Volumes of Pyramids and Cones

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Learning goals

Explain the one-third volume factor using an exact prism dissection
Compare parallel section areas using similarity and Cavalieri's principle
Calculate pyramid and cone volumes with the actual perpendicular height
Recover missing dimensions and verify them against geometric information

How the sections shrink

Sections parallel to a pyramid’s base shrink in a predictable way. Take a pyramid with base area $12$ square units and perpendicular height $8$ units. A section $2$ units below the apex is one quarter of the full height away from that apex. Every section length is therefore one quarter of its corresponding base length, by similar triangles along the sides. Its area uses the square of that factor.

A=12\left(\frac28\right)^2.

This is $3/4$ square unit. A section $4$ units below the apex instead has half the base lengths and area $3$ square units. Equal vertical steps do not give equal area changes.

The section is parallel to the base. Measure both distances from the apex; the small axial triangle explains the length fraction.

The numbers $12$ and $8$ were only an example. For a pyramid with positive base area $B$ and perpendicular height $h$ , let $t$ be the perpendicular distance from the apex down to a base-parallel section. Similar triangles give section-to-base length factor $t/h$ , so

A(t)=B\left(\frac th\right)^2.

Here $t$ runs from $0$ at the apex to $h$ at the base. This works even when the apex leans away from the base center. Every ray from the apex meets the cutting plane at the same fraction of its route to the base, so the whole section is a smaller similar copy of the base.

Recall Cavalieri’s principle: two ordinary solids between the same parallel planes have equal volumes if their section areas agree in every plane parallel to those planes. Two pyramids with equal base areas and equal perpendicular heights pass that test. At every distance $t$ below their apex levels, both section areas are the same squared fraction of their common base area. Their sections need not have the same shape.

We have established this equality of volumes without knowing a pyramid-volume formula. We will use it to justify the equal pieces in a dissection.

Check your understanding

A pyramid has base area $75$ square units and height $15$ units. What is the area of its base-parallel section $6$ units below the apex?

12

square units B

30

square units C

3

square units D

45

square units

Where one third comes from

A triangular prism can be cut into three pyramids of equal volume. A tetrahedron is a pyramid with a triangular base. Its four vertices can be named in any order; choosing a different face as its base does not change the solid.

For a triangular prism with base area $18$ square units and height $7$ units, the prism volume is $126$ cubic units. Once we justify the three equal pieces, each must occupy $42$ cubic units. The equality of the pieces needs a reason, not a guess from the picture.

The pieces are separated only for identification. In the original prism, cutting through ABF and AEF produces ABCF, ABEF, and ADEF.

For a numerical check of the comparison, take a right prism whose base triangle is right at $B$ , with $AB$ and $BC$ both $6$ units long. Its prism height is $7$ units. Both side faces $BCFE$ and $ABED$ are $6$ by $7$ rectangles. Each diagonal splits a $42$ square-unit face into two bases of area $21$ square units.

In the first pair, common apex $A$ is $6$ units from the face plane. In the second pair, common apex $F$ is also $6$ units from its face plane. Halfway from either apex to its chosen base, each section has area $21/4$ square units. At any other fraction of the height, both sections use that same fraction squared. The first two pieces therefore have equal volumes, and so do the second and third. The whole prism has volume $126$ cubic units, shared equally among the three pieces.

Three equal-volume tetrahedra fill a triangular prism#

Let $ABC$ and $DEF$ be the two triangular bases of any nondegenerate prism, with corresponding vertices $A,D$ ; $B,E$ ; and $C,F$ . The joining edges are parallel and equal. The prism may be right or oblique.

First cut along triangle $ABF$ . One side is tetrahedron $ABCF$ . The remainder is a pyramid with apex $F$ and parallelogram base $ABED$ . Cut that base along diagonal $AE$ , extending the cut to $F$ . This gives tetrahedra $ABEF$ and $ADEF$ . The cuts cover the prism with no gaps and no overlapping interiors.

A diagonal halves a parallelogram into equal areas. Use the named common apex outside each face plane to compare the volumes.

Compare $ABCF$ and $ABEF$ using bases $BCF$ and $BEF$ . These are the two equal-area halves of parallelogram $BCFE$ , split by $BF$ . Both tetrahedra have apex $A$ , so their perpendicular heights to that same face plane agree. The matching-section argument just proved gives equal volumes.

Now compare $ABEF$ and $ADEF$ using bases $ABE$ and $ADE$ . These are equal-area halves of parallelogram $ABED$ , split by $AE$ . Their common apex is $F$ , so their heights to that face plane agree too. The same section argument gives equal volumes.

Thus all three pieces have equal volumes. The prism volume is $Bh$ , where $B$ is the area of $ABC$ and $h$ is the perpendicular separation of the prism’s base planes. Tetrahedron $ABCF$ has that same base and height. Each piece takes one third of the prism, so

V_{ABCF}=\frac13Bh.

Any triangular pyramid with positive base area $B$ and height $h$ can be compared with this tetrahedron by matching every parallel section. It therefore has the same volume. Finally, split a simple polygonal base into triangles and join each to the pyramid’s apex. Those triangular pyramids have common height $h$ ; adding their volumes gives one third of the whole base area times $h$ .

For any pyramid with a simple polygonal base of positive area and an apex outside its base plane,

V=\frac13Bh.

The height is the perpendicular distance from the apex to the base plane, even if the perpendicular foot is outside the base. A slanted edge is not that distance.

Worked example 1 Use the dissection

Problem. A triangular prism has base area $18$ square centimeters and perpendicular height $7$ cm. It is cut into the three pieces described above. Find the prism volume and the volume of each piece.

Solution. The prism repeats its base area through the full height.

\begin{aligned} V_{\text{prism}}&=18(7)\\ V_{\text{prism}}&=126\text{ cm}^3. \end{aligned}

The dissection proof establishes three equal volumes, so divide by three.

V_{\text{piece}}=42\text{ cm}^3.

The three volumes add back to $126$ cubic centimeters.

Check your understanding

A triangular prism of volume $99$ cubic units is cut into these three tetrahedra. What is the volume of each?

49.5

cubic units B

297

cubic units C

33

cubic units D

11

cubic units

Find the whole base area first

The $B$ in the formula is an area, not a base edge length. For a triangular base, first use its own base and perpendicular triangle height. That triangle height lies in the base plane and differs from the pyramid’s height.

Worked example 2 Separate the two heights

Problem. A pyramid has a trapezoidal base with parallel sides $6$ cm and $12$ cm. Their perpendicular separation within the base is $5$ cm. The apex is $8$ cm perpendicularly from the base plane. Find the pyramid’s volume.

The 5 cm measurement belongs to the trapezoid; the 8 cm measurement belongs to the solid.

Solution. Average the two parallel sides, then multiply by their separation.

\begin{aligned} B&=\frac{6+12}{2}(5)\\ B&=45\text{ cm}^2. \end{aligned}

Use the solid’s $8$ cm perpendicular height for volume.

\begin{aligned} V&=\frac13(45)(8)\\ V&=120\text{ cm}^3. \end{aligned}

Check your understanding

A pyramid has a right triangular base with legs $9$ cm and $4$ cm. Its perpendicular height is $5$ cm. What is its volume?

60

cubic cm B

90

cubic cm C

30

cubic cm D

18

cubic cm

Carry the result to cones

A cone’s parallel sections shrink by the same length fraction as a pyramid’s. Compare a cone and a pyramid, each with base area $36$ square units and perpendicular height $12$ units. At $4$ units below either apex, the length fraction is one third, so each section has area $4$ square units. Halfway down, each section has area $9$ square units.

The corresponding section shapes differ, but their areas agree. The two solids share base area and perpendicular height.

Those two cuts illustrate the comparison; they do not prove it. At every distance $t$ below the apex level, both section areas are

A(t)=36\left(\frac t{12}\right)^2.

Place their bases in one plane and their apexes in a second parallel plane. Every intervening parallel plane gives equal section areas, so Cavalieri gives equal volumes. Both volumes are $144$ cubic units.

For any positive base area $B$ and perpendicular height $h$ , the same reasoning uses $B(t/h)^2$ in both solids. It works for right and oblique cones and pyramids. Moving the apex sideways changes the section positions, but not that area at each height.

For a circular cone of positive radius $r$ , substitute the disk area for $B$ .

\begin{aligned} B&=\pi r^2,\\ V&=\frac13\pi r^2h. \end{aligned}

Worked example 3 Start from a diameter

Problem. A right circular cone has base diameter $14$ cm and perpendicular height $9$ cm. Find its exact volume.

The given 14 cm diameter spans the whole circular base; the radius is half that distance.

Solution. Halve the diameter to obtain radius $7$ cm. Square that radius for the base area, then use one third of base area times height.

\begin{aligned} V&=\frac13\pi(7)^2(9)\\ V&=147\pi\text{ cm}^3. \end{aligned}

Keeping $\pi$ gives an exact answer.

Check your understanding

A cone has radius $3$ cm and perpendicular height $11$ cm. What is its exact volume?

99\pi

cubic cm B

33\pi

cubic cm C

11\pi

cubic cm D

9\pi

cubic cm

Recover the perpendicular height

For a right circular cone, the slant height $\ell$ is the straight distance along the side from the apex to the base rim. The apex lies directly above the base center, so the radius, perpendicular height, and slant height form a right triangle. The slant height is its hypotenuse.

A pyramid’s edge or side-face altitude is also a different measurement from the perpendicular height of the solid. The next lesson uses slanted measurements to find surface areas. Here they help us recover the height needed for volume.

Worked example 4 Read the axial right triangle

Problem. A right circular cone has radius $8$ cm and slant height $17$ cm. Find its perpendicular height and exact volume.

The dashed interior height meets the radius at a right angle. The 17 cm side is the hypotenuse.

Solution. Subtract the radius square from the slant-height square.

\begin{aligned} h^2&=17^2-8^2\\ h^2&=225\\ h&=15\text{ cm}. \end{aligned}

Now use that perpendicular height.

\begin{aligned} V&=\frac13\pi(8)^2(15)\\ V&=320\pi\text{ cm}^3. \end{aligned}

Missing dimensions can also come from the volume measurement. For positive base area, solving $V=Bh/3$ for the height gives

h=\frac{3V}{B}.

Worked example 5 Check a recovered height against the geometry

Problem. An oblique pyramid has base area $54$ square centimeters and volume $288$ cubic centimeters. Its apex is $S$ , and $H$ is the perpendicular foot on the base plane. A point $Q$ in that plane is $12$ cm from $H$ , and $SQ$ is $20$ cm. Find the perpendicular height and check whether both measurements support it.

This side section shows the actual perpendicular foot H. The segment SQ follows a slanted route to the base plane.

Solution. Recover the perpendicular height from the volume.

\begin{aligned} h&=\frac{3(288)}{54}\\ h&=16\text{ cm}. \end{aligned}

Because $SH$ is perpendicular to the base plane, it is perpendicular to $HQ$ . Check the right triangle.

\begin{aligned} 16^2+12^2&=400,\\ 20^2&=400. \end{aligned}

The geometric measurement supports the recovered height. Substituting it into the volume rule gives $288$ cubic centimeters again. The $20$ cm slanted segment would not give that volume.

Check your understanding

A pyramid with a square base of side $6$ cm has volume $84$ cubic centimeters. What is its perpendicular height?

14

cm B

7

cm C

21

cm D

7/3

Volumes of Pyramids and Cones

Learning goals

How the sections shrink

Where one third comes from

Find the whole base area first

Carry the result to cones

Recover the perpendicular height

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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