AMC 10 · 2008 · #18

Grade 10 geometry-3d

Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid

volume-pyramidpythagorean-theoremarea-trianglessystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 3 insights

Problem

A pyramid has a square base of known area and two side faces of known areas. Find its volume.

Pick an answer.

(A)

392

(B)

$196\sqrt {6}$

(C)

$392\sqrt {2}$

(D)

$392\sqrt {3}$

(E)

784

How to solve

Strategy Convert to Algebra

The volume formula needs only one number that is not handed to us: the height h. Tool #7 (Subproblems) splits the job into (a) translate each given area into a length, and (b) recover h from those lengths. Tool #13 (Convert to Algebra) does the recovery: drop coordinates on the square, call the apex (p, q, h), and write each triangle's height as a distance from E to a line. Two of the three coordinates fall out of those expressions, leaving two equations in q and h. Tool #17 (Spatial) is what makes the algebra safe — it shows the missing coordinate p can be anything, so we must not assume where E hangs, and we never do. Tool #1 (Diagram) supplies the picture that makes the two right triangles visible.

1STEP 1

Read the side length off the base

The base area gives the side length and the gap between opposite edges.

AB = CD = √(196) = 14, dist(line AB, line CD) = 14

2STEP 2

Turn each area into a distance to a line

Each face area becomes a distance from the apex to a base edge.

d₁ = (2 · 105)/14 = 15, d₂ = (2 · 91)/14 = 13

3STEP 3

Coordinates show which numbers matter

Coordinates show only two numbers matter.

dist(E, AB) = √(q² + h²), dist(E, CD) = √((q-14)² + h²)

4STEP 4

Subtract the two equations

Subtracting the two equations gives the height.

q² + h² = 15² ; (q-14)² + h² = 13² → 28q - 196 = 56 → q = 9 → h = √(225 - 81) = 12

5STEP 5

Show such a pyramid really exists

An explicit apex shows the pyramid really exists.

E = (7, 9, 12): √(9²+12²) = 15, √(5²+12²) = 13, 9 + 5 = 14

6STEP 6

Apply the pyramid volume formula

The volume formula gives 784, choice (E).

V = 1/3 · 196 · 12 = 784

Answer

784

First, an apex has to be possible at all. Think of the plane cut perpendicular to AB: it meets line AB at one point P and line CD at one point Q, with PQ = 14, EP = 15, EQ = 13. Those three lengths satisfy the triangle inequality (15 - 13 = 2 ＜ 14 ＜ 28 = 15 + 13), so a genuine non-degenerate triangle — and hence a genuine pyramid — exists; had the areas been, say, 105 and 7, no apex could exist. Second, the numbers check internally: h = 12 is smaller than both 15 and 13, as any leg must be, and the two right triangles 9-12-15 and 5-12-13 have legs 9 + 5 = 14, exactly the distance between the lines. Third, the choices correspond to heights 6, 3√(6), 6√(2), 6√(3), 12; the system pinned h uniquely at 12, so only 784 survives. Worth flagging: the pyramid itself is not determined by the given data — the apex may slide freely parallel to AB — yet every such pyramid has height 12, so the volume the question asks for is still well defined.

💡Key takeaway

Each triangle's area tells you how far the tip is from one edge's line; two such distances and the 14 between the edges pin the tip's height at 12, and one third of 196 × 12 is the volume.

Read the side length off the base
Turn each area into a distance to a line
Coordinates show which numbers matter
Subtract the two equations
Show such a pyramid really exists
Apply the pyramid volume formula