Volumes of Prisms, Cylinders, Pyramids, Cones, and Spheres

14.4: Volumes of Prisms, Cylinders, Pyramids, Cones, and Spheres. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 4 of 63 parts Practice Flashcards

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Prisms and cylinders

Objective: Multiply base area by perpendicular height.

Prerequisite skills: circle area, fractions, and multiplying dimensions. Any prism has constant cross-sectional area B through its perpendicular height h, giving volume Bh. A cylinder has circular base B=πr². Rectangular prisms use B=lw. Common error: using circumference as the cylinder base area. Restrictions and interpretation: Heights in volume formulas are perpendicular; radius is half a supplied diameter; use cubic units.

Formulas

$V_prism=Bh; V_cylinder=πr²h$

Worked examples

Try itA cylinder has radius 23 and perpendicular height 24. If V=Kπ, what is K?Show answer

The base area is π(23)²=529π. Multiply by height 24: V=12696π cubic units, so K=12696. Therefore the requested answer is 12696.

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Prisms and cylinders

Objective: Multiply base area by perpendicular height.

Formulas

$V_prism=Bh; V_cylinder=πr²h$

Worked examples

Try itA cylinder has radius 23 and perpendicular height 24. If V=Kπ, what is K?Show answer

The base area is π(23)²=529π. Multiply by height 24: V=12696π cubic units, so K=12696. Therefore the requested answer is 12696.

Pyramids and cones

Objective: Apply one third of base area times height.

A pyramid and a cone each have volume Bh/3, where h is perpendicular to the base plane. For a cone the base area is πr². A slant height cannot replace h in this formula. Common error: omitting the factor one third or using slant height. Restrictions and interpretation: Heights in volume formulas are perpendicular; radius is half a supplied diameter; use cubic units.

Formulas

$V_pyramid=Bh/3; V_cone=πr²h/3$

Worked examples

Try itA pyramid has base area 75 and perpendicular height 6. What is its volume?Show answer

V=Bh/3=75×6/3=150 cubic units. The stated 6 is the perpendicular height, not an edge. Therefore the requested answer is 150.

Spheres and inverse volume

Objective: Use the sphere formula and solve a dimension constraint.

A sphere has volume 4πr³/3. A hemisphere has half that volume. If a volume is given and a radius is requested, isolate r³ before taking its nonnegative cube root. Cubic dependence makes doubling the radius multiply volume by eight. Common error: using the surface-area formula 4πr² for a volume. Restrictions and interpretation: Heights in volume formulas are perpendicular; radius is half a supplied diameter; use cubic units. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$V_sphere=4πr³/3$

Worked examples

Try itA sphere has volume 497664π cubic units. What is its radius?Show answer

Set 4πr³/3=497664π. Cancel π and multiply by 3/4: r³=373248. The positive cube root is r=72. Therefore the requested answer is 72.

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