Solid Geometry

What you need to know

• Volume and surface. Prism or cylinder: $V = Bh$, where $B$ is the area of the base and $h$ the height. Pyramid or cone: $V = \frac13 Bh$. Sphere of radius $r$: $V = \frac43\pi r^3$, surface area $4\pi r^2$. Example: with radius $3$ and height $4$, a cylinder holds $36\pi$ and a cone $12\pi$.
• Scaling. Multiplying every length by $k$ multiplies areas by $k^2$ and volumes by $k^3$. A cut parallel to the base of a cone or pyramid leaves a smaller cone or pyramid of the same shape. Example: doubling a cube's edges makes its surface area $4$ times and its volume $8$ times as big.
• Diagonals. A cube with edge $s$ has face diagonal $s\sqrt2$ and space diagonal (corner to opposite corner, through the inside) $s\sqrt3$. An $a\times b\times c$ box has space diagonal $\sqrt{a^2+b^2+c^2}$. Example: a $2\times3\times6$ box has diagonal $\sqrt{4+9+36} = 7$.
• Slicing a cube. A flat cut through a cube can give a triangle, a four-sided shape (square, rectangle, rhombus, parallelogram, trapezoid), a pentagon, or a hexagon. It never has more than six sides, since a cube has six faces. Cutting through the midpoints of six edges gives a regular hexagon.
• Paths on the surface. The shortest path along the outside becomes a straight line when you unfold the solid flat (into a net).

How AMC 10 tests it

• Early: the volume or surface area of a box, or how the surface area changes when unit cubes are removed from a corner (no change: $3$ faces are lost and $3$ are uncovered) or from an edge (it goes up by $2$).
• Water poured between containers, or a partly filled cone standing on its tip. Set the volumes equal and use $k^3$ scaling.
• Problems 15–22: a sphere or cylinder inside a cone (similar triangles in the cross-section through the axis), or an ant's shortest path on a box.

Standard approaches

1. Draw the flat cross-section through the axis (the center line). Most solid problems become a triangle problem there.
2. Use scaling before computing: similar solids with length ratio $k$ have volume ratio $k^3$.
3. For odd shapes, start from a full prism or box and subtract pyramids or corner tetrahedra (triangular pyramids).
4. For surface paths, unfold the net and use the Pythagorean Theorem. Try several unfoldings and keep the shortest.

Worked example

An inverted cone (vertex down) has height $12$ and base radius $4$. It contains water to a depth of $9$. The water is poured into an empty cylinder of radius $3$. What is the depth of the water in the cylinder?

(A) $2$ (B) $\dfrac52$ (C) $3$ (D) $\dfrac{27}{8}$ (E) $4$

Solution. Step 1. The water forms a smaller cone of the same shape. Its height is $9$ out of $12$, so every length is scaled by $\frac{9}{12} = \frac34$, and its radius is $4\cdot\frac34 = 3$.

Step 2. The water's volume is
$$\frac13\pi\cdot 3^2\cdot 9 = 27\pi.$$
(Check by scaling: the full cone holds $\frac13\pi\cdot 4^2\cdot 12 = 64\pi$, and $64\pi\cdot\left(\frac34\right)^3 = 27\pi$.)

Step 3. In the cylinder, water of depth $h$ has volume $\pi\cdot 3^2\cdot h = 9\pi h$. Setting $9\pi h = 27\pi$ gives $h = 3$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Pitfalls

• Dropping the $\frac13$ for a cone or pyramid, or using it for a cylinder.
• Scaling volume by $k$ or $k^2$ instead of $k^3$ when a cone is partly filled. Example: a cone standing on its tip, filled to half its height, holds $\frac18$ of its volume, not $\frac12$. (Standing on its base, it would hold $\frac78$.)
• Mixing up a cone's slant height (along the slope) with its height (straight up). Example: radius $3$ and height $4$ give slant height $5$.

Traps that recur

• Using the sphere's diameter as the slant height, or forgetting the middle term Rr in the frustum volume formula. (2014 AMC 10B #23)
• Assuming that filling to half the height fills half the container, so the rest takes about the same 35 minutes again and a guess is pushed toward the 70 in (A). (2025 AMC 10B #19)
• Counting edges as 12 times 4 = 48 without halving for the two faces each edge borders. (2023 AMC 10A #18)
• Assuming each semicircle is half a great circle of radius 2, which gives 8pi = pisqrt(64), not among the choices. (2023 AMC 10B #20)