles are given. For each: (1) State if there are sufficient or insufficient conditions to show the triangles are identical, and (2) explain your reasoning.

Answer:

a. The triangles are identical by the three sides condition. △ABC ↔ △SRT

b. The triangles are identical by the two angles and included side condition. The marked side is between the given angles. △MNO ↔ △RQP

Question 4.

In the following diagram, the length of one side of the smaller shaded square is \(\frac{1}{3}\) the length of square ABCD. What percent of square ABCD is shaded? Provide all evidence of your calculations.

Answer:

Let x be the length of the side of the smaller shaded square. Then AD = 3x ; the length of the side of the larger shaded square is 3x – x = 2x.

Area_{ABCD} = (3x)^{2} = 9x^{2}

Area_{Large Shaded} = (2x)^{2} = 4x^{2}

Area_{Small Shaded} = (x)^{2} = x^{2}

Area_{Shaded} = 4x^{2} + x^{2} = 5x^{2}

Percent Area_{Shaded} = \(\frac{1}{2}\)(100%) = 55 (\(\frac{5}{9}\)) %

Question 5.

Side \(\overline{E F}\) of square DEFG has a length of 2 cm and is also the radius of circle F. What is the area of the entire shaded region? Provide all evidence of your calculations.

Answer:

Area_{Circle F} = (π)(2 cm)^{2} = 4π cm^{2}

Area_{\(\frac{3}{4}\)Circle F} = \(\frac{3}{4}\) (4π cm^{2}) = 3π cm^{2}

Area_{DEFG} = (2 cm)(2 cm) = 4 cm^{2}

Area_{Shaded Region} = 4 cm^{2} + 3π cm^{2}

Area_{Shaded Region} ≈ 13.4 cm^{2}

Question 6.

For his latest design, a jeweler hollows out crystal cube beads (like the one in the diagram) through which the chain of a necklace is threaded. If the edge of the crystal cube is 10 mm, and the edge of the square cut is 6 mm, what is the volume of one bead? Provide all evidence of your calculations.

Answer:

VolumeLarge Cube = (10 mm)^{3} = 1,000 mm^{3}

VolumeHollow = (10 mm)(6 mm)(6 mm) = 360 mm^{3}

VolumeBead = 1,000 mm3 – 360 mm^{3} = 640 mm^{3}

Question 7.

John and Joyce are sharing a piece of cake with the dimensions shown in the diagram. John is about to cut the cake at the mark indicated by the dotted lines. Joyce says this cut will make one of the pieces three times as big as the other. Is she right? Justify your response.

Answer:

VolumeTrapezoidal Prism = (\(\frac{1}{2}\)) (5 cm + 2.5 cm)(6 cm)(10 cm) = 225 cm^{3}

VolumeTriangular Prism = (\(\frac{1}{2}\)) (2.5 cm)(6 cm)(10 cm) = 75 cm^{3}

Joyce is right; the current cut would give 225 cm^{3} of cake for the trapezoidal prism piece and 75 cm^{3} of cake for the triangular prism piece, making the larger piece 3 times the size of the smaller piece (\(\frac{225}{75}\)) = 3).

Question 8.

A tank measures 4 ft. in length, 3 ft. in width, and 2 ft. in height. It is filled with water to a height of 1.5 ft. A typical brick measures a length of 9 in., a width of 4.5 in., and a height of 3 in. How many whole bricks can be added before the tank overflows? Provide all evidence of your calculations.

Answer:

Volume in tank not occupied by water:

V = (4 ft.)(3 ft.)(0.5 ft.) = 6 ft^{3}

Volume_{Brick} = (9 in.)(4.5 in.)(3 in.) = 121.5 in^{3}

Conversion (in^{3} to ft^{3}): (121.5 in^{3} )(\(\frac{1 f t^{3}}{12^{3}-1 n^{3}}\)) = 0.0703125 ft^{3}

Number of bricks that fit in the volume not occupied by water: (\(\frac{6 \mathrm{ft}^{3}}{0.07037 .25 \mathrm{ft}^{3}}\)) = 85 (\(\frac{1}{3}\))

Number of whole bricks that fit without causing overflow: 85

Question 9.

Three vertical slices perpendicular to the base of the right rectangular pyramid are to be made at the marked locations: (1) through \(\overline{A B}\), (2) through\(\overline{C D}\), and (3) through vertex E. Based on the relative locations of the slices on the pyramid, make a reasonable sketch of each slice. Include the appropriate notation to indicate measures of equal length.

Answer:

Sample response:

Question 10.

Five three-inch cubes and two triangular prisms have been glued together to form the composite three-dimensional figure shown in the diagram. Find the surface area of the figure, including the base. Provide all evidence of your calculations.

Answer:

19 square surfaces: 19(3 in.)^{2} = 171 in^{2}

4 triangular surfaces: (4) (\(\frac{1}{2}\)) (3 in.)(4 in.) = 24 in^{2}

3 × 5 rectangular surface: (3 in.)(5 in.) = 15 in^{2}

3 × 4 rectangular surface: (3 in.)(4 in.) = 12in^{2}

6 × 5 rectangular surface: (6 in.)(5 in.) = 30 in^{2}

6 × 4 rectangular surface: (6 in.)(4 in.) = 24 in^{2}

Total surface area: 171 in^{2} + 24 in^{2} + 15 in^{2} + 12 in^{2} + 30 in^{2} + 24 in^{2} = 276 in^{2}