AMC 10 Step by Step

Solid Geometry

Volume, surface area, cubes, cones, spheres, cylinders, cross-sections, nets

48
primary-topic problems (3.7% of all)
20
more as a secondary topic
Where it appears
11
P1-10
7
P11-15
14
P16-20
16
P21-25

What you need to know

  • Volume and surface. Prism or cylinder: V=BhV = Bh, where BB is the area of the base and hh the height. Pyramid or cone: V=13BhV = \frac13 Bh. Sphere of radius rr: V=43πr3V = \frac43\pi r^3, surface area 4πr24\pi r^2. Example: with radius 33 and height 44, a cylinder holds 36π36\pi and a cone 12π12\pi.
  • Scaling. Multiplying every length by kk multiplies areas by k2k^2 and volumes by k3k^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 44 times and its volume 88 times as big.
  • Diagonals. A cube with edge ss has face diagonal s2s\sqrt2 and space diagonal (corner to opposite corner, through the inside) s3s\sqrt3. An a×b×ca\times b\times c box has space diagonal a2+b2+c2\sqrt{a^2+b^2+c^2}. Example: a 2×3×62\times3\times6 box has diagonal 4+9+36=7\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: 33 faces are lost and 33 are uncovered) or from an edge (it goes up by 22).
  • Water poured between containers, or a partly filled cone standing on its tip. Set the volumes equal and use k3k^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 kk have volume ratio k3k^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 1212 and base radius 44. It contains water to a depth of 99. The water is poured into an empty cylinder of radius 33. What is the depth of the water in the cylinder?

(A) 22 (B) 52\dfrac52 (C) 33 (D) 278\dfrac{27}{8} (E) 44

Solution. Step 1. The water forms a smaller cone of the same shape. Its height is 99 out of 1212, so every length is scaled by 912=34\frac{9}{12} = \frac34, and its radius is 4⋅34=34\cdot\frac34 = 3.

Step 2. The water's volume is
13π⋅32⋅9=27π. \frac13\pi\cdot 3^2\cdot 9 = 27\pi.
(Check by scaling: the full cone holds 13π⋅42⋅12=64π\frac13\pi\cdot 4^2\cdot 12 = 64\pi, and 64π⋅(34)3=27π64\pi\cdot\left(\frac34\right)^3 = 27\pi.)

Step 3. In the cylinder, water of depth hh has volume π⋅32⋅h=9πh\pi\cdot 3^2\cdot h = 9\pi h. Setting 9πh=27π9\pi h = 27\pi gives h=3h = 3.

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

Pitfalls

  • Dropping the 13\frac13 for a cone or pyramid, or using it for a cylinder.
  • Scaling volume by kk or k2k^2 instead of k3k^3 when a cone is partly filled. Example: a cone standing on its tip, filled to half its height, holds 18\frac18 of its volume, not 12\frac12. (Standing on its base, it would hold 78\frac78.)
  • Mixing up a cone's slant height (along the slope) with its height (straight up). Example: radius 33 and height 44 give slant height 55.

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)

Problems, easiest first

Open any of them to try it here; the worked solution is on the problem's own page.

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