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Calculating Volumes of Compound Objects
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Homework
  • Review Individual Solutions to the Assessment Task (10 minutes)
    • Unless you have time during the lesson, schedule this task for the next lesson. Some teachers ask students to work through this task for homework.

      • Return students' individual papers from the Glasses assessment task, along with a fresh copy of the task sheet..
        • Read through your original responses, and think about what you have learned in this lesson.
        • Suppose you were to work on another volume task tomorrow. What advice would you give yourself?
        • Using what you have learned, try to improve your solution.
      • If a student is satisfied that he or she has completed the task satisfactorily, ask the student try some of the questions on the extension sheet, Glasses: Extension Questions.
      • Question 3 on this sheet is much more difficult than the previous questions, so it should provide an appropriate challenge for students who have succeeded on the other parts of this task.
  • Solutions: Glasses
        1. The volume of Glass 1 = π × 3 × 3 × 6 = 54π = 170 cm3.
        2. The volume of the hemisphere = (4π × 33) ÷ 6 = 18π = 56.5.
          The volume of the cylinder = π × 32 × 3 = 27π = 84.8.
          Total volume of Glass 2 = 45π = 141 cm3.
        3. Using the Pythagorean Theorem, the height of the cone is √72 − 32 = √40 = 2√10.
          The volume of Glass 3 = (π × 32 × 2√10) ÷ 3 = 6√10π = 60 cm3.
      1. The volume of liquid in the half-full Glass 2 is 141 ÷ 2 = 70.5.

        The volume of liquid in the cylinder = 70.5 − 56.5 = 14.

        14 = π × 32 × height in cylinder.

        Height in cylinder = 14 ÷ 9π = 0.5.

        The total height = 3 + 0.5 = 3.5 cm.

      2. Glass 4 is composed from a cylinder and cone. While we do not yet have enough information to deduce the formula, it is possible to rule out three of the formulas on grounds that the dimensions of the formulas are incorrect:

        ⅙πdh    Only two lengths are multiplied so this has the dimension of an area.

        ⅙πd2h2    Four lengths are multiplied, so this is not a volume either.

        Both ⅙πd2h and ⅙πdh2 involve multiplying together three lengths, but ⅙πdh2 involves the square of the height and so, cannot be correct.

        The correct formula is therefore ⅙πd2h.

  • Solutions: Glasses Extension Problems
      1. This question is intended to encourage the discussion of dimensional analysis.

        When lengths are combined by addition we obtain another length.

        If two lengths are multiplied we obtain an area.

        If three are multiplied we obtain a volume.

      2. The volume of the Glass = volume of cylinder + volume of cone

        = π(d2)2(h2) + ⅓π(d2)2(h2)

        = 43π(d2)2(h2)

        = ⅙πd2h.

      3. When Glass 3 is half full, it will hold 30 cm3 (from Q1c).

        If the height of liquid is h and the radius of the top of the liquid is r then ⅓πr2h = 30.

        So r2h = 90.    (1)

        By similar triangles:

        The ratio height of bowl : radius of bowl = h : r = 2 √10 :3

        This means that r = 0.47h    (2)

        Substituting (2) in (1):

        0.225 h3 = 29, and so h = 5.1 cm.

        The height of liquid will be 5.1 cm.

  • Analysis of Student Responses to Discuss
    • Logan's method

      Logan uses the correct formula for the volume of a cone. He identifies that the correct height to substitute into that formula is the perpendicular height of the cone. He uses the Pythagorean Theorem to figure out the perpendicular height but forgets to find the square root of b2 = 40. This leads to an incorrect solution.

      Logan should recalculate the cone volume using 2√10 for the perpendicular height.

      Isaac's method

      Isaac uses the correct formula for the volume of a cone. He uses an incorrect measure. He uses the slant height rather than the perpendicular height. He should first calculate the perpendicular height of the cone using the Pythagorean Theorem, and then substitute that value into the formula.

      Yasmin's method

      Yasmin has assumed that halving the volume halves the height of the liquid in the glass. This is a false assumption. In order to figure out the correct solution, Yasmin should first halve the total volume of the glass, and then subtract the volume of the hemisphere from the half volume. She will then know the volume of liquid in the cylinder. She can calculate the height of the liquid, and add that on to the height of liquid in the hemisphere.

      Brianna's method

      Brianna calculates half the volume of the glass. She then subtracts the volume of the hemisphere. This shows how much of the volume of the cylinder is taken up when the glass is half full. Brianna needs to complete her solution by working out the height of a cylinder of radius 3 cm that has this volume. That height, added to the height of the hemisphere (3 cm), is the height of liquid in the half-full glass.

      Brianna also used the equals sign in a non-standard way.

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