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Archdiocese of New York Practice Items

MathematicsGrade 6Teacher Sample Packet

Unit 5

NY MATH_TE_G6_U5.indd 1 1/29/14 3:39 PM

1Mathematics Assessment, Unit 5

1. Horatio’s patio is shaped like an isosceles trapezoid. He wants to paint it but is unsure of how much paint to purchase. Pictured below are the measurements for Horatio’s patio.

5 feet

13 feet

18 feet

12 feet

After reconstructing the trapezoid, what is the total area that needs to be covered with paint?

A 72 square feet

B 276 square feet

C 336 square feet

D 4,680 square feet

Key BMeasured CCLS: 6.G.A.1

Standard Description: Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.

Commentary: The item measures standard 6.G.A.1 because to determine the answer, the student will need to know the area formulas for triangles, rectangles and/or trapezoids and be able to use them in conjunction with one another to be able to find the total area of the figure.

Extended RationaleA This answer is incorrect. It was found by using an incorrect formula. While the math is correct, the formula used

was the perimeter formula, not the area formula. B This is the correct answer. It is a result of finding the area of the rectangle (18 × 12 = 216), and the area of the

triangles (12 × 5 × 12 × 2 = 60). Those areas (216 + 60) are added together for the result of 276 square feet.

C This answer is incorrect. It is derived from attempting to use the area formulas for rectangles and triangles but

forgetting that the area formula for a triangle is to be halved (12 bh).

D This is an incorrect answer that is formed by multiplying all of the numbers given and the dividing by the number of shapes (3) given. The result indicates a deficit in either knowing the formulas required and/or utilizing those formulas.


2 Grade 6

Key DMeasured CCLS: 6.G.A.2

Standard Description: Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = lwh and V = bh to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.

Commentary: The item measures standard 6.G.A.2 because to determine the answer, the student will need to use the

volume formula for a rectangular prism to determine how many 19 inch cubes would fit into the box.

Extended Rationale

A This is an incorrect answer that began by correctly multiplying all 3 dimensions (39 × 79 × 29 = 429 ). However, rather

than divide that result (429 ) by 19 (the measurement of the unit cubes), the student just divided by 9 and got an

answer of 4 69 and then rounded up.

B This is in incorrect answer that can be found by adding all the numerators (3 + 7 + 2 = 12). While the addition is

done correctly, the numbers should have been multiplied rather than added.

C This incorrect answer is a result of multiplying 2 of the numbers (3 × 7 = 21) and then adding the third number

(21 + 2 = 23) instead of multiplying all 3.

D This is the correct answer. It is found by multiplying all the given dimensions (39 × 79 × 29) to find the volume (429 )

and then dividing by the unit cube (19) which results in finding the number of cubes that would fit inside the

rectangular solid. 429 ÷ 19 → 42

9 × 91 = 42

2. The box below was filled with packing material.

inches79

inches29

inches39

If the packing material was 19 inch cubes, how many of those cubes would fit into the box?

A 5 cubes

B 12 cubes

C 23 cubes

D 42 cubes



Key DMeasured CCLS: 6.G.A.3

Standard Description: Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems.

Commentary: The item measures standard 6.G.A.3 because to determine the answer, the student will need to be able to use coordinates to find the length of a side using those coordinate points.

Extended RationaleA This is an incorrect answer. It is likely a result of not being able to correctly graph the given coordinates and/or

using the x-coordinates (instead of the y-coordinates) to find the distance between Point B and Point D.B This is an incorrect answer. It is derived from using the correct coordinates (y-coordinates) but not subtracting the

signed numbers correctly. 3 – (–8) = 3 + 8 = 11, 3 – (–8) ≠ 3 – 8 ≠ 5.C This is an incorrect answer. It is a result of finding the distance between Point A and Point C instead of Point B and

Point D. The distance formula was correctly utilized, but using the incorrect coordinates resulted in the incorrect answer.

D This is the correct answer. It is found by finding the distance between the y-coordinates of Point B and Point D, or finding the difference between 3 and – 8. 3 – (– 8) = 11.

3. Dr. Skillman was recording some star locations. On her chart were the points A (5, –1), B (–4, 3), C (3, –1) and D (–4, –8). What is the length between Points B and Point D? Use the coordinate plane below to help solve the problem.

y

x–10 –8 –6 –4 –2 2 4 6 8 10

–2

2

4

6

8

10

–4

–6

–8

–10

A 0 units

B 5 units

C 8 units

D 11 units


4 Grade 6

Key AMeasured CCLS: 6.G.A.4

Standard Description: Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.

Commentary: The item measures standard 6.G.A.4 because to determine the answer, the student will need to be able to use nets to find the surface area of a triangular prism.

Extended RationaleA This is the correct answer. It can be found by recognizing that a triangular solid requires 5 lateral sides (3 connecting

the angled pieces of the triangle + 2 pieces that form the “ends” of the prism). And that 3 of the sides form rectangles while 2 sides consist of triangles.

B This is an incorrect answer. It can be achieved by not recognizing that a triangular prism is made from triangles. The focus here seems to be prism only, which is typically thought of in terms of rectangles.

C This is an incorrect answer. It is a result of not understanding that a triangular prism has more than 1 rectangle and that triangles should be opposite each other because they represent the ends of the prism.

D This is an incorrect answer. It is found when it is thought that a triangular prism only contains surface areas that are triangular in shape. Rectangles are missing.

4. Ray was hired to wrap gifts in a store. He sometimes had to construct a box for the gift. If he needed to create a box that could hold a triangular wedge of cheese, which of these nets could be used to create a triangular prism?

A

C

B

D



Measured CCLS: 6.G.A.1

Standard Description: Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.

Commentary: The item measures standard 6.G.A.1 because to find the answers, the student must know the area formula for both right triangles and trapezoids. Furthermore, the student needs to find a measurement (height) of the triangle to be able to correctly find the area of the trapezoid.

Extended Rationale: As indicated in the rubric, student responses will be rated on whether the student provides evidence of a strong understanding of formulas for right triangles and trapezoids. In addition, the student will need to (1) recognize which formula to use, (2) find and substitute in any values given, and (3) correctly utilize the formulas.

Once the proper numbers have been ascertained, correct units (linear, square or cubic) must be used. Because there are both lengths and areas given and/or determined, both yards and square yards should be used as labels. The best responses will contain supporting evidence consistent with the answers given.

SAMPLE STUDENT RESPONSES AND SCORES APPEAR ON THE FOLLOWING PAGES:

5. Below is the shape of a rectangular floor found in an auditorium. Sections A and C will have maroon carpeting, while Section B will have navy carpeting.

40 yards

Section B(navy)

Section A Section C

If the total area for all 3 sections is 400 square yards, and Sections A and C have a base of 5 yards each, what is the area for Section C?

How much navy carpeting will be required for Section B?

Show your work.

Answer


6 Grade 6


40 yards

Section B(navy)

Section A Section C



Show your work.

Answer

Score Point 2 (out of 2 points)

The work provided for these answers shows logical and correct thinking. The student understands the steps required to answer the questions and also the tools necessary for completing those steps (formulas, substitution, solving equations).

To find the area of Section C, one must find the height of Section A or C (because the heights are equal). To find the height of section C, one needs to know that the area of the entire rectangle is 400 yds2. Since the width of the rectangle is 40 yds, the height must be 10 yds. Given b = 5 and h = 10, use A = 1

2 bh to find the area of either Section A or C. A = 1

2 (5) (10) = 25 yds2.

Area of C = 25 yds2

To find the carpeting required for section B, use the formula 12 h (b1 + b2).

The height is 10, and the two bases are 30 (40 – 5 – 5 = 30) and 40 yards.

A = 12 h (b1 + b2) = 1

2 (10) (30 + 40) = 350 yds2

25 square yards, 350 square yards




40 yards

Section B(navy)

Section A Section C



Show your work.

Answer


The work provided supports a comprehensive understand of what the problem is asking and the tools necessary to find those answers. In addition, this work is evidence of a systematic approach to recognizing the facts present, using the required tools and then accurately computing the numbers as necessary.

To find the area of Section C, one can find the area of Section A. Sections A and C have the same area (because the heights and bases are equal). To find the height of section A, one needs to know that the area of the entire rectangle is 400 yds2. Since the width of the rectangle is 40 yds, the height must be 10 yds. Given b = 5 and h = 10, use A = 1

2 bh to find that the areas of both Section A and C are 25 yds2 each. Area of C = 25 yds2

To find the carpeting required for section B, take the areas of Sections A and C (25 + 25 = 50) and subtract those areas from the total area (400). The result (400 – 50) is the area of Section B alone, or 350 yds2.

Area of Section B = Total Area – (Area of Section A + Area of Section C ) = 400 – 50 = 350 yds2



8 Grade 6


The work provided shows a decent understanding of area formulas. In the first part, the student masters solving for a variable using the correct formula for the area of a triangle. However, in the second part, the student neglected to subtract the area of both triangles to find the remaining area of the trapezoid.


40 yards

Section B(navy)

Section A Section C



Show your work.

Answer

To find the area of Section C, one must find the height of Section A or C (because the heights are equal). To find the height of section C, one needs to know that the area of the entire rectangle is 400 yds2. Since the width of the rectangle is 40 yds, the height must be 10 yds. Given b = 5 and h = 10, use A = 1

2 bh to find the area of either Section A or C. A = 1

2 (5) (10) = 25 yds2.

Area of Section C = 25 yds2

To find the area of Section B, take the total area given (400 yds2) and then subtract the area of Section C found in the first part of the problem. 400 – 25 = 375 yds2.

Area of Section B = 375 yds2





The work shown on the first part of the question shows some knowledge of formulas, their purpose and substitution. However, since the incorrect formula for the area of a triangle was used, finding the correct answer was not likely. (The student neglected to find half of base times height for the triangle’s area.) For the second part, the correct formula for the area of trapezoid was used and therefore the correct answer was ascertained.


40 yards

Section B(navy)

Section A Section C



Show your work.

Answer

To find the area of Section C, find the height and base of that triangle. The base is given (5). The height is determined by taking the overall area (400) and dividing it by the known base (40). The result is a height of 10. To find the area of Section C, use the formula A = bh, or (5) (10) = 50.


To find the carpeting required for section B, use the formula 12 h (b1 + b2).

The height is 10, and the two bases are 30 (40 – 5 – 5 = 30) and 40 yards.

A = 12 h (b1 + b2) = 1

2 (10) (30 + 40) = 350 yds2



10 Grade 6


This work shows evidence of a limited understanding of formulas, particularly which formula to use when finding the area of triangles and trapezoids. The mathematical computations were done competently but when beginning with the wrong formula, the answers are almost always doomed from the beginning. The likely failures are true for both parts of the problem.


40 yards

Section B(navy)

Section A Section C



Show your work.

Answer

To find the area of Section C, find the height and base of that triangle. The base is given (5). The height is determined by taking the overall area (400) and dividing it by the known base (40). The result is a height of 10. To find the area of Section C, use the formula A = bh, or (5) (10) = 50.


To find the area of Section B, take the total area given (400 yds2) and then subtract the areas of Sections A and C found in the first part of the problem. 400 – 50 – 50 = 300 yds2.

Area of Section B = 375 yds2





This work shows evidence of deficiencies regarding formulas and logical thinking. The formula for a rectangle was used instead of the formula for the area of a triangle. In addition, area was labeled as a linear unit (instead of a squared unit). Also, the student misunderstood that the total area of the rectangle was the same as Section B.


40 yards

Section B(navy)

Section A Section C



Show your work.

Answer

To find the area of Section C, find the height and base of that triangle. The base is given (5). The height is determined by taking the overall area (400) and dividing it by the known base (40). The result is a height of 10. To find the area of Section C, use the formula A = bh, or (5)(10) = 50.

Area of Section C = 50 yds

The area for Section B is given in the original wording of the problem. It is 400 square yards, therefore no work has to be done, other than to correctly read the given problem.

Area of Section B = 400 yards2

50 yards, 400 square yards


12 Grade 6

Teaching Tips for Item 5

6.G.A.1

Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.

Study:• ClueWords

º Area—the 2 dimensional space that something takes up; it is often found using multiplication of bases and heightsº Height—the perpendicular distance between the top of a figure to its baseº Base—the “bottom” dimension of a figure

• Factsº AreaofSectionA=AreaofSectionCº Area of all sections (rectangle) = 400 square yardsº Base of the triangle = 5 yardsº Bases of the trapezoid = 30 and 40 yardsº Answersaretobe:areaofSectionCandareaofSectionB

• Discoveryº Area of Section A → Area of Section Bº Formula for the area of a triangleº AreaofSectionAand/orCº Formula for the area of a trapezoid

• Askyourself:º Have I ever seen a problem similar to this one?º If so, how is it similar?º What did I need to do?

Strategize:• GamePlan

º FindtheheightofSectionA(orC)º Determine formula to be used (A = 12 bh)

º Use that area to find the area of Section Bº Determineformulatobeused(Totalareaofrectangle–AreasofSectionsAandC)orusetheTrapezoidFormula

[A = 12 h(b1 + b2)])

• Strategiesº Formula for the area of a triangle.º Formula for the area of a trapezoid.º Solve equations for unknown variables.

Solve:• Useyourstrategiestosolvetheproblem.

• FormulafortheareaofatriangleisA = 12 bh

• Formula for the area of a rectangle is A = bh

• Area is 400 and base is 40 so height is 10.

• A = 12 bh = 12 (5)(10) = 25 yds2

Ruminate:• Lookoveryoursolution.• Doesitseemprobable?• Didyouanswerthequestion?

• Areyoucertain?• Didyouanswerusingthelanguageinthequestion?• Istheanswerinthesameunits?


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