Volume and surface area

This chapter applies to all AMC 8/10/12 test takers.

Some AMC geometry problems involve three-dimensional (3D) shapes. You usually won’t be given the formulas for volume and surface area, so you’ll want to have the common ones memorized. While these quantities can be derived using calculus, AMC problems won’t require calculus. Using the standard formulas is the most direct approach.

A helpful way to organize volume formulas is to think in terms of a base area and a height:

If a solid has the same cross-section all the way up (like a cube, rectangular prism, or cylinder), its volume is the area of the base times the height.
If a solid “tapers” (like a cone or pyramid), its volume is the area of the base times the height, multiplied by a fraction.
A sphere is different because it doesn’t have a base.

Cube

Cube

$Volume = s^{3}$

$Surface Area = 6 s^{2}$

:::

Rectangular prism

Rectangular Prism

$Volume = lw h$

$Surface Area = 2 lw + 2 w h + 2 l h$

:::

Cylinder

Cylinder

$Volume = π r^{2} h$

$Surface Area = 2 π r^{2} + 2 π r h$

:::

Cone

Cylinder

Notice that $L$ is not the vertical height of the cone. $L$ is the slant length.

$Volume = \frac{1}{3} (π) r^{2} h$

$Surface Area = π r^{2} + π r L$

:::

Sphere

Sphere

$Volume = \frac{4}{3} π r^{3}$

$Surface Area = 4 π r^{2}$

:::

Square pyramid

Square pyramid

In these formulas, $s$ is the side length of the square base. If the base is a rectangle, use the product of its two side lengths for the base area instead. Like a cone, a pyramid has a slant length, which you use when finding surface area. The vertical height of a pyramid is also called the altitude.

$Volume = \frac{1}{3} (area of base) h = \frac{1}{3} s^{2} h$

$Surface Area = s^{2} + 2 s L$

:::

Example: The question below is from 2018 AMC 10B

In the rectangular parallelepiped shown, $A B$ = $3$ , $BC$ = $1$ , and $CG$ = $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BC H E$ and apex $M$ ?

A. $1$
B. $\frac{4}{3}$
C. $\frac{3}{2}$
D. $\frac{5}{3}$
E. $2$

(spoiler)

Answer: E. $2$

We want the pyramid’s volume, so we need:

the area of the base $BC H E$
the perpendicular height from $M$ to the plane of $BC H E$

First, find the base area. The base $BC H E$ is a rectangle with side lengths $BC$ and $BE$ . We’re given $BC = 1$ , so we just need $BE$ .

Using the Pythagorean theorem on the appropriate right triangle in the prism, $BE = 13$ . So the base area is:

$(area of base) = BC \cdot BE = 1 \cdot 13 = 13$

Next, find the pyramid’s height. Using right triangles in the prism (again via the Pythagorean theorem), the perpendicular height from $M$ to the base plane works out to be $\frac{6}{13}$ .

Now substitute into the pyramid volume formula:

$Volume = \frac{1}{3} (area of base) h = \frac{1}{3} (13) (\frac{6}{13}) = 2$

Tetrahedron

Tetrahedron

A tetrahedron is a pyramid with a triangular base instead of a square base. If the base and all faces are equilateral triangles, it’s a regular tetrahedron. Many tetrahedra are not regular, so you’ll need to find the base area using whatever triangle information is given. For a regular tetrahedron, the surface area is 4 times the area of one face. For a non-regular tetrahedron, find the surface area by adding the areas of all four faces.

$Volume = \frac{1}{3} (area of base) (height)$

:::

Common themes

The diagonal of a cube is side length times $3$ .
The diagonal of any rectangular prism is $l^{2} + w^{2} + h^{2}$ .
You may have more complex-looking shapes, but if you can split them up into more basic shapes, you are more likely to find the answer.
You may also be asked about a cross-section of a 3D shape. A cross-section is simply a 2D plane that cuts through the shape.
The slant length must always be greater than the vertical height.
If the ratio of the sides of similar solids is just $a : b$ , then the ratio of their surface areas would be $a^{2} : b^{2}$ and the ratio of their volumes would be $a^{3} : b^{3}$ .
Regular polyhedra like cubes and regular tetrahedra can be circumscribed by a sphere.

Volume and Surface Area Formulas

Memorize standard formulas for cube, rectangular prism, cylinder, cone, sphere, square pyramid, tetrahedron
Volume often = (base area) × (height); cones/pyramids multiply by 1/3
Surface area formulas typically sum areas of all faces/surfaces

Cube

Volume: $s^{3}$
Surface area: $6 s^{2}$

Rectangular Prism

Volume: $lw h$
Surface area: $2 lw + 2 w h + 2 l h$

Cylinder

Volume: $π r^{2} h$
Surface area: $2 π r^{2} + 2 π r h$

Cone

Volume: $\frac{1}{3} π r^{2} h$
Surface area: $π r^{2} + π r L$
- $L$ = slant height (not vertical height)

Sphere

Volume: $\frac{4}{3} π r^{3}$
Surface area: $4 π r^{2}$

Square Pyramid

Volume: $\frac{1}{3} s^{2} h$ (or $\frac{1}{3} (base area) h$ )
Surface area: $s^{2} + 2 s L$
- $L$ = slant height

Tetrahedron

Volume: $\frac{1}{3} (area of base) (height)$
For regular tetrahedron: surface area = 4 × (area of one face)

Common 3D Geometry Themes

Cube diagonal: $s 3$
Rectangular prism diagonal: $l^{2} + w^{2} + h^{2}$
Complex shapes: break into basic shapes for calculation
Cross-section: 2D plane cutting through 3D shape
Slant height always > vertical height
Similar solids: side ratio $a : b$ leads to surface area ratio $a^{2} : b^{2}$ , volume ratio $a^{3} : b^{3}$
Regular polyhedra (cube, regular tetrahedron) can be circumscribed by a sphere

Volume and surface area

This chapter applies to all AMC 8/10/12 test takers.

A helpful way to organize volume formulas is to think in terms of a base area and a height:

If a solid has the same cross-section all the way up (like a cube, rectangular prism, or cylinder), its volume is the area of the base times the height.
If a solid “tapers” (like a cone or pyramid), its volume is the area of the base times the height, multiplied by a fraction.
A sphere is different because it doesn’t have a base.

Cube

Cube

$Volume = s^{3}$

$Surface Area = 6 s^{2}$

:::

Rectangular prism

Rectangular Prism

$Volume = lw h$

$Surface Area = 2 lw + 2 w h + 2 l h$

:::

Cylinder

Cylinder

$Volume = π r^{2} h$

$Surface Area = 2 π r^{2} + 2 π r h$

:::

Cone

Cylinder

Notice that $L$ is not the vertical height of the cone. $L$ is the slant length.

$Volume = \frac{1}{3} (π) r^{2} h$

$Surface Area = π r^{2} + π r L$

:::

Sphere

Sphere

$Volume = \frac{4}{3} π r^{3}$

$Surface Area = 4 π r^{2}$

:::

Square pyramid

Square pyramid

$Volume = \frac{1}{3} (area of base) h = \frac{1}{3} s^{2} h$

$Surface Area = s^{2} + 2 s L$

:::

Example: The question below is from 2018 AMC 10B

In the rectangular parallelepiped shown, $A B$ = $3$ , $BC$ = $1$ , and $CG$ = $2$ . Point $M$ is the midpoint of $\overline{FG}$ . What is the volume of the rectangular pyramid with base $BC H E$ and apex $M$ ?

A. $1$
B. $\frac{4}{3}$
C. $\frac{3}{2}$
D. $\frac{5}{3}$
E. $2$

(spoiler)

Answer: E. $2$

We want the pyramid’s volume, so we need:

the area of the base $BC H E$
the perpendicular height from $M$ to the plane of $BC H E$

First, find the base area. The base $BC H E$ is a rectangle with side lengths $BC$ and $BE$ . We’re given $BC = 1$ , so we just need $BE$ .

Using the Pythagorean theorem on the appropriate right triangle in the prism, $BE = 13$ . So the base area is:

$(area of base) = BC \cdot BE = 1 \cdot 13 = 13$

Next, find the pyramid’s height. Using right triangles in the prism (again via the Pythagorean theorem), the perpendicular height from $M$ to the base plane works out to be $\frac{6}{13}$ .

Now substitute into the pyramid volume formula:

$Volume = \frac{1}{3} (area of base) h = \frac{1}{3} (13) (\frac{6}{13}) = 2$

Tetrahedron

Tetrahedron

$Volume = \frac{1}{3} (area of base) (height)$

:::

Common themes

The diagonal of a cube is side length times $3$ .
The diagonal of any rectangular prism is $l^{2} + w^{2} + h^{2}$ .
You may have more complex-looking shapes, but if you can split them up into more basic shapes, you are more likely to find the answer.
You may also be asked about a cross-section of a 3D shape. A cross-section is simply a 2D plane that cuts through the shape.
The slant length must always be greater than the vertical height.
If the ratio of the sides of similar solids is just $a : b$ , then the ratio of their surface areas would be $a^{2} : b^{2}$ and the ratio of their volumes would be $a^{3} : b^{3}$ .
Regular polyhedra like cubes and regular tetrahedra can be circumscribed by a sphere.