Basic 3D geometry

An SAT Math micro-topic under Area and volume (Geometry and trigonometry). Free to read — no account needed.

Three-dimensional figures are handled with a small toolkit of formulas, so the work is mostly about choosing the right one and supplying the missing measurement.
For a cube of edge ss, the six identical square faces give a surface area of 6s26s^2, and the volume is s3s^3. For a rectangular box with length ll, breadth bb and width ww, there are three pairs of rectangular faces, so the surface area is 2lb+2bw+2lw2lb + 2bw + 2lw and the volume is lbwlbw.
For curved solids: a cylinder of radius rr and height hh has volume πr2h\pi r^2 h; a sphere of radius rr has surface area 4πr24\pi r^2; and a pyramid has volume 13×base area×h\frac{1}{3} \times \text{base area} \times h.
A very common two-step pattern is to back out a length from a given quantity and then use it. If a cube has surface area 5454, then 6s2=546s^2 = 54 gives s=3s = 3, and its volume is 33=273^3 = 27. Recognising which formula links what you are given to what you are asked is the whole skill.

Worked examples

A cube has surface area 9696 square inches.
From 6s2=966s^2 = 96 we get s2=16s^2 = 16, so s=4s = 4 inches.
Its volume is then s3=43=64s^3 = 4^3 = 64 cubic inches.
A sphere has diameter 1414, so its radius is 77. Using surface area 4πr24\pi r^2 with π=227\pi = \frac{22}{7}: 4×227×72=6164 \times \frac{22}{7} \times 7^2 = 616 square units.
A cylinder has radius 44 and height 1010. Its volume is πr2h=π×42×10=160π\pi r^2 h = \pi \times 4^2 \times 10 = 160\pi.

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