Apply Volume Formulas - 8th Grade Math

A school has a cylindrical water tank with radius $2\text{ m}$ and height $5\text{ m}$ . About how much water can it hold?
Answer: $20\pi\text{ m}^3\ (\approx 63\text{ m}^3)$ This question tests applying the correct volume formula for a cylinder, $V=\pi r^2 h$ . With $r=2$ m and $h=5$ m, $V=\pi(2^2)(5)=\pi(4)(5)=20\pi \approx 62.8$ $m^3$ . Common errors include using the cone formula (one third of this value), using the diameter instead of the radius, or making an exponent mistake. Squaring the radius before multiplying by the height, then by pi, gives the correct volume of $20\pi$ cubic meters, about 63 cubic meters.
A spherical ornament has diameter $10\text{ ft}$ . What is its volume, rounded to the nearest cubic foot? (Use $\pi\approx 3.14$ .)
Answer: $\approx 524\text{ ft}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as sphere, use corresponding formula with given dimensions (convert diameter to radius: r=10/2=5 ft), calculate (exponents first, multiply, approximate π≈3.14, round). Example: sphere r=5 ft using V=(4/3)π(125)≈(4/3)*3.14*125≈523.33, rounded to 524 ft³. Correct formula selection and calculation yield ≈524 ft³ for this sphere. Error like using diameter as radius (r=10, V=(4/3)π*1000≈4186.67), wrong exponent, arithmetic error, or incorrect rounding. Steps: (1) identify shape (sphere), (2) gather dimensions (r=5), (3) select formula ((4/3)πr³), (4) substitute, (5) calculate (5³=125, 125*4/3≈166.67, *3.14≈523.33, round to 524), (6) units (cubic: ft³).
Which shape has the greater volume? Shape 1: a cylinder with radius $3\text{ cm}$ and height $4\text{ cm}$ Shape 2: a sphere with radius $3\text{ cm}$
Answer: They have equal volume. This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For comparison, cylinder V=π(3²)(4)=36π, sphere V=(4/3)π(3³)=36π, so equal volumes. Common errors include wrong formulas (cylinder as (1/3)πr²h), exponent mix-ups (sphere r²), or not calculating both. Steps: (1) identify shapes, (2) gather dimensions, (3) select formulas, (4) substitute and calculate each (36π both), (5) compare numerically, (6) conclude equality. Comparing requires computing both volumes fully to see they match.
A cone-shaped paper cup has a diameter of 8 cm and a height of 12 cm. If the cup is filled to exactly $\frac{3}{4}$ of its height with water, what fraction of the total cup volume is occupied by water?
Answer: The water occupies exactly $\frac{27}{64}$ of the total cup volume This involves similar cones. The water forms a smaller cone with height $\frac{3}{4} \times 12 = 9$ cm. By similar triangles, if the original radius is 4 cm, the water cone's radius is $\frac{9}{12} \times 4 = 3$ cm. Volume scales as the cube of the linear scale factor. The scale factor is $\frac{3}{4}$ , so the volume fraction is $(\frac{3}{4})^3 = \frac{27}{64}$ . Choice C incorrectly assumes volume scales linearly with height. Choice B uses $(\frac{3}{4})^2$ . Choice D is $\frac{1}{2}$ , which doesn't relate to the given fraction.
A cylindrical water tank has radius $2\text{ m}$ and height $5\text{ m}$ . About how much water can it hold? (Give your answer in terms of $\pi$ and in cubic meters.)
Answer: $20\pi\text{ m}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=2 m and h=5 m, V=π(2²)(5)=π(4)(5)=20π m³, matching choice B. Common errors include using cone formula for 20/3 π, doubling radius incorrectly, arithmetic like 2²=2, or non-cubic units. Steps: (1) identify cylinder, (2) gather r=2 m, h=5 m, (3) select V=πr²h, (4) substitute values, (5) calculate 20π, (6) add m³. This represents the tank's capacity in exact terms with π.
A classroom has a cylindrical container for storing rice. The container has radius $3\text{ cm}$ and height $10\text{ cm}$ . What is the volume of the container? (Give the answer in terms of $\pi$ .)
Answer: $90\pi\text{ cm}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cylinder, use corresponding formula with given dimensions (radius given, no need to convert), calculate (exponents first, multiply, leave exact in terms of π). Example: cylinder r=3 cm, h=10 cm using V=π(3²)(10)=π(9)(10)=90π cm³. Correct formula selection and calculation yield 90π cm³ for this cylinder. Error like not squaring radius (π*3*10=30π), using wrong formula like for cone ((1/3)π*9*10=30π), arithmetic error (3²=6), or units not cubed (cm vs cm³). Steps: (1) identify shape (cylinder with circular base and height), (2) gather dimensions (r=3, h=10), (3) select formula (πr²h), (4) substitute (plug values: r=3, h=10), (5) calculate (exponents: 3²=9, multiply: π×9×10=90π), (6) units (cubic: cm³).
A snow cone is shaped like a cone with radius $4\text{ in}$ and height $9\text{ in}$ . What is the volume of the cone? (Give the answer in terms of $\pi$ .)
Answer: $48\pi\text{ in}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cone, use corresponding formula with given dimensions (radius given, no need to convert), calculate (exponents first, multiply, leave exact in terms of π). Example: cone r=4 in, h=9 in using V=(1/3)π(4²)(9)=(1/3)π(16)(9)=48π in³. Correct formula selection and calculation yield 48π in³ for this cone. Error like wrong formula (cylinder π*16*9=144π, missing 1/3), using diameter as radius (if misread), wrong exponent (r instead of r²), arithmetic error (16*9=128), or units not cubed. Steps: (1) identify shape (cone with circular base and point), (2) gather dimensions (r=4, h=9), (3) select formula ((1/3)πr²h), (4) substitute, (5) calculate (16*9=144, 144/3=48, times π), (6) units (cubic: in³).
A playground ball is a sphere with radius $6\text{ m}$ . What is the volume of the sphere? Give your answer in terms of $\pi$ and in cubic meters.
Answer: $288\pi\text{ m}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this sphere with r=6 m, V=(4/3)π(6³)=(4/3)π(216)=288π m³, matching choice B. Common errors include using r² instead of r³ (like 36π times something), forgetting the 4/3 factor for 216π, dividing instead of multiplying, or incorrect cubing like 6³=216. Steps: (1) identify sphere, (2) gather r=6 m, (3) select V=(4/3)πr³, (4) substitute value, (5) calculate 288π, (6) add m³. Spheres use only radius, no height, due to their uniform symmetry.
A snow cone is shaped like a cone with radius $4\text{ in}$ and height $9\text{ in}$ . What is the volume of the cone? (Give your answer in terms of $\pi$ or as an approximation.)
Answer: $48\pi\text{ in}^3\ (\approx 151\text{ in}^3)$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cone with r=4 in and h=9 in, V=(1/3)π(4²)(9)=(1/3)π(16)(9)=48π≈150.8 in³. Common errors include forgetting the 1/3 factor (using cylinder formula), using diameter instead of radius, wrong exponent like r³, arithmetic mistakes (16*9=128), or missing cubic units. Steps: (1) identify shape as cone (circular base tapering to point), (2) gather dimensions (r=4 in, h=9 in), (3) select formula (1/3)πr²h, (4) substitute values, (5) calculate (1/3)*144π=48π, (6) add cubic units in³. Remember the 1/3 distinguishes cone from cylinder to avoid formula confusion.
A clear tube is a cylinder with diameter $12\text{ cm}$ and height $7\text{ cm}$ . What is the volume of the tube? (Give your answer in terms of $\pi$ or as an approximation.)
Answer: $252\pi\text{ cm}^3\ (\approx 791\text{ cm}^3)$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with d=12 cm so r=6 cm, h=7 cm, V=π(6²)(7)=π(36)(7)=252π≈791.7 cm³. Common errors include using diameter as radius (r=12), adding 1/3 like cone, exponent mistakes (6²=36 not 72), arithmetic errors (36*7=242), or wrong units. Steps: (1) identify cylinder, (2) gather r=6 cm from d=12, h=7 cm, (3) select πr²h, (4) substitute, (5) calculate 36*7*π=252π, (6) add cubic units cm³. Converting diameter correctly is key for accurate radius.
A science club is building a cylindrical container with radius $3\text{ cm}$ and height $10\text{ cm}$ . What is the volume of the cylinder? Give your answer in terms of $\pi$ and in cubic centimeters.
Answer: $90\pi\text{ cm}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=3 cm and h=10 cm, V=π(3²)(10)=π(9)(10)=90π cm³, matching choice B. Common errors include using diameter instead of radius, wrong exponent like r³ for cylinder, arithmetic mistakes such as 3²=6, or forgetting cubic units. Steps: (1) identify cylinder, (2) gather r=3 cm, h=10 cm, (3) select V=πr²h, (4) substitute values, (5) calculate 90π, (6) add cm³. Remember, leaving in terms of π keeps it exact, as required here.
A science class uses a cylindrical container to store sand. The container has radius $3\text{ cm}$ and height $10\text{ cm}$ . What is the volume of the container? (Give your answer in terms of $\pi$ or as an approximation.)
Answer: $90\pi\text{ cm}^3\ (\approx 283\text{ cm}^3)$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=3 cm and h=10 cm, V=π(3²)(10)=π(9)(10)=90π≈282.7 cm³. Common errors include using the cone formula by mistake (adding 1/3), treating radius as diameter, incorrect squaring (like 3²=6), or forgetting to cube the units (cm instead of cm³). Steps: (1) identify shape as cylinder (circular base and constant height), (2) gather dimensions (r=3 cm, h=10 cm), (3) select formula πr²h, (4) substitute values, (5) calculate 9*10*π=90π, (6) add cubic units cm³. Always double-check arithmetic and formula selection to avoid mistakes like confusing cylinder with cone.
Compare volumes: Which has the greater volume? - Cylinder: radius $3$ units, height $4$ units - Sphere: radius $3$ units
Answer: They have equal volume. This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). Cylinder V=π(3²)(4)=36π, sphere V=(4/3)π(3³)=(4/3)π(27)=36π, so equal, matching C. Common errors: wrong formulas like cone for cylinder, or miscalculating r³=27. Steps: (1) identify shapes, (2) gather dimensions, (3) select formulas, (4) calculate both, (5) compare 36π=36π, (6) conclude equal. Comparisons require computing each volume fully.
A playground ball is a sphere with radius $6\text{ m}$ . What is the volume of the ball? (Give the answer in terms of $\pi$ .)
Answer: $288\pi\text{ m}^3$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as sphere, use corresponding formula with given dimensions (radius given), calculate (exponents first, multiply, leave exact in terms of π). Example: sphere r=6 m using V=(4/3)π(6³)=(4/3)π(216)=288π m³. Correct formula selection and calculation yield 288π m³ for this sphere. Error like wrong formula (using r² instead of r³, like (4/3)π*36=48π), arithmetic error (6³=216 mistaken as 36), or forgetting the 4/3 factor. Steps: (1) identify shape (sphere, perfect roundness), (2) gather dimensions (r=6), (3) select formula ((4/3)πr³), (4) substitute, (5) calculate (6³=216, 216*4/3=288, times π), (6) units (cubic: m³).
Which formula correctly gives the volume of a cylinder with radius $r$ and height $h$ ?
Answer: $V=\pi r^2 h$ This question tests applying volume formulas: cylinder $V=\pi r^2 h$ , cone $V=\frac{1}{3} \pi r^2 h$ , sphere $V=\frac{4}{3} \pi r^3$ , selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $\pi r^2$ times height $h$ giving $V=\pi r^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3} \pi r^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3} \pi r^3$ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: $r=d/2$ ), calculate (exponents first, multiply, approximate $\pi \approx 3.14$ or leave exact). The correct cylinder formula is $V=\pi r^2 h$ , matching choice B, representing base area times height. Common errors include adding 1/3 like cone, using $r^3$ like sphere, or surface area like $2 \pi r h$ . Steps: (1) identify cylinder by uniform height, (2) recall $\pi r^2 h$ , (3) select from options, (4) avoid extras like fractions unless tapered. Formulas derive from geometry, so visualize shapes.
Which formula correctly gives the volume of a cone with radius $r$ and height $h$ ?
Answer: $V=\tfrac{1}{3}\pi r^2h$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). The correct cone formula is V=(1/3)πr²h, matching choice C, as it accounts for the tapering volume. Common errors include confusing with cylinder (no 1/3), sphere (r³), or made-up like πr³h. Steps: (1) identify cone by its point and base, (2) recall need for 1/3 factor, (3) select (1/3)πr²h, (4) distinguish from others. Memorize formulas to avoid mixing cone and cylinder.
A sphere has volume $288\pi\text{ m}^3$ . What is the radius of the sphere?
Answer: $r=6\text{ m}$ This question tests applying volume formulas: cylinder $V=\pi r^2 h$ , cone $V=\frac{1}{3} \pi r^2 h$ , sphere $V=\frac{4}{3} \pi r^3$ , selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $\pi r^2$ times height h giving $V=\pi r^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3} \pi r^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3} \pi r^3$ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: $r=d/2$ ), calculate (exponents first, multiply, approximate $\pi\approx3.14$ or leave exact). For $V=288\pi$ , $(4/3)\pi r^3=288\pi$ so $r^3=288\times(3/4)=216$ , $r=6 \text{ m}$ ( $6^3=216$ ), matching B. Common errors: ignoring 4/3, getting r=9 (wrong inverse), or confusing with other formulas. Steps: (1) identify sphere, (2) set $(4/3)\pi r^3=288\pi$ , (3) solve $r^3=216$ , (4) cube root to r=6, (5) check units m. Reverse calculations test formula understanding.
A spherical ornament has volume $288\pi\text{ cm}^3$ . What is the radius of the sphere?
Answer: $r=6\text{ cm}$ This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). To find r for sphere V=288π, set (4/3)πr³=288π, divide by π, multiply by 3/4, r³=216, r=6 cm. Common errors include using wrong formula (like πr²h), incorrect solving (forgetting cube root), or arithmetic mistakes ( $216^{1/3}$ =6). Steps: (1) identify sphere, (2) use V=(4/3)πr³, (3) set equal to 288π, (4) solve r³=288*(3/4)=216, (5) r=∛216=6, (6) check units cm. Solving backwards requires isolating r and taking cube root accurately.
A model planet is a sphere with diameter $10\text{ ft}$ . What is its volume? (Give your answer in terms of $\pi$ or as an approximation.)
Answer: $\frac{500}{3}\pi\text{ ft}^3\ (\approx 524\text{ ft}^3)$ This question tests applying volume formulas: cylinder $V=πr^2 h$ , cone $V=\frac{1}{3}πr^2 h$ , sphere $V=\frac{4}{3}πr^3$ , selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $πr^2$ times height $h$ giving $V=πr^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3}πr^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3}πr^3$ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: $r=d/2$ ), calculate (exponents first, multiply, approximate $π≈3.14$ or leave exact). For this sphere with d=10 ft so r=5 ft, $V=\frac{4}{3}π(5^3)=\frac{4}{3}π(125)=\frac{500}{3}π≈523.6 \text{ ft}^3$ . Common errors include using diameter as radius ( $r=10$ ), forgetting to halve diameter, wrong exponent ( $r^2$ ), calculation mistakes ( $5^3=125$ not 25), or incorrect units. Steps: (1) identify shape as sphere, (2) gather dimensions (r=5 ft from d=10), (3) select formula $\frac{4}{3}πr^3$ , (4) substitute values, (5) calculate ( $\frac{4}{3}*125π=\frac{500}{3}π$ ), (6) add cubic units ft³. Always convert diameter to radius to avoid doubling errors.
A cylindrical pencil holder has diameter $10\text{ cm}$ and height $8\text{ cm}$ . What is its volume? (Give the answer in terms of $\pi$ .)
Answer: $200\pi\text{ cm}^3$ This question tests applying volume formulas: cylinder $V=πr^2 h$ , cone $V=\frac{1}{3}πr^2 h$ , sphere $V=\frac{4}{3}πr^3$ , selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area $πr^2$ times height $h$ giving $V=πr^2 h$ (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving $V=\frac{1}{3}πr^2 h$ (tapers to point reducing volume to one-third), sphere is $V=\frac{4}{3}πr^3$ (radius cubed, no height—symmetric). Apply: identify shape as cylinder, use formula with dimensions (convert diameter to radius: $r=10/2=5$ cm), calculate (exponents first, multiply, leave exact). Example: cylinder $r=5$ cm, $h=8$ cm using $V=π(25)(8)=200π$ cm³. Correct formula selection and calculation yield $200π$ cm³. Error like using diameter as radius ( $r=10$ , $π*100*8=800π$ ), wrong formula, or arithmetic ( $5^2=20$ mistaken). Steps: (1) identify cylinder, (2) gather $r=5$ , $h=8$ , (3) select $πr^2 h$ , (4) substitute, (5) calculate ( $25*8=200$ , times $π$ ), (6) units cm³.

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