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AMC 12 focus

3D Geometry Basics

Volume and surface area formulas for the standard solids, plus the cross-section trick that turns a 3D problem into a 2D one.

Volume and surface area

Cube: V=s3, SA=6s2Rectangular prism: V=lwh, SA=2(lw+lh+wh)\text{Cube:}\ V=s^3,\ SA=6s^2 \qquad \text{Rectangular prism:}\ V=lwh,\ SA=2(lw+lh+wh)
Cylinder: V=πr2h, SA=2πr2+2πrh\text{Cylinder:}\ V=\pi r^2 h,\ SA=2\pi r^2+2\pi rh
Cone: V=13πr2h, SA=πr2+πr  (=slant height)\text{Cone:}\ V=\dfrac13\pi r^2 h,\ SA = \pi r^2 + \pi r \ell \ \ (\ell = \text{slant height})
Sphere: V=43πr3, SA=4πr2\text{Sphere:}\ V=\dfrac43\pi r^3,\ SA = 4\pi r^2

Cross-sections and projections

A common AMC 12 move is to slice a 3D solid with a plane and reduce the problem to 2D geometry on that cross-section. Some cross-sections are worth recognizing immediately: a plane through a sphere's center gives a great circle; a plane through a cube parallel to a face gives a square; a plane cutting a cube diagonally can give a hexagon.

Similarly, an orthogonal projection of a 3D figure onto a plane (imagine shining a light straight through it) is often easier to reason about than the solid itself, especially for volume-comparison or shadow-area problems.

Worked example

Problem: Find the radius of the largest sphere that fits inside a cube with side length 6.

Solution: The largest inscribed sphere is tangent to all six faces, so its diameter equals the cube's side length:

2r=6    r=32r = 6 \;\Longrightarrow\; r = 3

Practice problems

1.Find the volume of a cylinder with radius 3 and height 10.

2.Find the surface area of a sphere with radius 5.

3.A cone has radius 6 and height 8. Find its volume and its slant height.

4.A rectangular prism has dimensions 4, 5, 6. Find its total surface area.