Area and Volume Formulas: Geometry Math Test Answer Key – PDF Download

Question 1: Circle measures (2 points)

(a) \(C=\pi d=16\pi\approx 50.27\text{ in}\).

(b) \(r=8\), so \(A=\pi(8)^{2}=64\pi\approx 201.06\text{ in}^{2}\).

Answers: \(16\pi\text{ in}\); \(64\pi\text{ in}^{2}\).

Scoring: 1 point for the circumference, 1 point for the area (the radius must be 8, not 16).

Question 2: Rearranged sectors (2 points)

(a) The base is half the circumference: \(\frac{2\pi(12)}{2}=12\pi\text{ cm}\).

(b) The height is the radius, \(12\text{ cm}\). Area \(=12\pi\times 12=144\pi\text{ cm}^{2}\), which confirms \(A=\pi r^{2}\).

Answers: \(12\pi\text{ cm}\); \(144\pi\text{ cm}^{2}\).

Scoring: 1 point for the base; 1 point for the area and the formula.

Question 3: Slanted cylinder (2 points)

(a) The right cylinder has volume \(\pi(3)^{2}(10)=90\pi\text{ cm}^{3}\), and the slanted cylinder has the same volume.

(b) Cavalieri's principle: both solids are \(10\text{ cm}\) tall and every horizontal cross section is a circle of radius \(3\text{ cm}\) (area \(9\pi\text{ cm}^{2}\)), so the volumes are equal.

Answer: \(90\pi\text{ cm}^{3}\).

Scoring: 1 point for \(90\pi\); 1 point for naming Cavalieri's principle with equal heights and equal cross-sectional areas.

Question 4: Oil drum (2 points)

(a) \(r=12\), so \(V=\pi(12)^{2}(36)=5184\pi\text{ in}^{3}\) (about \(16{,}286.02\text{ in}^{3}\)).

(b) \(16{,}286.02: 231\approx 70.50\), so about \(70.5\) gallons.

Answers: \(5184\pi\text{ in}^{3}\); about \(70.5\) gallons.

Scoring: 1 point for using the radius 12 and getting \(5184\pi\); 1 point for the correct conversion to gallons.

Question 5: Pyramid height (2 points)

(a) \(150=\frac{1}{3}(36)h=12h\), so \(h=12.5\text{ m}\).

(b) The prism has \(V=Bh=36\times 12.5=450\text{ m}^{3}\), which is 3 times the pyramid.

Answers: \(12.5\text{ m}\); \(450\text{ m}^{3}\).

Scoring: 1 point for the height; 1 point for the prism volume.

Question 6: Pile of sand (2 points)

(a) \(r=5\), so \(V=\frac{1}{3}\pi(25)(6)=50\pi\approx 157.08\text{ ft}^{3}\).

(b) \(157.08\times 100=15{,}708\) pounds.

Answers: \(50\pi\approx 157.08\text{ ft}^{3}\); about \(15{,}708\) pounds.

Scoring: 1 point for the volume (radius 5, factor \(\frac{1}{3}\)); 1 point for the weight.

Question 7: Volume of a sphere (2 points)

\(V=\frac{4}{3}\pi(9)^{3}=\frac{4}{3}\pi(729)=972\pi\approx 3{,}053.63\).

Answer: \(972\pi\approx 3{,}053.63\text{ cm}^{3}\).

Scoring: 1 point for the correct formula and substitution; 1 point for \(972\pi\) and the rounded value.

Question 8: Hemisphere and cylinder (3 points)

(a) \(V=\pi(6)^{2}(6)=216\pi\text{ cm}^{3}\).

(b) \(V=\frac{1}{3}\pi(36)(6)=72\pi\text{ cm}^{3}\).

(c) \(216\pi-72\pi=144\pi\). The hemisphere is half a sphere: \(\frac{1}{2}\cdot\frac{4}{3}\pi(6)^{3}=\frac{2}{3}\pi(216)=144\pi\). The volumes are equal, as Cavalieri's principle predicts.

Answers: \(216\pi\); \(72\pi\); \(144\pi\text{ cm}^{3}\).

Scoring: 1 point for the cylinder; 1 point for the cone; 1 point for the subtraction and the matching hemisphere volume.

Question 9: Cone candles (3 points)

(a) \(\pi(4)^{2}(9)=144\pi\text{ cm}^{3}\).

(b) \(\frac{1}{3}\pi(3)^{2}(6)=18\pi\text{ cm}^{3}\).

(c) \(144\pi: 18\pi=8\) candles.

Answers: \(144\pi\text{ cm}^{3}\); \(18\pi\text{ cm}^{3}\); 8 candles.

Scoring: 1 point for each correct part.

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