G6 U7 Constructed Response Rubric
Grade 6 Unit 7 Constructed Response Geometry Description Task
Common Core State Standard for Mathematical Content (MC)
Standards for Mathematical Practice (MP)
1. Draw and Find the Area of Polygons on the Coordinate Grid
6.G.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. 6.G.3 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.
MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 MP.8
2. Find the Volume of Rectangular Prisms
6.G.2 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 = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 MP.8
6.G.4 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.
MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 MP.8
3. Find the Surface Area of Solid Figures Using Nets
Note to Teacher: The following scoring rubric should be used as a guide to determine points given to students for each question answered. Students are required to show the process through which they arrived at their answers for every question involving problem solving. For questions involving a written answer, full points should be given to answers that are written in complete sentences which address each component of the questions being asked. Copyright © Swun Math Grade 6 Unit 7 Constructed Response Rubric, Page 1
Scoring Rubric Question 1. a. Student plots and labels Mike’s house. See sample coordinate grid.
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b. Student plots and labels the points correctly. See sample coordinate grid.
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c. Student gives the correct coordinates. (-7,-5)
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d. Student gives the correct answer. 28 blocks
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2. a. Student plots the points correctly and gives the correct missing vertex. See sample coordinate grid. (-5, -1) b. Student gives the correct answer. 32 ft2 c. Student plots the points correctly and gives the correct missing vertex. See sample coordinate grid. (1, 1½)
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d. Student gives the correct answer and shows work. 29 ft2 Work Sample: Area of the rectangle (from 2b) = 32 ft2 Area of the parallelogram: A = bh = (3×1) = 3 ft2 32 ft2 – 3 ft2 = 29 ft2 3. a. Student gives the correct answer and an accurate explanation. Wording may vary. Sample response: Juanita’s error was that she did not use the correct formula. The area of a triangle is half of the area of a rectangle. She should have solved the problem like this: A = ½ (b × h) = ½ (6 units × 6½ units) = ½ (39 units2) = 19½ units2
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b. Student gives the correct answer, shows work, and gives an accurate explanation. Wording may vary. Sample response: I divided the hexagon into four pieces, found the area of each piece, and then added them together to find the total area. (continued on the next page)
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Copyright © Swun Math Grade 6 Unit 7 Constructed Response Rubric, Page 2
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Total
Area A: A = ½bh = ½(2×2) = ½(4) = 2 units2 Area B: A = bh = (9×2) = 18 units2 Area C: A = bh = (5×6) = 30 units2 Area D: A = ½bh = ½(2×8) = ½(16) = 8 units2 Total Area = 2 + 18 + 30 + 8 = 58 units2 4. a. Student gives the correct answer and shows work: ⅛ of a cubic inch ½ in × ½ in × ½ in = ⅛ in3
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b. Student gives the correct answer, an accurate explanation, and draws a model to support the answer. Wording may vary. Sample response and model: Yes, Johnson is correct. If the volume of a ½-inch cube is ⅛ in3, from my knowledge of fractions I know it would 1 in ½ in take 8 of the cubes to make a full cubic inch ½ in because ⅛ + ⅛ + ⅛ + ⅛ + ⅛ + ⅛ + ⅛ + ⅛ = 1. I also know that the length of each side of a ½ in ½-inch cube is ½ inch. Two faces of a ½-inch 1 in cube is needed to measure the 1-inch cube ½ in on all sides. My model shows how three ½ in ½ in faces of each ½-inch cube are part of the 1 inch measurement of length, width, and 1 in height of the 1-inch cube. 5. Students may use either V=bh or V=lwh to solve these problems. See work samples.
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a. Student gives the correct answer and shows work. 641¼ in3 Work sample: Using ½-inch unit cubes, the ballot box is 27 unit cubes long, 19 unit cubes wide, and 10 unit cubes high. V = bh Volume of a ½-inch unit cube: ⅛ in3 Area of the base layer = length × width Area of the base layer = 27 unit cubes × 19 unit cubes = 513 unit cubes Volume of the base layer = 513 unit cubes × ⅛ in3 = 64⅛ in3 V = bh V = 64⅛ in3 × 10 [layers] = 641¼ in3
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b. Student gives the correct answer and shows work. 197/32 ft3 Work sample: V = lwh = 10¼ ft × 2½ ft × ¾ ft = 197/32 ft3
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c. Student gives the correct answer and shows work. 1,880¼ in3 Work sample: V = lwh = 25⅞ in × 18⅙ in × 4 in = 1,880¼ in3
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d. Student gives the correct answer and shows work. 514.71 in3 Work sample: V = lwh = 13.3 in × 8.6 in × 4.5 in = 514.71 in3
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6. a. Student gives the correct answer and shows work. 70⅜ ft2 Work sample: Surface Area = 2(Al) + 2(Am) + 2(As) Area of large rectangle (Al) = 10¼ ft × 2½ ft = 25⅝ ft2 Area of medium rectangle (Am) = 10¼ ft × ¾ ft = 711/16 ft2 Area of small rectangle (As) = 2½ ft × ¾ ft = 1⅞ ft2 Surface Area = 2(25⅝ ft2) + 2(711/16 ft2) + 2(1⅞ ft2) Surface Area = 51¼ ft2 + 15⅜ ft2 + 3¾ ft2 = 70⅜ ft2 b. Student sketches the net, gives the correct answer, and shows work. 44⅞ ft2
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2
2 ft ¼ ft
2 ft ¼ ft
9¾ ft
Work sample: Surface Area = 2(Al) + 2(Am) + 2(As) Area of large rectangle (Al) = 9¾ ft × 2 ft = 19½ ft2 Area of medium rectangle (Am) = 9¾ ft × ¼ ft = 27/16 ft2 Area of small rectangle (As) = 2 ft × ¼ ft = ½ ft2 Surface Area = 2(19½ ft2) + 2(27/16 ft2) + 2(½ ft2) Surface Area = 39 ft2 + 4⅞ ft2 + 1 ft2 = 44⅞ ft2 7. Student gives the correct answer and shows work. 29.52 in2 Work sample: Surface Area = 3(Ar) + 2(At) Area of rectangle (Ar) = 7.8 in × 1.2 in = 9.36 in2 Area of triangle (At) = ½(1.2 in × 1.2 in) = .72 in2 Surface Area = 3(9.36 in2) + 2(.72 in2) Surface Area = 28.08 in2 + 1.44 in2 = 29.52in2
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8. a. Student gives the correct answer: pyramid b. Student gives the correct answer and shows work. 23¾ yd2 Work sample: Surface Area = As + 4(At) Area of square (As) = (2½ yd)2 = 6¼ yd2 Area of triangle (At) = ½(2½ yd × 3½ yd) = 4⅜ yd2 Surface Area = 6¼ yd2 + 4(4⅜ yd2) Surface Area = 6¼ yd2 + 17½ yd2 = 23¾ yd2 9. Student gives the correct answer and an accurate explanation, and solves the problem correctly. No, Vebjorn’s work is not correct. Instead of finding the surface area of the cube, he found the volume. He should have found the area of one square (¾ m × ¾ m = 9/16 m2), then multiplied the area of one square by all six squares (6 × 9/16 m2 = 33/8 m2). The surface area of the cube is 33/8 m2. Total
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Common Core State Standard for Mathematical Content (MC)
Standards for Mathematical Practice (MP)
1. Draw and Find the Area of Polygons on the Coordinate Grid
6.G.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. 6.G.3 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.
MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 MP.8
2. Find the Volume of Rectangular Prisms
6.G.2 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 = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 MP.8
6.G.4 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.
MP.1 MP.2 MP.3 MP.4 MP.6 MP.7 MP.8
3. Find the Surface Area of Solid Figures Using Nets
Note to Teacher: The following scoring rubric should be used as a guide to determine points given to students for each question answered. Students are required to show the process through which they arrived at their answers for every question involving problem solving. For questions involving a written answer, full points should be given to answers that are written in complete sentences which address each component of the questions being asked. Copyright © Swun Math Grade 6 Unit 7 Constructed Response Rubric, Page 1
Scoring Rubric Question 1. a. Student plots and labels Mike’s house. See sample coordinate grid.
Points 1
b. Student plots and labels the points correctly. See sample coordinate grid.
1
c. Student gives the correct coordinates. (-7,-5)
1
d. Student gives the correct answer. 28 blocks
1
2. a. Student plots the points correctly and gives the correct missing vertex. See sample coordinate grid. (-5, -1) b. Student gives the correct answer. 32 ft2 c. Student plots the points correctly and gives the correct missing vertex. See sample coordinate grid. (1, 1½)
2 1 1
d. Student gives the correct answer and shows work. 29 ft2 Work Sample: Area of the rectangle (from 2b) = 32 ft2 Area of the parallelogram: A = bh = (3×1) = 3 ft2 32 ft2 – 3 ft2 = 29 ft2 3. a. Student gives the correct answer and an accurate explanation. Wording may vary. Sample response: Juanita’s error was that she did not use the correct formula. The area of a triangle is half of the area of a rectangle. She should have solved the problem like this: A = ½ (b × h) = ½ (6 units × 6½ units) = ½ (39 units2) = 19½ units2
1
b. Student gives the correct answer, shows work, and gives an accurate explanation. Wording may vary. Sample response: I divided the hexagon into four pieces, found the area of each piece, and then added them together to find the total area. (continued on the next page)
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Copyright © Swun Math Grade 6 Unit 7 Constructed Response Rubric, Page 2
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Total
Area A: A = ½bh = ½(2×2) = ½(4) = 2 units2 Area B: A = bh = (9×2) = 18 units2 Area C: A = bh = (5×6) = 30 units2 Area D: A = ½bh = ½(2×8) = ½(16) = 8 units2 Total Area = 2 + 18 + 30 + 8 = 58 units2 4. a. Student gives the correct answer and shows work: ⅛ of a cubic inch ½ in × ½ in × ½ in = ⅛ in3
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b. Student gives the correct answer, an accurate explanation, and draws a model to support the answer. Wording may vary. Sample response and model: Yes, Johnson is correct. If the volume of a ½-inch cube is ⅛ in3, from my knowledge of fractions I know it would 1 in ½ in take 8 of the cubes to make a full cubic inch ½ in because ⅛ + ⅛ + ⅛ + ⅛ + ⅛ + ⅛ + ⅛ + ⅛ = 1. I also know that the length of each side of a ½ in ½-inch cube is ½ inch. Two faces of a ½-inch 1 in cube is needed to measure the 1-inch cube ½ in on all sides. My model shows how three ½ in ½ in faces of each ½-inch cube are part of the 1 inch measurement of length, width, and 1 in height of the 1-inch cube. 5. Students may use either V=bh or V=lwh to solve these problems. See work samples.
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a. Student gives the correct answer and shows work. 641¼ in3 Work sample: Using ½-inch unit cubes, the ballot box is 27 unit cubes long, 19 unit cubes wide, and 10 unit cubes high. V = bh Volume of a ½-inch unit cube: ⅛ in3 Area of the base layer = length × width Area of the base layer = 27 unit cubes × 19 unit cubes = 513 unit cubes Volume of the base layer = 513 unit cubes × ⅛ in3 = 64⅛ in3 V = bh V = 64⅛ in3 × 10 [layers] = 641¼ in3
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b. Student gives the correct answer and shows work. 197/32 ft3 Work sample: V = lwh = 10¼ ft × 2½ ft × ¾ ft = 197/32 ft3
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(Continued on the next page.) Copyright © Swun Math Grade 6 Unit 7 Constructed Response Rubric, Page 3
c. Student gives the correct answer and shows work. 1,880¼ in3 Work sample: V = lwh = 25⅞ in × 18⅙ in × 4 in = 1,880¼ in3
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d. Student gives the correct answer and shows work. 514.71 in3 Work sample: V = lwh = 13.3 in × 8.6 in × 4.5 in = 514.71 in3
1
6. a. Student gives the correct answer and shows work. 70⅜ ft2 Work sample: Surface Area = 2(Al) + 2(Am) + 2(As) Area of large rectangle (Al) = 10¼ ft × 2½ ft = 25⅝ ft2 Area of medium rectangle (Am) = 10¼ ft × ¾ ft = 711/16 ft2 Area of small rectangle (As) = 2½ ft × ¾ ft = 1⅞ ft2 Surface Area = 2(25⅝ ft2) + 2(711/16 ft2) + 2(1⅞ ft2) Surface Area = 51¼ ft2 + 15⅜ ft2 + 3¾ ft2 = 70⅜ ft2 b. Student sketches the net, gives the correct answer, and shows work. 44⅞ ft2
1
2
2 ft ¼ ft
2 ft ¼ ft
9¾ ft
Work sample: Surface Area = 2(Al) + 2(Am) + 2(As) Area of large rectangle (Al) = 9¾ ft × 2 ft = 19½ ft2 Area of medium rectangle (Am) = 9¾ ft × ¼ ft = 27/16 ft2 Area of small rectangle (As) = 2 ft × ¼ ft = ½ ft2 Surface Area = 2(19½ ft2) + 2(27/16 ft2) + 2(½ ft2) Surface Area = 39 ft2 + 4⅞ ft2 + 1 ft2 = 44⅞ ft2 7. Student gives the correct answer and shows work. 29.52 in2 Work sample: Surface Area = 3(Ar) + 2(At) Area of rectangle (Ar) = 7.8 in × 1.2 in = 9.36 in2 Area of triangle (At) = ½(1.2 in × 1.2 in) = .72 in2 Surface Area = 3(9.36 in2) + 2(.72 in2) Surface Area = 28.08 in2 + 1.44 in2 = 29.52in2
Copyright © Swun Math Grade 6 Unit 7 Constructed Response Rubric, Page 4
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8. a. Student gives the correct answer: pyramid b. Student gives the correct answer and shows work. 23¾ yd2 Work sample: Surface Area = As + 4(At) Area of square (As) = (2½ yd)2 = 6¼ yd2 Area of triangle (At) = ½(2½ yd × 3½ yd) = 4⅜ yd2 Surface Area = 6¼ yd2 + 4(4⅜ yd2) Surface Area = 6¼ yd2 + 17½ yd2 = 23¾ yd2 9. Student gives the correct answer and an accurate explanation, and solves the problem correctly. No, Vebjorn’s work is not correct. Instead of finding the surface area of the cube, he found the volume. He should have found the area of one square (¾ m × ¾ m = 9/16 m2), then multiplied the area of one square by all six squares (6 × 9/16 m2 = 33/8 m2). The surface area of the cube is 33/8 m2. Total
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