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Volume of Cylinders, Cones and Spheres: Grade 8 Math Test Answer Key – PDF Download

    Volume of Cylinders, Cones and Spheres – Grade 8 math test answer key

    This answer key gives a worked solution and a point-by-point scoring guide for each of the 8 questions. Award partial credit as described, and accept equivalent forms such as \(\frac{4}{3}\pi\cdot 729\) written before simplifying. The total is 20 points.

    Grade 8Common Core: 8.G.C.9

    Question 1: Volume of a cylinder (2 points)

    \(V=\pi r^{2}h=\pi(4)^{2}(9)=144\pi\).

    \(144\pi\text{ ft}^{3}\)

    r = 4 fth = 9 ftV = π(16)(9)= 144π ft³
    Volume of the cylinder.

    Scoring: 1 point for the correct formula and substitution; 1 point for \(144\pi\) with cubic units.

    Question 2: Volume of a cone (2 points)

    The radius is \(14: 2=7\text{ cm}\). \(V=\frac{1}{3}\pi(7)^{2}(15)=\frac{1}{3}\pi\cdot 735=245\pi\).

    \(245\pi\text{ cm}^{3}\)

    r = 7 cmh = 15 cmV = ⅓π(49)(15)= 245π cm³
    The radius is half the diameter.

    Scoring: 1 point for using \(r=7\); 1 point for \(245\pi\) with cubic units.

    Question 3: Volume of a sphere (2 points)

    1. \(V=\frac{4}{3}\pi(9)^{3}=\frac{4}{3}\pi\cdot 729=972\pi\text{ in}^{3}\).
    2. \(972\cdot 3.14=3{,}052.08\), which rounds to \(3{,}052.1\text{ in}^{3}\).
    r = 9 inV = ⁴⁄₃π(729)= 972π in³≈ 3,052.1 in³
    Volume of the sphere.

    Scoring: 1 point for part (a); 1 point for part (b).

    Question 4: Missing height of a cylinder (2 points)

    \(1008\pi=\pi(6)^{2}h=36\pi h\), so \(h=1008: 36=28\).

    \(h=28\text{ cm}\)

    Scoring: 1 point for the equation \(1008\pi=36\pi h\); 1 point for \(h=28\text{ cm}\).

    Question 5: Missing height of a cone (2 points)

    \(1200\pi=\frac{1}{3}\pi(10)^{2}h=\frac{100}{3}\pi h\). Multiply both sides by 3: \(3600=100h\), so \(h=36\).

    \(h=36\text{ m}\)

    Scoring: 1 point for a correct equation; 1 point for \(h=36\text{ m}\).

    Question 6: Cone and cylinder together (3 points)

    1. \(V=\pi(5)^{2}(12)=300\pi\text{ in}^{3}\).
    2. \(V=\frac{1}{3}\pi(5)^{2}(12)=100\pi\text{ in}^{3}\).
    3. \(300\pi: 100\pi=3\), so 3 cones.
    r = 5 inh = 12 inCylinderr = 5 inh = 12 inCone
    Cylinder: 300π in³. Cone: 100π in³.

    Scoring: 1 point for each part (a), (b), (c).

    Question 7: Planter and soil (3 points)

    1. \(V=\pi(1.5)^{2}(2)=\pi\cdot 2.25\cdot 2=4.5\pi\approx 4.5\cdot 3.14=14.13\). About \(14.13\text{ ft}^{3}\).
    2. \(14.13: 2=7.065\). Seven bags are not enough, so she needs 8 bags.

    Scoring: (a) 1 point for the substitution, 1 point for 14.13 with units; (b) 1 point for rounding up to 8.

    Question 8: Ice cream cone (4 points)

    1. \(\frac{1}{3}\pi(3)^{2}(10)=30\pi\text{ in}^{3}\).
    2. \(\frac{1}{2}\cdot\frac{4}{3}\pi(3)^{3}=\frac{1}{2}\cdot 36\pi=18\pi\text{ in}^{3}\).
    3. \(30\pi+18\pi=48\pi\text{ in}^{3}\).
    4. \(48\cdot 3.14=150.72\), so about \(150.7\text{ in}^{3}\).
    r = 3 inh = 10 incone: 30πhemisphere: 18πtotal: 48π
    The two parts added.

    Scoring: 1 point for each part (a), (b), (c), (d).

    Total

    20 points

    See every chapter on the Grade 8 math page.

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