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Circumference and Area of Circles: Grade 7 Math Test Answer Key – PDF Download

    Circumference and Area of Circles – Grade 7 math test answer key

    Use this key to score the test out of 20 points. Each solution uses \(\pi=3.14\) unless the question asks for \(\pi\) in the answer, and the scoring note explains how partial credit is given. Accept any correct method, and award full credit when the answer has the right units.

    Grade 7Common Core: 7.G.B.4

    Question 1: Radius, diameter, and pi (2 points)

    (a) \(d=2\times 9=18\text{ cm}\).

    (b) \(C=2\pi r=2\times 9\times \pi=18\pi\text{ cm}\).

    Scoring: 1 point for each part.

    Question 2: Circumference of a round table (2 points)

    \(C=\pi d=3.14\times 15=47.1\).

    \(47.1\text{ ft}\)

    Scoring: 1 point for the correct formula and substitution; 1 point for \(47.1\text{ ft}\).

    Question 3: Area of a circle (2 points)

    \(A=\pi r^{2}=3.14\times 12^{2}=3.14\times 144=452.16\).

    \(452.16\text{ ft}^{2}\)

    12 ft
    Solved: A = 3.14 times 144 = 452.16 square feet.

    Scoring: 1 point for squaring the radius (144); 1 point for \(452.16\text{ ft}^{2}\) with square units.

    Question 4: From circumference to area (3 points)

    \(d=C: \pi=37.68: 3.14=12\), so \(r=6\text{ in}\).

    \(A=3.14\times 6^{2}=3.14\times 36=113.04\).

    Radius \(6\text{ in}\); area \(113.04\text{ in}^{2}\)

    Scoring: 1 point for the diameter \(12\); 1 point for the radius \(6\); 1 point for \(113.04\text{ in}^{2}\).

    Question 5: Why the area formula works (2 points)

    The base is half the circumference: \(\frac{1}{2}\times 2\pi r=\pi r\). The height is the radius, \(r\).

    Area = base \(\times \) height \(=\pi r\times r=\pi r^{2}\).

    Scoring: 1 point for base \(\pi r\) and height \(r\); 1 point for the product \(\pi r^{2}\).

    Question 6: Semicircle (3 points)

    \(r=5\). Area: \(\frac{1}{2}\times 3.14\times 25=39.25\text{ ft}^{2}\).

    Perimeter: half the circumference \(\frac{1}{2}\times 3.14\times 10=15.7\), plus the diameter \(10\): \(25.7\text{ ft}\).

    diameter 10 ftr = 5 ft
    Solved: radius 5 ft.

    Scoring: 1 point for the radius 5; 1 point for the area \(39.25\text{ ft}^{2}\); 1 point for the perimeter \(25.7\text{ ft}\) including the diameter.

    Question 7: Comparing pizzas (3 points)

    Small: \(r=6\), \(A=3.14\times 36=113.04\). Large: \(r=8\), \(A=3.14\times 64=200.96\).

    Difference: \(200.96-113.04=87.92\).

    \(87.92\text{ in}^{2}\) more

    Scoring: 1 point for each correct area; 1 point for the difference \(87.92\text{ in}^{2}\).

    Question 8: Running track (3 points)

    The two semicircles form one circle with diameter \(20\) and radius \(10\).

    (a) Perimeter: two straight sides \(60\) plus \(C=3.14\times 20=62.8\): \(122.8\text{ m}\).

    (b) Area: rectangle \(30\times 20=600\) plus circle \(3.14\times 100=314\): \(914\text{ m}^{2}\).

    30 m30 m20 m
    Solved: two semicircles make one circle of radius 10 m.

    Scoring: 1 point for recognizing the circle of radius 10; 1 point for \(122.8\text{ m}\); 1 point for \(914\text{ m}^{2}\).

    Total

    20 points

    See every chapter on the Grade 7 math page.

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