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Pyramids and Cones: Volume and Surface Area

Find the volume and surface area of pyramids and cones, use the slant height, and work backwards to a missing dimension.

Practice this skill 25 min Pyramids and cones Leer en español
Key idea

What you'll learn

A pyramid comes to a point above a flat base, like the Great Pyramid in Egypt or the roof of a gazebo. A cone is the round version: a circle base that rises to a point, like an ice-cream cone, a traffic cone, or a funnel.

In this lesson you will learn to:

  • find the volume of any pyramid or cone (how much it holds),
  • find the surface area (how much material covers it), including the slant height,
  • tell the height and the slant height apart, and find one from the other,
  • work backwards to find a missing height, radius, or side when the volume or surface area is given.

Why it matters on the GED: Target Q.5 (3-D figures) appears on almost every GED Math test, usually as 1–3 questions. Pyramid and cone items are popular because they test several things at once: reading a figure, choosing between height and slant height, using the formula sheet, and often the Pythagorean theorem. The good news: both formulas are printed on the formula sheet, so your job is to plug in the right numbers.

Real life: How much mulch is in a cone-shaped pile? How much canvas does a tent need? How much does a funnel hold? Same math.

Words to know

Key words

  • pyramid — a solid with a polygon base and triangular sides that meet at one point (a square pyramid has a square base).
  • cone — a solid with a circular base and one curved side that narrows to a point.
  • apex (vertex) — the top point of a pyramid or cone.
  • base (B) — the flat bottom face. In formulas, B means the area of the base (a 6-by-6 square base has B = 36).
  • height (h) — the straight, vertical distance from the apex down to the base, at a right angle to the base (it goes through the inside of the solid).
  • slant height (s) — the distance from the apex down the outside surface to the edge of the base (for a pyramid, to the middle of a base edge).
  • radius (r) — the distance from the center of a circle to its edge (diameter 10 → radius 5).
  • perimeter (p) — the distance around the base (a square with side 10 has p = 40).
  • lateral area — the area of the sides only, without the base (the triangles of a pyramid, the curved part of a cone).
  • surface area (SA) — the total area of every outside surface, in square units (ft², cm²).
  • volume (V) — the space inside, in cubic units (ft³, cm³).

All Math words in the glossary

Square pyramid. The base is a square with side b. The height h is a dashed line from the apex straight down to the center of the base, meeting it at a right angle. The slant height s runs from the apex down the middle of a triangular face to the midpoint of a base edge. A dashed segment of length b/2 joins the center of the base to that midpoint, so h, b/2 and s form a right triangle with s as the hypotenuse. h s base edge b b b/2

In a pyramid, the height h, half of a base edge \tfrac{b}{2}, and the slant height s make a right triangle. The slant height is the hypotenuse.

Volume: one-third of a prism or cylinder

Picture a box and a pyramid with the same base and the same height. If you filled the pyramid with water and poured it into the box, it would take exactly 3 pyramids to fill the box. So:

\text{pyramid volume} = \tfrac{1}{3} \times \text{(box with same base and height)} = \tfrac{1}{3}Bh

The same is true for a cone and a cylinder:

\text{cone volume} = \tfrac{1}{3} \times \text{(cylinder)} = \tfrac{1}{3}\pi r^2 h

How to find B for a pyramid. B is the area of whatever shape the base is:

Base shapeBase area BExample
square, side bb^2side 6 → B = 36
rectangle l \times wlw8 by 6 → B = 48
triangle\tfrac{1}{2} \times \text{base} \times \text{height}base 6, height 4 → B = 12

Example: a square pyramid with base side 6 and height 10: V = \tfrac{1}{3}(36)(10) = 120 cubic units.

A box 6 by 6 by 10 drawn with dashed edges, and a square pyramid with the same 6 by 6 base and the same height 10 inside it. The box holds 36 times 10 = 360 cubic units; the pyramid holds one-third of that, 120 cubic units. 6 6 h = 10 Box (prism): V = 6 × 6 × 10 = 360 Pyramid, same base and same height: V = ⅓ × 360 = 120 3 pyramids fill the box.

For a cone, the base is a circle, so B = \pi r^2. Example: r = 3, h = 7: V = \tfrac{1}{3}\pi(9)(7) = 21\pi \approx 66.0.

A cone with radius 3 and height 7 drawn inside a dashed cylinder with the same radius and height. The cone holds one-third of the cylinder: V = one-third times pi times 9 times 7 = 21 pi, about 66.0 cubic units. r = 3 h = 7 Cylinder: π(3²)(7) = 63π Cone = ⅓ of it: ⅓π(9)(7) = 21π ≈ 66.0 cubic units

Units: volume is always in cubic units (in³, ft³, m³). If the answer choices are in square units, they are not volume.

Height vs. slant height

This is the #1 place students lose points.

  • The height h goes straight down inside the solid, from the apex to the center of the base. It is used for volume.
  • The slant height s runs down the outside surface. It is the height of each triangular face of a pyramid. It is used for surface area.

They form a right triangle, so you can always find one from the other with the Pythagorean theorem (a^2 + b^2 = c^2, also on the formula sheet):

SolidRight triangleFormula
square pyramid, base edge bh, \tfrac{b}{2}, ss^2 = h^2 + \left(\tfrac{b}{2}\right)^2
cone, radius rh, r, ss^2 = h^2 + r^2

Example (pyramid): base edge 16, height 15. Half the edge is 8, so s^2 = 15^2 + 8^2 = 289 and s = 17.

Square pyramid with base edge 16 and height 15. The height (dashed, inside), half the base edge (8) and the slant height (along the front face) form a right triangle, shown separately on the right: 15 squared plus 8 squared = 289, so the slant height is 17. h = 15 s = 17 8 16 15 8 17 the right triangle inside the pyramid

Example (cone, going the other way): r = 9, s = 15. Then h^2 = 15^2 - 9^2 = 144 and h = 12.

Cone with radius 9 and slant height 15; the height is unknown. The height, radius and slant height form a right triangle, shown on the right: 15 squared minus 9 squared = 144, so the height is 12. r = 9 h = ? s = 15 h = 12 9 15 15² − 9² = 144

Quick sanity check: the slant height is the hypotenuse, so it is always longer than the height. If you get s < h, something went wrong.

Learn the common Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, and their multiples (6-8-10, 9-12-15). GED problems use them often.

Surface area: triangles plus base, or cone side plus base

Pyramid. A square pyramid has 4 identical triangles plus a square base. Each triangle has base b and height s (the slant height), so its area is \tfrac{1}{2}bs. Four of them give 4 \times \tfrac{1}{2}bs = \tfrac{1}{2}(4b)s = \tfrac{1}{2}ps, where p is the perimeter. Add the base:

SA = \tfrac{1}{2}ps + B

Example: base 10, slant height 13. p = 40, B = 100. SA = \tfrac{1}{2}(40)(13) + 100 = 260 + 100 = 360 square units.

Net of a square pyramid: a 10 by 10 square base (area 100) with four triangles, each with base 10 and height equal to the slant height 13, so each triangle is 65. Surface area = 4 times 65 plus 100 = 360 square units. s = 13 base B = 100 10 65 Each triangle: ½(10)(13) = 65 4 triangles: 4 × 65 = 260 (= ½ps = ½(40)(13)) Base: 10 × 10 = 100 SA = 260 + 100 = 360

Cone. The curved side has area \pi r s, and the circular base has area \pi r^2:

SA = \pi r s + \pi r^2

Example: r = 6, s = 10. SA = 60\pi + 36\pi = 96\pi \approx 301.6 square units.

Cone with radius 6 and slant height 10 split into its two surfaces: the curved side, area pi times 6 times 10 = 60 pi, and the circular base, area pi times 6 squared = 36 pi. Total 96 pi, about 301.6 square units. s = 10 r = 6 6 base circle: π(6²) = 36π curved side: π(6)(10) = 60π SA = 60π + 36π = 96π ≈ 301.6 square units

Read what is covered. A tent with no floor, a paper cup with an open top, or "the lateral area" means leave out the base: use only \tfrac{1}{2}ps or \pi r s.

Rectangular-base pyramids have two different slant heights (one for the long sides, one for the short sides). The GED almost always uses square bases for surface area, but if it is rectangular, add each triangle separately.

Working backwards: find a missing dimension

Sometimes the GED gives the volume or surface area and asks for a length. Write the formula, substitute everything you know, and undo the operations one at a time.

Missing height (pyramid or cone). Multiply both sides by 3, then divide by the base area:

h = \frac{3V}{B}

Example: rectangular pyramid, base 8 by 6, V = 192. B = 48, so h = \dfrac{3(192)}{48} = 12.

Rectangular pyramid with a base 8 by 6 and an unknown height h. The volume is 192. Base area B = 48, so h = 3 times 192 divided by 48 = 12. h = ? 8 6 V = 192 B = 8 × 6 = 48 h = 3V ÷ B = 3(192) ÷ 48 = 576 ÷ 48 = 12

Missing radius (cone). Solve for r^2, then take the square root:

r^2 = \frac{3V}{\pi h}, \quad r = \sqrt{\frac{3V}{\pi h}}

Example: V \approx 603.2, h = 9. r^2 = \dfrac{3(603.2)}{9\pi} \approx 64, so r = 8.

Missing side of a square base. Find B with B = \dfrac{3V}{h}, then b = \sqrt{B}. Example: V = 300, h = 9: B = 100, so b = 10.

Missing slant height from surface area. Subtract the base, then divide: s = \dfrac{SA - B}{\tfrac{1}{2}p}.

How changes affect volume. Because r is squared, doubling the radius of a cone multiplies its volume by 4. Doubling the height multiplies it by 2. Doubling both multiplies it by 8.

Formula

Formulas (all on the GED formula sheet)

\text{Pyramid: } V = \tfrac{1}{3}Bh \qquad SA = \tfrac{1}{2}ps + B \text{Cone: } V = \tfrac{1}{3}\pi r^2 h \qquad SA = \pi r s + \pi r^2
  • B = area of the base, h = height (straight up, inside), p = perimeter of the base, s = slant height (along the outside), r = radius of the base.
  • \pi \approx 3.14, or use the \pi key.

All four formulas are on the GED formula sheet, as is the Pythagorean theorem a^2 + b^2 = c^2. You do not need to memorize them, but you do need to know: B is an area, h and s are different, and r is half the diameter.

Worked example

Example 1 (easy): volume of a square pyramid

A square pyramid has base edges of 9 in and a height of 10 in. Find its volume.

  1. Base area: B = 9 \times 9 = 81\text{ in}^2.
  2. Formula: V = \tfrac{1}{3}Bh = \tfrac{1}{3}(81)(10).
  3. 81 \times 10 = 810, and 810 \div 3 = 270.
Example 1: square pyramid with base edges 9 inches and height 10 inches (dashed, inside). Base area 81 square inches; volume one-third of 81 times 10 = 270 cubic inches. h = 10 in 9 in 9 in B = 9 × 9 = 81 in² V = ⅓(81)(10) = 270 in³

Answer: V = 270\text{ in}^3

Check: a box 9 × 9 × 10 holds 810 in³; one-third of 810 is 270 ✓.

Worked example

Example 2: surface area of a cone from its height

A solid cone has radius 5 cm and height 12 cm. Find its total surface area.

Example 2: solid cone with radius 5 cm and height 12 cm (dashed, inside). The slant height s along the outside is unknown: s squared = 12 squared + 5 squared = 169, so s = 13 cm. r = 5 cm h = 12 cm s = ? s² = 12² + 5² = 169 s = 13 cm
  1. Surface area needs the slant height. s^2 = h^2 + r^2 = 12^2 + 5^2 = 144 + 25 = 169, so s = 13 cm.
  2. Curved side: \pi r s = \pi(5)(13) = 65\pi.
  3. Base: \pi r^2 = \pi(25) = 25\pi.
  4. Total: 65\pi + 25\pi = 90\pi \approx 282.7.

Answer: 90\pi \approx 282.7\text{ cm}^2

Check: 5-12-13 is a Pythagorean triple, and s = 13 is longer than h = 12 ✓.

Worked example

Example 3 (real world): glass for a greenhouse

A greenhouse is a square pyramid with a base 12 ft on each side and a height of 8 ft. The four sides are glass; the floor is dirt. How many square feet of glass are needed?

Example 3: greenhouse shaped like a square pyramid, base 12 feet on each side, height 8 feet. Half the base edge is 6 feet, so the slant height is 10 feet. The four glass sides are shaded; the dirt floor is not covered. h = 8 ft s = ? 6 ft 12 ft 12 ft glass = 4 triangles (no floor) s² = 8² + 6² = 100 s = 10 ft
  1. Half of a base edge: 12 \div 2 = 6 ft.
  2. Slant height: s^2 = 8^2 + 6^2 = 64 + 36 = 100, so s = 10 ft.
  3. Perimeter: p = 4 \times 12 = 48 ft.
  4. Only the sides are glass, so use the lateral area: \tfrac{1}{2}ps = \tfrac{1}{2}(48)(10) = 240.

Answer: 240 ft² of glass

Check: one triangle is \tfrac{1}{2}(12)(10) = 60; four triangles = 240 ✓.

Worked example

Example 4 (work backwards): height of a paper cup

A cone-shaped paper cup has a radius of 3 in and holds about 37.7 in³ of water. How tall is the cup?

  1. Formula: V = \tfrac{1}{3}\pi r^2 h, so 37.7 = \tfrac{1}{3}\pi(9)h.
  2. Simplify: \tfrac{1}{3}(9) = 3, so 37.7 = 3\pi h.
  3. Divide by 3\pi \approx 9.425: h = 37.7 \div 9.425 \approx 4.0.
Example 4: cone-shaped paper cup, point down, with radius 3 inches at the rim and unknown height h. It holds about 37.7 cubic inches; the height works out to about 4 inches. r = 3 in h = ? V ≈ 37.7 in³ 37.7 = ⅓π(9)h = 3πh h ≈ 4 in

Answer: about 4 in tall

Check: \tfrac{1}{3}\pi(9)(4) = 12\pi \approx 37.7 ✓.

Worked example

Example 5 (GED-hard): volume from the slant height

A monument is a square pyramid with base edges of 14 m and a slant height of 25 m. Find its volume.

Example 5: square pyramid monument with base edges 14 m and slant height 25 m; the height h is unknown. The height, half the base edge (7 m) and the slant height form a right triangle: 25 squared minus 7 squared = 576, so h = 24 m. h = ? s = 25 m 7 m 14 m h = 24 7 25 25² − 7² = 576
  1. Volume needs the height, not the slant height. Half of a base edge is 14 \div 2 = 7 m.
  2. h^2 = s^2 - 7^2 = 625 - 49 = 576, so h = \sqrt{576} = 24 m.
  3. Base area: B = 14 \times 14 = 196\text{ m}^2.
  4. V = \tfrac{1}{3}Bh = \tfrac{1}{3}(196)(24) = 196 \times 8 = 1{,}568.

Answer: V = 1{,}568\text{ m}^3

Check: 7^2 + 24^2 = 49 + 576 = 625 = 25^2 ✓ (7-24-25 is a triple), and h = 24 < s = 25 ✓.

Calculator tip

On the TI-30XS MultiView

  • One-third: type 1 n/d 3 ▶ (the right arrow leaves the fraction), then the rest. Or simply multiply everything and divide by 3 at the end — this is often faster and avoids fraction mistakes.
  • Squares: use the x² key. For \tfrac{1}{3}\pi(9^2)(12): π × 9 x² × 12 ÷ 3 enter → 324\pi; press toggle → 1017.87…
  • Square roots: 2nd then x² gives \sqrt{\ }. For a slant height: 2nd x² 15 x² + 8 x² enter → 17.
  • Exact vs. decimal: if the answer choices contain \pi (like 96\pi), do not press the π key — find the number in front of \pi by hand. If the choices are decimals, use the π key rather than 3.14 unless the question says to use 3.14.
  • Use parentheses when you divide by a product: 603.2 \div (3\pi) must be typed 603.2 ÷ ( 3 π ).
  • Pythagorean triples (3-4-5, 5-12-13, 8-15-17) let you skip the calculator entirely.
Common mistake

Common mistakes

  • Forgetting the \tfrac{1}{3}. \tfrac{1}{3}(36)(10) = 120, not 36 \times 10 = 360. Without the \tfrac{1}{3} you have found a prism or cylinder.
  • Mixing up height and slant height. Volume uses h (inside); surface area uses s (outside). If a figure shows both, each number is used in a different formula.
  • Using a side length for B. For a square base of side 6, B = 36, not 6.
  • Using the diameter as the radius. Diameter 3 means r = 1.5. Using 3 makes the volume 4 times too big.
  • Using the whole base edge in the slant-height triangle. The triangle uses half the base edge: s^2 = h^2 + \left(\tfrac{b}{2}\right)^2.
  • Forgetting the square root when solving for r or s. r^2 = 64 means r = 8, not 64.
Test strategy

GED test strategy

  • Circle what is asked: volume (cubic units) or surface area (square units)? Total or lateral? This tells you which formula and which length to use.
  • Label the figure with r, h, and s before you calculate. Cross out any number you will not use.
  • Estimate first. A cone is one-third of a cylinder. If your cone answer is bigger than the cylinder \pi r^2 h, it is wrong.
  • Use the distractors. Wrong choices are usually "forgot ⅓" (3 times too big), "used diameter" (4 times too big), or "used the other height". If one choice is exactly 3 times another, the smaller one is often correct for a cone or pyramid.
  • Plug answers back in for work-backwards questions: put each choice into the formula and see which one gives the stated volume.
  • Keep \pi until the end and round only at the last step, to the place the question asks for.

Check yourself

Pyramids and cones

A glass paperweight is shaped like a square pyramid. Its base is 6 in on each side, and its height is 10 in. What is the volume of the paperweight?

Answer choices
Hint

Find the area of the square base first. Then use the pyramid volume formula from the formula sheet.

Check yourself

Pyramids and conesFill in

A cone has a radius of 3 cm and a height of 7 cm. What is the volume of the cone, rounded to the nearest tenth of a cubic centimeter?

cm³

Type your answer in the box. Fractions like 3/4 are fine.

Hint

The cone formula is on the formula sheet: V = \tfrac{1}{3}\pi r^2 h. Square the radius before you do anything else.

Summary

  • Pyramid: V = \tfrac{1}{3}Bh and SA = \tfrac{1}{2}ps + B. Cone: V = \tfrac{1}{3}\pi r^2 h and SA = \pi r s + \pi r^2 — all on the formula sheet.
  • A pyramid or cone holds one-third of the prism or cylinder with the same base and height.
  • Height (inside, for volume) ≠ slant height (outside, for surface area). Connect them with s^2 = h^2 + \left(\tfrac{b}{2}\right)^2 or s^2 = h^2 + r^2.
  • Leave out the base when the object has no floor or an open top.
  • Work backwards with h = \dfrac{3V}{B} and r = \sqrt{\dfrac{3V}{\pi h}} — don't forget the square root.
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