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Volumes of Pyramids and Cones: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A polygonal base

A pyramid has base area $28.8$ square centimeters and perpendicular height $5$ cm. Find its volume.

Use the area of the base and the perpendicular height.

Text description of this figure

A pyramid has base area 28.8 square centimeters and a long-dashed perpendicular height of 5 cm. A right-angle mark identifies its foot on the base plane. Short-dashed edges are hidden.
1. Hint 1
  Compare the pyramid with a prism having the same base area and height.
2. Hint 2
  The pyramid occupies one third of that prism volume.
Answer

$48$ cubic centimeters.

Full solution

The matching prism has volume
$V_{\text{prism}}=28.8(5).$
This is 144 cubic centimeters.

Take one third for the pyramid.
$\begin{aligned}V&=144/3\\V&=48\text{ cm}^3.\end{aligned}$

Answer

$48$ cubic centimeters.

Key idea

The one-third factor applies to the whole base area.
Problem 2 The circular base

A circular cone has diameter $9$ cm and perpendicular height $8$ cm. Find its exact volume in terms of pi.

The stated diameter spans twice the labeled radius.

Text description of this figure

A circular cone has given base diameter 9 cm and perpendicular height 8 cm. The drawing labels the radius r without evaluating it. The rear half of the base and interior height are dashed, with a right-angle mark at the center.
1. Hint 1
  The disk-area formula needs a radius rather than a diameter.
2. Hint 2
  Halve the diameter, square the radius, then use one third of base area times height.
Answer

$54\pi$ cubic centimeters.

Full solution

The radius is half the diameter, or 4.5 cm.

The circular base area is
$B=20.25\pi\text{ cm}^2.$

Use the perpendicular height.
$\begin{aligned}V&=\frac13(20.25\pi)(8)\\V&=54\pi\text{ cm}^3.\end{aligned}$

Answer

$54\pi$ cubic centimeters.

Key idea

Halve a given diameter before squaring for base area.
Problem 3 Two perpendicular heights

A pyramid has a right triangular base with perpendicular legs $6$ cm and $10$ cm. Its apex is $7.5$ cm perpendicularly from the base plane. Find its volume.

This true base view shows the two legs used to find its area.

Text description of this figure

The true view of the right triangular base has perpendicular legs 6 cm and 10 cm. The separate solid height, stated in the task, is 7.5 cm. The right angle in the figure belongs to the base triangle.
1. Hint 1
  Find the area of the triangular base first.
2. Hint 2
  The triangle height belongs to the base area calculation; the 7.5 cm distance belongs to the solid volume.
Answer

$75$ cubic centimeters.

Full solution

The base is a right triangle.
$\begin{aligned}B&=\frac12(6)(10)\\B&=30\text{ cm}^2.\end{aligned}$

Use the separate solid height.
$\begin{aligned}V&=\frac13(30)(7.5)\\V&=75\text{ cm}^3.\end{aligned}$

Answer

$75$ cubic centimeters.

Key idea

Keep the base triangle height separate from the pyramid height.
Problem 4 Two apex positions

A right pyramid and an oblique pyramid each have base area $40$ square centimeters and perpendicular height $12$ cm. The bases need not have the same shape.

Find the volume of each pyramid.

Find the area of each base-parallel section $3$ cm above the base plane, and explain why the lean does not affect either result.
1. Hint 1
  For volume, both pyramids have the same base area and perpendicular height.
2. Hint 2
  For the section, measure its distance from the apex, not from the base.
3. Hint 3
  Apply the squared length fraction to each base area.
Answer

Each volume is $160$ cubic cm; each section area is $22.5$ square cm. The apex position does not change these values.

Full solution

Both pyramids use the same two factors in the volume relation.
$\begin{aligned}V&=\frac13(40)(12)\\V&=160\text{ cm}^3.\end{aligned}$

The section is 9 cm below the apex level because it is 3 cm above the base in a solid of height 12 cm.

Its area uses the square of the apex-distance fraction.
$\begin{aligned}A&=40\left(\frac9{12}\right)^2\\A&=22.5\text{ cm}^2.\end{aligned}$

Moving the apex sideways shifts section positions.
At every height their areas still use the same squared fraction of the base areas.
Cavalieri therefore also explains the equal whole volumes.

Answer

Each volume is $160$ cubic cm; each section area is $22.5$ square cm. The apex position does not change these values.

Key idea

A lean changes section positions while preserving the area at each corresponding height.
Problem 5 An unknown opening

A circular cone has volume $78\pi$ cubic centimeters and perpendicular height $6.5$ cm. Find its radius and diameter.
1. Hint 1
  The volume relation first gives the square of the radius.
2. Hint 2
  After canceling pi, isolate the radius square and take its positive square root.
Answer

Radius $6$ cm; diameter $12$ cm.

Full solution

Substitute the supplied volume and height.
$78\pi=\frac13\pi r^2(6.5).$

Cancel pi and solve for the squared radius.
$\begin{aligned}r^2&=\frac{3(78)}{6.5}\\r^2&=36.\end{aligned}$

The radius is the positive square root, 6 cm.
The diameter is twice that length, or 12 cm.

Check the volume.
$\frac13\pi(6)^2(6.5)=78\pi\text{ cm}^3.$

Answer

Radius $6$ cm; diameter $12$ cm.

Key idea

Recover a squared radius before taking the positive root.
Problem 6 A measured side

A right circular cone has radius $7$ cm and slant height $25$ cm. Find its perpendicular height and exact volume.

The slanted 25 cm side is the hypotenuse.

Text description of this figure

A right circular cone has radius 7 cm and slant height 25 cm. The interior dashed segment h is perpendicular to the base radius. Its rear base arc is hidden; no height value or volume is shown.
1. Hint 1
  The radius and perpendicular height meet at a right angle.
2. Hint 2
  Use the slant height as the hypotenuse, then substitute the recovered perpendicular height into the volume relation.
Answer

Height $24$ cm; volume $392\pi$ cubic centimeters.

Full solution

The right-triangle relation gives
$\begin{aligned}h^2&=25^2-7^2\\h^2&=576\\h&=24\text{ cm}.\end{aligned}$

Use that perpendicular height.
$\begin{aligned}V&=\frac13\pi(7)^2(24)\\V&=392\pi\text{ cm}^3.\end{aligned}$

The height and radius squares add to 625, which matches the given slant-height square.

Answer

Height $24$ cm; volume $392\pi$ cubic centimeters.

Key idea

The slant height supplies a right-triangle constraint, not the volume height.
Problem 7 Locate the slice

A pyramid has base area $150$ square centimeters and perpendicular height $20$ cm. A plane parallel to its base cuts a section of area $54$ square centimeters.

Find the section distance below the apex.

Find the section distance above the base plane.

This axial view locates the section relative to the apex and base plane. The diagram is schematic and not to scale.

Text description of this figure

The pyramid has base area 150 square centimeters and perpendicular height 20 cm. The highlighted base-parallel section has area 54 square centimeters. Its perpendicular distance below the apex is t; no requested distance is evaluated.
1. Hint 1
  Area ratios of similar parallel sections are squares of length ratios.
2. Hint 2
  Take the positive square root of the section-to-base area ratio.
3. Hint 3
  Use the fraction of the whole perpendicular height, then find the complementary distance.
Answer

Below the apex: $12$ cm. Above the base plane: $8$ cm.

Full solution

The area ratio is
$\frac{54}{150}=\frac9{25}.$
Its positive square root gives the section-to-base length fraction, three fifths.

Multiply that fraction by the full height to obtain the apex distance.
$\begin{aligned}t&=\frac35(20)\\t&=12\text{ cm}.\end{aligned}$

Subtract from the full height for the distance above the base plane.
$20-12=8\text{ cm}.$

Answer

Below the apex: $12$ cm. Above the base plane: $8$ cm.

Key idea

The area fraction must be square-rooted before locating a parallel section.
Problem 8 Filling a conical mold

A mold has a right circular conical cavity with inside radius $3.6$ cm and perpendicular depth $12.5$ cm. It is filled exactly to the rim.

Find the volume of material needed, exactly in terms of pi in cubic centimeters, then to the nearest tenth of a milliliter. Use one cubic centimeter as one milliliter.
1. Hint 1
  Use the inside cavity dimensions to measure the amount of material.
2. Hint 2
  Calculate the exact cone volume first; then approximate pi and convert the unit.
Answer

$54\pi$ cubic cm, or $54\pi$ mL; approximately $169.6$ mL.

Full solution

The inside base disk area is
$B=12.96\pi\text{ cm}^2.$

Use the perpendicular depth.
$\begin{aligned}V&=\frac13(12.96\pi)(12.5)\\V&=54\pi\text{ cm}^3.\end{aligned}$

One cubic centimeter is one milliliter, so the numerical value in milliliters is also fifty-four times pi.

Only now approximate.
$V\approx169.6\text{ mL}.$
This is the amount needed to fill the stated ideal cavity to its rim.

Answer

$54\pi$ cubic cm, or $54\pi$ mL; approximately $169.6$ mL.

Key idea

Use inside dimensions and preserve exact values until the final rounding.
Problem 9 The section comparison

Numerical case: a pyramid has base area $50$ square centimeters and perpendicular height $10$ cm. Find the area of a base-parallel section $4$ cm below its apex.

General case: a pyramid has positive base area B and positive height h. A base-parallel plane is perpendicular distance t below its apex, with t between zero and h. Derive its section area in terms of B, h, and t.

Use this result to explain why any two pyramids with equal base areas and equal perpendicular heights have equal volumes, even when their base shapes differ. Do not assume the pyramid-volume formula.

The numerical case measures both distances from the apex.

Text description of this figure

For the numerical case, the pyramid has base area 50 square centimeters and height 10 cm. The plane section is parallel to the base, at perpendicular distance 4 cm below the apex. Its area is not evaluated. The axial triangle illustrates the length comparison used in the general case.
1. Hint 1
  Compare distances from the apex to obtain a section-to-base length fraction.
2. Hint 2
  A parallel section is similar to its own base, so area uses the square of that fraction.
3. Hint 3
  For two pyramids, compare corresponding areas at every height between common parallel boundary planes.
Answer

Numerical area $8$ square cm. General area $B(t/h)^2$ . Equal base areas and heights give equal section areas at every height, so the volumes are equal.

Full solution

Numerical case: similar triangles along the sides give section-to-base length fraction four tenths.
$\begin{aligned}A&=50\left(\frac4{10}\right)^2\\A&=8\text{ cm}^2.\end{aligned}$

General case: the same parallel-plane argument gives length fraction t divided by h for every corresponding section length.
The numerical values played no special role.

A similar plane region has the square of the length fraction in area.
$A(t)=B\left(\frac th\right)^2.$
At the apex the section has zero area; at the base it has area B.

For the volume comparison, place both bases in one plane and the apexes in a second parallel plane.
Equal heights make these common boundary planes possible.

At every intervening level both sections have the same area because both B and the distance fraction agree.
Cavalieri gives equal volumes without using a pyramid-volume formula.

Answer

Numerical area $8$ square cm. General area $B(t/h)^2$ . Equal base areas and heights give equal section areas at every height, so the volumes are equal.

Key idea

Matching every corresponding section can establish equal volumes before a volume formula is known.
Problem 10 A missing base edge

A pyramid of volume $600$ cubic centimeters has apex S. Its perpendicular foot on the base plane is H. A point Q in that plane satisfies SQ of $29$ cm and HQ of $21$ cm.

Its base is right triangle KLM, right at K, with KL of $12$ cm and unknown KM.

Find the perpendicular height SH and the base edge KM. Check the recovered dimensions in the volume relation.

The upper panel determines the solid height; the lower panel shows the base measurements.

Text description of this figure

A pyramid has volume 600 cubic centimeters. Its side-section triangle SQH is right at H, with SQ 29 cm, HQ 21 cm, and unknown perpendicular SH. Its separate true base KLM is right at K, with KL 12 cm and KM unknown. The two panels describe different planes.
1. Hint 1
  First find the perpendicular height from the given side section.
2. Hint 2
  Use that height and the volume to recover the entire triangular base area.
3. Hint 3
  The area of the right triangular base is half the product of its perpendicular legs.
Answer

Height SH is $20$ cm; base edge KM is $15$ cm. The recovered dimensions give volume $600$ cubic cm.

Full solution

SH is perpendicular to every base-plane line through H, so triangle SHQ is right at H.
$\begin{aligned}SH^2&=29^2-21^2\\SH^2&=400\\SH&=20\text{ cm}.\end{aligned}$

Recover the base area from the volume.
$\begin{aligned}B&=\frac{3(600)}{20}\\B&=90\text{ cm}^2.\end{aligned}$

Let x be KM in centimeters.
The right triangular base gives
$\begin{aligned}\frac12(12)x&=90\\6x&=90\\x&=15.\end{aligned}$

The recovered base legs give area 90 square centimeters.
Together with the actual solid height, they reproduce the volume.
$\begin{aligned}V&=\frac13(90)(20)\\V&=600\text{ cm}^3.\end{aligned}$

The side-section check also holds because 400 plus 441 is 841, the square of 29.

Answer

Height SH is $20$ cm; base edge KM is $15$ cm. The recovered dimensions give volume $600$ cubic cm.

Key idea

Separate the base geometry from the perpendicular solid height before checking a recovered dimension.

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