Do the volume formulas change for an oblique cone or a tilted pyramid?

No. By Cavalieri's principle, only the cross-sectional areas and the total height matter, and tilting the apex sideways does not change either one. Use V = 1/3Bh with h as the perpendicular distance from the apex to the base plane, not the length of the tilted axis.

When do I use slant height instead of perpendicular height?

Slant height belongs to lateral surface area, such as r for a cone. Volume always uses the perpendicular height. If a problem gives you slant height and radius, use r^2 + h^2 = ^2 to solve for h first.

How do I find the height of a cone when I already know its volume?

Substitute into V = 1/3 r^2 h and solve for h, giving h = 3V/ r^2. For a pyramid the same rearrangement gives h = 3V/B. The factor of 3 in the numerator is where most algebra errors show up, so multiply by 3 before dividing.

Why is a sphere's volume 4/3 r^3 and not something involving a height?

A sphere has no independent height; its size is completely determined by the radius, and the distance across it is just 2r. Comparing a hemisphere to a cylinder with a cone removed shows the hemisphere equals 2/3 r^3, and doubling it gives the full sphere. Notice the radius is cubed, because volume requires three factors of length.

GEOM-10.4

Volume: Pyramids, Cones & Spheres

Learn to compute volumes of pyramids, cones, spheres, and hemispheres, see why pointy solids carry a factor of one third, and avoid slant-height mistakes.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Volume: Pyramids, Cones & Spheres, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Volume: Pyramids, Cones & Spheres free

What this lesson covers

You already know that a prism's volume is base area times height, and that a cylinder works the same way. Now the solids get pointy and round. A cone sitting inside a cylinder of the same base and height fills exactly one third of it — not most of it, not half of it, one third. A sphere squeezed inside the smallest cylinder that holds it takes up exactly two thirds of that cylinder.

This lesson gives you the four formulas you need, an informal but honest explanation of where the one third and the four thirds come from, and a careful look at the single mistake that ruins more pyramid and cone problems than anything else: using slant height where the formula demands perpendicular height. By the end you should be able to pour concrete into a conical form, fill a spherical tank, or hollow out a composite solid and get the right number of cubic units.

Why Pointy Solids Carry a Factor of One Third

Take a cube with edge length

s

and pick one corner. Draw segments from that corner to every other vertex. The cube splits into exactly three congruent square pyramids, each having one face of the cube as its base and the cube's edge as its perpendicular height. Since the three pyramids fill the cube and are congruent, each has volume

\frac{1}{3}s^3

, which is

\frac{1}{3}

times the base area

s^2

times the height

s

. That single dissection is the cleanest reason to believe the one third.

To know it works for every pyramid and cone, not just that special one, use Cavalieri's principle: if two solids have equal heights and every horizontal slice at the same level has the same area, they have the same volume. Slice a cone at height

h

above the base and you get a circle whose radius has been scaled down; slice any pyramid with the same base area and same total height at that level and the cross section shrinks by the same ratio. Matching slices means matching volumes, so all cones and pyramids obey the same rule.

A classroom demonstration makes it concrete: fill a hollow cone with rice and empty it into a cylinder with the identical base circle and identical height. It takes three full cones to fill the cylinder, every time. So

V_{\text{pyramid}} = \frac{1}{3}Bh \qquad V_{\text{cone}} = \frac{1}{3}\pi r^2 h

where

B

is the area of the base and

h

is the perpendicular height. The common misconception is that the fraction depends on how steep the point is. It does not. A tall skinny cone and a wide flat cone both take exactly one third of their matching cylinder.

The Four Formulas and What Each Letter Means

Solid	Volume formula	What to measure
Pyramid	$V=\frac{1}{3}Bh$	$B$ = area of the base polygon, $h$ = perpendicular height from apex to base plane
Cone	$V=\frac{1}{3}\pi r^2 h$	$r$ = base radius, $h$ = perpendicular height
Sphere	$V=\frac{4}{3}\pi r^3$	$r$ = radius (half the diameter)
Hemisphere	$V=\frac{2}{3}\pi r^3$	$r$ = radius of the flat circular face

The letter

h

always means the perpendicular distance from the apex straight down to the plane of the base. The slant height, usually written

\ell

, runs along the outside surface from the apex to the edge of the base. Those two are different lengths, and slant height belongs to surface area, not volume.

In a right cone,

r

h

, and

\ell

form a right triangle, so

r^2 + h^2 = \ell^2

. If a problem hands you

\ell

and asks for volume, your first move is the Pythagorean Theorem to recover

h

. For a regular pyramid, the same triangle uses the apothem of the base instead of

r

a^2 + h^2 = \ell^2

.

Two more traps. First, a diameter is not a radius; if a ball is 12 inches across, then

r=6

and

r^3=216

, not

1728

. Second, cube your units. Volume answers end in cubic units, and if you switch from feet to inches you multiply by

12^3 = 1728

, not by 12.

Oblique pyramids and cones — where the apex leans off to one side — use the exact same formulas, because Cavalieri's principle only cares about cross-sectional area at each height, not about whether the axis is vertical.

Spheres and Hemispheres

The sphere formula

V=\frac{4}{3}\pi r^3

can be justified without calculus using Cavalieri's principle. Compare a hemisphere of radius

r

with a second solid: a cylinder of radius

r

and height

r

with a cone of radius

r

and height

r

drilled out of it, apex pointing down.

Slice both at height

y

above the flat base. The hemisphere's cross section is a circle whose radius satisfies

x^2 = r^2 - y^2

, so its area is

\pi(r^2-y^2)

. The drilled cylinder's cross section is a ring: the big circle has area

\pi r^2

and the cone's hole at that level has radius

y

, area

\pi y^2

, leaving

\pi r^2 - \pi y^2

. The two areas match at every level, so the volumes match:

V_{\text{hemisphere}} = \pi r^2 \cdot r - \frac{1}{3}\pi r^2 \cdot r = \frac{2}{3}\pi r^3

Double it for a full sphere and you get

\frac{4}{3}\pi r^3

. A useful consequence, known to Archimedes: a sphere occupies exactly two thirds of the smallest cylinder that contains it, since that cylinder has volume

\pi r^2(2r) = 2\pi r^3

.

Where students go wrong: mixing the sphere volume

\frac{4}{3}\pi r^3

with the sphere surface area

4\pi r^2

. Check the exponent against the units you want. Volume needs three factors of length, so the radius must be cubed. Also remember that a hemisphere has no height variable of its own — its height equals its radius, so you never need a separate

h

Composite Solids, Scaling, and Real Objects

Real objects are rarely one clean shape. A grain silo is a cylinder with a cone on top. An ice cream cone with a scoop is a cone plus a hemisphere. A funnel is a cone with a smaller cone removed. The strategy is always the same: break the solid into standard pieces, compute each volume separately, then add or subtract. Keep answers in terms of

\pi

until the very last step so rounding errors do not pile up.

Watch for shared dimensions. When a hemisphere caps a cylinder, the hemisphere's radius equals the cylinder's radius, and the hemisphere adds a length of

r

to the total outside height. So if a tank is 20 feet tall overall with a 3-foot radius hemispherical top, the cylindrical part is only 17 feet tall.

Scaling matters too. If you enlarge a solid by a linear scale factor

k

, every length multiplies by

k

, surface area multiplies by

k^2

, and volume multiplies by

k^3

. Doubling the radius of a spherical balloon makes it hold eight times as much air, which surprises people every time.

Modeling problems usually go one step past the volume. Typical follow-ups include converting cubic inches to fluid capacity, multiplying volume by a density to get mass, or dividing a container's volume by a scoop's volume to count how many scoops fit. Read the units in the question and make your final answer match them. If a problem says a sand pile weighs 100 pounds per cubic foot, compute the cubic feet first, then multiply — never mix the multiplication into the volume formula, because that is where sign-of-reasonableness checks get lost.

Key terms

Perpendicular height (altitude).: The shortest distance from the apex of a pyramid or cone to the plane containing the base, measured along a segment perpendicular to that plane. This is the $h$ in every volume formula.
Slant height.: For a right cone, the distance from the apex to a point on the edge of the base measured along the lateral surface; for a regular pyramid, the height of a triangular face. Used for surface area, never directly for volume.
Base area ( $B$ ).: The area of the polygon or circle that forms the base of a pyramid or cone. For a cone, $B=\pi r^2$ ; for a regular polygon base, $B=\frac{1}{2}aP$ with apothem $a$ and perimeter $P$ .
Cavalieri's principle.: If two solids have the same height and every pair of corresponding cross sections at the same level has equal area, then the solids have equal volume.
Apex.: The single point where the lateral faces of a pyramid meet, or the tip of a cone.
Hemisphere.: Exactly half of a sphere, cut by a plane through the center. Its volume is $\frac{2}{3}\pi r^3$ and its height equals its radius.
Oblique solid.: A pyramid or cone whose apex is not directly above the center of the base. Volume formulas are unchanged, but the perpendicular height is not the length of the axis.
Composite solid.: A three-dimensional figure formed by joining or removing standard solids; its volume is found by adding or subtracting the pieces.

Worked example

A dessert consists of a right circular cone filled completely with sorbet and topped by a solid hemisphere of sorbet resting on the cone's opening. The cone's opening has a diameter of 6 centimeters and the cone's slant height is 10.44 centimeters, which rounds from an exact value. The cone's perpendicular height is exactly 10 centimeters. Find the total volume of sorbet, exactly and rounded to the nearest tenth of a cubic centimeter.

Step 1: Identify the radius. The diameter is 6 centimeters, so

r = 3

centimeters. Both the cone and the hemisphere share this radius, since the hemisphere sits exactly on the cone's circular opening.

Step 2: Confirm the height to use. The formula needs the perpendicular height,

h = 10

centimeters. The slant height 10.44 is a check, not the number you plug in:

\sqrt{3^2 + 10^2} = \sqrt{109} \approx 10.44

, which matches. Using 10.44 in place of 10 would inflate the answer.

Step 3: Cone volume.

V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2(10) = \frac{1}{3}\pi (9)(10) = 30\pi

Step 4: Hemisphere volume.

V_{\text{hemi}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3)^3 = \frac{2}{3}\pi (27) = 18\pi

Step 5: Add the pieces.

V_{\text{total}} = 30\pi + 18\pi = 48\pi

Step 6: Approximate at the end.

48\pi \approx 150.796

, so the dessert holds about 150.8 cubic centimeters of sorbet.

Sanity check: the cylinder that would enclose the cone has volume

\pi(9)(10) = 90\pi

, and the cone is one third of that,

30\pi

. The scoop is smaller than the cone, which matches the picture.

Practice questions

A square pyramid has a base edge of 6 inches and a slant height of 5 inches measured up the middle of a triangular face. What is its volume?

$48$ cubic inches
$60$ cubic inches
$120$ cubic inches
$144$ cubic inches

Answer: $48$ cubic inches

The 5 inches is slant height, not perpendicular height, so you cannot substitute it directly. The apothem of the square base is half the edge,

3

inches. That apothem, the perpendicular height, and the slant height form a right triangle:

3^2 + h^2 = 5^2

, so

h^2 = 16

and

h = 4

. The base area is

B = 6^2 = 36

. Then

V = \frac{1}{3}(36)(4) = 48

cubic inches. The value 60 is the common wrong answer that comes from using the slant height as if it were the altitude, and 144 comes from forgetting the one third entirely.

A spherical water tank holds

972\pi

cubic feet when full. Find the radius of the tank, and then explain what happens to the capacity if a new tank is built with twice that radius.

Answer: The radius is 9 feet; doubling the radius to 18 feet multiplies the capacity by 8, giving $7776\pi$ cubic feet.

Set the formula equal to the given volume:

\frac{4}{3}\pi r^3 = 972\pi

. Divide both sides by

\pi

to get

\frac{4}{3}r^3 = 972

, then multiply by

\frac{3}{4}

r^3 = 729

. Taking the cube root gives

r = 9

feet. For the second part, volume scales by the cube of the linear scale factor. With

k = 2

, the new volume is

2^3 = 8

times the old one, so

8 \cdot 972\pi = 7776\pi

cubic feet. You can verify directly:

\frac{4}{3}\pi(18)^3 = \frac{4}{3}\pi(5832) = 7776\pi

. Students often guess the capacity merely doubles, which underestimates the tank by a factor of four.

A cylinder and a cone have the same radius and the same height. A student claims that if you melt the cone and pour it into the cylinder, the liquid will reach halfway up the cylinder. Is the student right? Justify your answer with the formulas.

Answer: No. The liquid rises to one third of the cylinder's height, not one half.

The cone's volume is

\frac{1}{3}\pi r^2 h

and the cylinder's is

\pi r^2 h

, so the cone contains exactly one third as much. When that liquid sits in the cylinder it forms a shorter cylinder of the same radius, so its height

y

satisfies

\pi r^2 y = \frac{1}{3}\pi r^2 h

. Dividing both sides by

\pi r^2

gives

y = \frac{1}{3}h

. The halfway guess would require the cone to be half the cylinder, which contradicts both the dissection of a cube into three pyramids and the rice-pouring demonstration.

FAQ

Do the volume formulas change for an oblique cone or a tilted pyramid?: No. By Cavalieri's principle, only the cross-sectional areas and the total height matter, and tilting the apex sideways does not change either one. Use $V = \frac{1}{3}Bh$ with $h$ as the perpendicular distance from the apex to the base plane, not the length of the tilted axis.
When do I use slant height instead of perpendicular height?: Slant height belongs to lateral surface area, such as $\pi r \ell$ for a cone. Volume always uses the perpendicular height. If a problem gives you slant height and radius, use $r^2 + h^2 = \ell^2$ to solve for $h$ first.
How do I find the height of a cone when I already know its volume?: Substitute into $V = \frac{1}{3}\pi r^2 h$ and solve for $h$ , giving $h = \frac{3V}{\pi r^2}$ . For a pyramid the same rearrangement gives $h = \frac{3V}{B}$ . The factor of 3 in the numerator is where most algebra errors show up, so multiply by 3 before dividing.
Why is a sphere's volume $\frac{4}{3}\pi r^3$ and not something involving a height?: A sphere has no independent height; its size is completely determined by the radius, and the distance across it is just $2r$ . Comparing a hemisphere to a cylinder with a cone removed shows the hemisphere equals $\frac{2}{3}\pi r^3$ , and doubling it gives the full sphere. Notice the radius is cubed, because volume requires three factors of length.

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