Chapter 18 — Three Dimensional Geometry
In this chapter
- 18.1 Planes, Surface Area, and Volume
- 18.2 Spheres
- 18.3 Cubes and Boxes
- 18.4 Prisms and Cylinders
- 18.5 Pyramids and Cones
- 18.6 Polyhedra & Euler’s Formula
- 18.7 How to Solve 3D Problems
- Volume & Surface Area Summary Table
- Exercises & Problems
Volume & Surface Area Summary Table
| Solid | Volume | Surface Area | Notes |
|---|---|---|---|
| Sphere | $\dfrac{4}{3}\pi r^3$ | $4\pi r^2$ | |
| Cube (edge $s$) | $s^3$ | $6s^2$ | Diagonal $= s\sqrt{3}$ |
| Box ($l \times w \times h$) | $lwh$ | $2(lw + wh + lh)$ | Space diagonal $= \sqrt{l^2+w^2+h^2}$ |
| Right Prism | $Bh$ | $2B + ph$ | $B$ = base area, $p$ = base perimeter |
| Right Cylinder | $\pi r^2 h$ | $2\pi r^2 + 2\pi rh$ | |
| Pyramid | $\dfrac{1}{3}Bh$ | $B + \text{lateral faces}$ | |
| Right Circular Cone | $\dfrac{1}{3}\pi r^2 h$ | $\pi r^2 + \pi r l$ | $l = \sqrt{r^2+h^2}$ (slant height) |
| Regular Pyramid | $\dfrac{1}{3}Bh$ | $B + \dfrac{1}{2}pl$ | $p$ = base perimeter, $l$ = slant height |
18.1 Planes, Surface Area, and Volume
Key Definitions:
- Parallel lines in space have exactly the same direction.
- Skew lines are non-parallel lines that never intersect (they exist in different planes).
- A plane is a flat surface extending infinitely in all directions (2D in 3D space).
- Any three noncollinear points determine a unique plane.
- Four or more points in the same plane are coplanar.
A line and a plane can have three relationships:
- The line lies in the plane.
- The line intersects the plane at exactly one point.
- The line is parallel to the plane (no intersection).
A line $\ell$ is perpendicular to a plane if every line in the plane through the intersection point is perpendicular to $\ell$.
- Total surface area: the area of all surfaces enclosing a 3D figure.
- Lateral surface area: surface area excluding the top and bottom bases.
- Volume: the amount of space enclosed within a figure.
18.2 Spheres
A sphere is the set of all points in space at a fixed distance $r$ (the radius) from a given point (the center).
\(\text{Volume} = \frac{4}{3}\pi r^3, \qquad \text{Surface Area} = 4\pi r^2\)
If a plane intersects a sphere, the intersection is:
- A point (if the plane is tangent), or
- A circle (in general). If the plane passes through the center, the circle is a great circle.
Example 18-1. A plane intersects a sphere of radius 8, forming a circle. If the center of the sphere is 5 units from the plane, find the radius of the circle.
Solution. Draw the center $O$ of the sphere, the center $B$ of the circle, and a radius $A$ of the circle. Since $OB \perp$ the plane, $\triangle OBA$ is a right triangle with $OA = 8$ (sphere radius) and $OB = 5$. By the Pythagorean Theorem:
\(AB = \sqrt{8^2 - 5^2} = \sqrt{39}\)
Example 18-2. Find the diameter of a sphere whose volume is $288\pi$.
Solution. $\frac{4}{3}\pi r^3 = 288\pi \implies r^3 = 216 \implies r = 6$. Diameter $= 2r = \boxed{12}$.
Exercise 18-1. A ball was floating in a lake when the lake froze. The ball was removed (without breaking ice), leaving a hole 24 cm across at the top and 8 cm deep. What was the radius of the ball? (AHSME 1987)
Exercise 18-2. Find the volume of a sphere with surface area 100.
18.3 Cubes and Boxes
Cubes
A cube has 6 congruent square faces, 12 edges of equal length $s$, and 8 vertices.
\[\text{Volume} = s^3, \qquad \text{Surface Area} = 6s^2\]Face diagonal $= s\sqrt{2}$, Space diagonal $= s\sqrt{3}$.
The space diagonal is found by applying the Pythagorean Theorem twice: first to find the face diagonal $AC = s\sqrt{2}$, then $EC = \sqrt{s^2 + 2s^2} = s\sqrt{3}$.
Example 18-3. In cube $ABCDEFGH$ with $AB = 4$, $M$ is the midpoint of edge $DE$. Find $BM$.
Solution. Draw face diagonal $BD$: $BD = 4\sqrt{2}$. Segment $DM = 2$ (half an edge). Triangle $BDM$ is a right triangle (the lateral edge is perpendicular to the face):
\(BM = \sqrt{BD^2 + DM^2} = \sqrt{32 + 4} = \sqrt{36} = \boxed{6}\)
Example 18-4. A cube of edge 6 is made of unit cubes, then all faces are painted. How many unit cubes have 0, 1, 2, or 3 painted faces?
Solution.
- 3 faces painted (corners): $8$
- 2 faces painted (edges, not corners): $12 \times 4 = 48$
- 1 face painted (face interiors): $6 \times 4^2 = 96$
- 0 faces painted (interior cube): $4^3 = 64$
Check: $8 + 48 + 96 + 64 = 216 = 6^3$. ✓
Example 18-5. Find the area of $\triangle BDE$ in cube $ABCDEFGH$ with $AB = 6$.
Solution. Each side of $\triangle BDE$ is a face diagonal: $BD = DE = BE = 6\sqrt{2}$. The triangle is equilateral, so:
\([BDE] = \frac{(6\sqrt{2})^2\sqrt{3}}{4} = \frac{72\sqrt{3}}{4} = \boxed{18\sqrt{3}}\)
Exercise 18-3. Find the volume of a cube with diagonal length 6.
Exercise 18-4. Edges $AB$, $AD$, $AE$ of a cube; $L$, $M$, $N$ are midpoints. Find $\angle LMN$.
Boxes (Rectangular Parallelepipeds)
A box has dimensions $l$, $w$, $h$ (length, width, height). Opposite faces are congruent rectangles.
\[\text{Volume} = lwh, \qquad \text{Surface Area} = 2(lw + wh + lh)\]\(\text{Space diagonal} = \sqrt{l^2 + w^2 + h^2}\)
Example 18-6. Find the volume of a box with diagonal $\sqrt{35}$ and two dimensions 1 and 5.
Solution. $35 = 1 + 25 + h^2 \implies h^2 = 9 \implies h = 3$. Volume $= 1 \cdot 5 \cdot 3 = \boxed{15}$.
Example 18-7. The areas of three faces of a rectangular solid are 24, 32, and 48. Find the volume.
Solution. Let dimensions be $x$, $y$, $z$. Then $xy = 24$, $xz = 32$, $yz = 48$. Multiply all three:
\[(xyz)^2 = 24 \cdot 32 \cdot 48 = 2^{12} \cdot 3^2\] \[xyz = 2^6 \cdot 3 = \boxed{192}\]This method avoids needing to find individual dimensions!
Exercise 18-5. How many 2-inch cubes are needed to fill a 4 in × 8 in × 10 in box?
Exercise 18-6. In rectangular parallelepiped $ABCDEFGH$, $AB = 4$, $BC = 3$, $CG = 9$, $BY = 3$, $DX = 5$. Find $XY$.
18.4 Prisms and Cylinders
A prism has two parallel congruent bases connected by lateral faces (parallelograms). A right prism has lateral edges perpendicular to the bases.
\(\text{Volume} = Bh \qquad (\text{base area} \times \text{height})\)
A (right circular) cylinder has circular bases of radius $r$ and height $h$:
\[\text{Volume} = \pi r^2 h\]\(\text{Lateral SA} = 2\pi rh, \qquad \text{Total SA} = 2\pi r^2 + 2\pi rh\)
Unrolling a cylinder. If you cut the curved surface along a vertical line and unroll it, you get a rectangle with dimensions $h \times 2\pi r$. This is why lateral SA $= 2\pi rh$.
Example 18-8. Show that the lateral surface area of a right prism equals the perimeter of the base times the altitude.
Proof. Each lateral face is a rectangle with one side equal to the altitude $h$ and the other equal to a side of the base. Summing: $x_1 h + x_2 h + \cdots + x_n h = (x_1 + x_2 + \cdots + x_n)h = ph$. $\square$
Example 18-9. An ant on the top edge of a cylinder (height 8 in, diameter 4 in) wants to reach the diametrically opposite point on the bottom edge. What is the shortest crawling distance? (MAΘ 1992)
Solution. Unroll the cylinder into a rectangle of width $2\pi r = 2\pi(2) = 4\pi$ and height 8. The opposite point $B$ is at the midpoint of one long side, so $CB = 4\pi/2 = 2\pi$. The path $AB$ on the surface corresponds to a straight line on the rectangle:
\(AB = \sqrt{8^2 + (2\pi)^2} = \sqrt{64 + 4\pi^2} = 2\sqrt{16 + \pi^2}\)
Exercise 18-7. Find the total surface area of a cylinder with height 5 and volume $45\pi$.
Exercise 18-8. What is the greatest distance between two points on a right circular cylinder with radius 4 and height 6?
18.5 Pyramids and Cones
Pyramids
A pyramid has one polygonal base and triangular lateral faces meeting at a common vertex (apex). The altitude $h$ is the perpendicular distance from the apex to the base.
\[\text{Volume} = \frac{1}{3}Bh\]A regular pyramid has a regular polygon base and vertex directly above the center. Its lateral faces are congruent isosceles triangles with common altitude called the slant height $l$.
\(\text{Lateral SA} = \frac{1}{2}pl \qquad (p = \text{base perimeter})\)
Example 18-10. Prove that the lateral surface area of a regular pyramid with base perimeter $p$ and slant height $l$ is $\frac{pl}{2}$.
Proof. Each lateral face is a triangle with base $s$ (a side of the regular polygon) and height $l$. With $n$ faces: total lateral SA $= n \cdot \frac{sl}{2} = \frac{(ns)l}{2} = \frac{pl}{2}$. $\square$
Cones
A (right circular) cone has a circular base of radius $r$, height $h$, and slant height $l = \sqrt{r^2 + h^2}$.
\[\text{Volume} = \frac{1}{3}\pi r^2 h\]\(\text{Lateral SA} = \pi r l, \qquad \text{Total SA} = \pi r^2 + \pi r l\)
Unrolling a cone: Cut along a slant height and unroll. The result is a sector of a circle with radius $l$ and arc length $2\pi r$.
Example 18-11. Find the total surface area of a cone with radius 5 and height 12.
Solution. Slant height: $l = \sqrt{25 + 144} = 13$. Total SA $= \pi(5)^2 + \pi(5)(13) = 25\pi + 65\pi = \boxed{90\pi}$.
Exercise 18-9. Prove that the lateral SA of a cone with radius $r$ and slant height $l$ is $\pi rl$. (Hint: unroll the surface and find the area of the resulting sector.)
Exercise 18-10. Find the volume of a cone whose vertex is the center of a sphere of radius 5 and whose base is the intersection of the sphere with a plane 3 units from the center.
18.6 Polyhedra
A polyhedron is a solid bounded by planar polygonal faces. A regular polyhedron (Platonic solid) has all faces congruent regular polygons with the same number meeting at each vertex.
The Five Platonic Solids
| Name | Face Shape | Faces | Vertices | Edges |
|---|---|---|---|---|
| Tetrahedron | Triangles | 4 | 4 | 6 |
| Hexahedron (Cube) | Squares | 6 | 8 | 12 |
| Octahedron | Triangles | 8 | 6 | 12 |
| Dodecahedron | Pentagons | 12 | 20 | 30 |
| Icosahedron | Triangles | 20 | 12 | 30 |
Euler’s Formula
For any convex polyhedron:
\[V - E + F = 2\]where $V$ = vertices, $E$ = edges, $F$ = faces.
Example 18-12. Find the volume of a regular tetrahedron with side length 1.
Solution. The base is an equilateral triangle with area $\frac{\sqrt{3}}{4}$. The centroid of the base is $\frac{2}{3}$ of the way from a vertex to the opposite midpoint. The median of the base has length $\frac{\sqrt{3}}{2}$, so the centroid is at distance $AO = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}$ from the midpoint. Actually, $AO = \frac{2}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{3}$.
From the Pythagorean Theorem (apex $B$ to centroid $O$):
\[BO = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}\]Volume:
\(V = \frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{6}}{3} = \frac{\sqrt{18}}{36} = \frac{3\sqrt{2}}{36} = \boxed{\frac{\sqrt{2}}{12}}\)
Example 18-13. Find the volume of a regular octahedron with side length 1.
Solution. Split the octahedron into two square-based pyramids. The square base $ABCD$ has side 1, so area $= 1$. The half-diagonal of the square is $\frac{\sqrt{2}}{2}$.
Height of each pyramid: $h = \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2}$.
Volume of octahedron $= 2 \cdot \frac{1}{3}(1)\left(\frac{\sqrt{2}}{2}\right) = \boxed{\frac{\sqrt{2}}{3}}$.
18.7 How to Solve 3D Problems
Strategy for 3D problems:
- Volume/SA of complex shapes: Dissect into simpler pieces (prisms, pyramids, etc.) and add/subtract.
- Lengths, angles, areas in 3D: Reduce to 2D cross-sections. Find a plane (or triangle) containing the desired quantity.
- Right triangles are your best friend in 3D — perpendicular edges create them everywhere.
- Unrolling curved surfaces (cylinders, cones) converts 3D shortest-path problems to 2D.
Remember: Three dimensional problems are disguised two-dimensional problems. Find the right cross-section and apply standard 2D geometry.
Problems to Solve for Chapter 18
Problem 303. The sum of all edge lengths of a cube is 144 inches. Find the length of a diagonal. (MATHCOUNTS 1989)
Problem 304. A 5 × 8 inch sheet is rolled into a cylinder of height 8, or height 5. Find the ratio of the volumes. (MATHCOUNTS 1989)
Problem 305. How many triangular faces does a pyramid with 10 edges have? (MATHCOUNTS 1992)
Problem 306. Regular hexagon $JKLMNO$ intersects the edges of a cube at their midpoints. Find the ratio of the hexagon’s area to the cube’s total surface area. (MAΘ 1990)
Problem 307. The surface area of a cube equals twice its volume. Find the diagonal length. (MATHCOUNTS 1988)
Problem 308. Find the radius of a right circular cone if its volume is 1.5 times its lateral SA and its radius is half its slant height. (MAΘ 1990)
Problem 309. A cube is inscribed in a sphere. Find the ratio of the sphere’s SA to the cube’s SA. (MAΘ 1992)
Problem 310. Height of a rectangular room is $h$; areas of two adjacent walls are $a$ and $b$. Find the floor area in terms of $a$, $b$, $h$. (MATHCOUNTS 1990)
Problem 311. A 6 × 3 × 12 cm box of liquid is poured onto water, forming a circular film 0.1 cm thick. Find the radius. (AHSME 1991)
Problem 312. A cylindrical box has radius 8 and height 3. Find the number of inches that can be added to either radius or height to give the same nonzero increase in volume. (AHSME 1951)
Problem 313. A paper cone, cut along its slant height and opened out, forms a semicircle of radius 10. Find the altitude of the cone. (MAΘ 1987)
Problem 314. Four vertices of a cube are vertices of a regular tetrahedron. Find the ratio of the cube’s SA to the tetrahedron’s SA. (AHSME 1980)
Problem 315. In unit cube $ABCDEFGH$, let $X$ be the center of face $ABCD$. Find $FX$. (MAΘ 1992)
Problem 316. A cube of edge $n$ is painted and cut into unit cubes. The number with one painted face equals the number with no painted faces. Find $n$. (AHSME 1985)
Problem 317. A cube of edge 3 has square holes of side 1 cut through center of each face to opposite face. Find the entire surface area including the inside. (AHSME 1982)
Problem 318. A cube of side 3 has unit cubes cut from each corner, then cubes of side 2 inserted. Find the surface area of the resulting solid. (MATHCOUNTS 1991)
Problem 319. In rectangular solid $ABCDEFGH$, $\angle DHG = 45°$ and $\angle FHB = 60°$. Find $\cos(\angle BHD)$. (AHSME 1982)
Problem 320. A truncated octahedron has 14 faces (6 squares, 8 hexagons). At each corner, 2 hexagons and 1 square meet. How many corners? (MATHCOUNTS 1984)
Problem 321. Find the volume of a regular octahedron whose vertices are the centers of the faces of a cube of edge 6. (MATHCOUNTS 1985)
Problem 322. A ball of radius $R$ is tangent to the floor and one wall. Find the radius of the largest sphere that fits between the ball, wall, and floor. (MAΘ 1992)
Problem 323. An inverted cone tank has base radius 16 ft and height 96 ft. Find the water height when the tank is 25% full. (MATHCOUNTS 1992)
Problem 324. A truncated icosahedron has 32 faces, 60 vertices, 90 edges. Faces are pentagons and hexagons; at each vertex 2 hexagons and 1 pentagon meet. How many hexagonal faces? (MATHCOUNTS 1988)
Problem 325. Find the distance from vertex $B$ to face $ACD$ in a regular tetrahedron of side 6.
Problem 326. Nine congruent spheres are packed in a unit cube: one at center, each other tangent to center sphere and three faces. Find the radius. (AHSME 1990)
Problem 327. A cone of radius 6 and height 6 is cut by a plane parallel to the base, 2 units from the base. Find the volume between the plane and the base.
Problem 328. Three cubes of edges 1, 2, 3 are stacked. Find the length of the segment $AB$ (connecting opposite corners of the stack) that lies inside the center cube. (MATHCOUNTS 1991)