AMC 10 · 2025 · #20

Grade 10 geometry-3d
Archetype Solid Volume by Decomposition └ Sub-archetype Sum of Primitive Solids
pythagorean-theoremvolume-prismvolume-pyramidspatial-visualization identify-subproblemsphysical-representation ↑ Prerequisites: pythagorean-theoremvolume-pyramid
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
A solid sits on a 13 × 8 rectangular base and rises to a straight top ridge of length 7 centered above the base. Two of its slanted faces are isosceles triangles with base 8 and equal sides 13; the other two are isosceles trapezoids with parallel sides 7 and 13 and slant legs 13. Find the volume of this solid.

Pick an answer.

(A)
416
(B)
520
(C)
528
(D)
676
(E)
832
How to solve
Strategy Identify Subproblems

The solid has no ready-made volume formula, so Tool #7 (Identify Subproblems) drives the whole plan: split the job into (a) find the missing ridge height, then (b) cut the solid into shapes we already know how to measure. Tool #17 (Visualize Spatial Relationships) supplies the mental picture — a tent-like ridge over a rectangle — and lets us drop the figure into coordinates so a slant edge becomes a 3D distance. Tool #1 (Draw a Diagram) fixes those coordinates on paper, which makes the height a clean 3D Pythagorean computation and makes the natural cutting planes visible: two vertical slices through the ridge ends carve the solid into a middle prism and two end pyramids.

1STEP 1

Find the ridge height

Set base corner O₁ at the origin and ridge end T₁=(3,4,H); the equal-side edge O₁T₁=13 gives 3²+4²+H²=169, so the ridge height is 12.

|O₁T₁|² = 3² + 4² + H² = 25 + H² = 13² = 169 → H² = 144 → H = 12
2STEP 2

Cut into a prism and two pyramids

Cut at x=3 and x=10 through the ridge ends: the middle has constant cross-section (a prism), and each end rises to one point (a pyramid).

V = V_prism + V_left pyramid + V_right pyramid
3STEP 3

Volume of the middle prism

The constant cross-section is a triangle of area 12\frac{1}{2}·8·12=48; multiply by the prism length 7 to get 336.

V_prism = (1/2 · 8 · 12) × 7 = 48 × 7 = 336
4STEP 4

Volume of the two end pyramids

Each end is a pyramid on a 3 × 8 base with apex height 12, volume 13\frac{1}{3}·24·12=96; the two together give 192.

V_pyramid = 1/3 (3 × 8)(12) = 1/3 (24)(12) = 96 → 2 V_pyramid = 192
5STEP 5

Add the pieces

Add the middle prism 336 and the two end pyramids 192 to get 528 — choice (C).

V = 336 + 192 = 528 → (C)
Answer
528
The solid is trapped inside the full 13 × 8 × 12 box of volume 1248, and it clearly fills well over a third of it, so 528 is in a believable range while 832 (two-thirds of the box) would be too fat for a shape that pinches to a line on top. A clean independent check uses the prismatoid formula V=h6\frac{h}{6}(A_bot+4A_mid+A_top) with height h=12: the bottom area is 13 · 8=104, the top (a segment) has area 0, and the mid-height cross-section is a 10 × 4 rectangle of area 40. This gives 126\frac{12}{6}(104+4 · 40+0)=2(264)=528, matching exactly and separating (C) 528 cleanly from its neighbor (B) 520.
💡Key takeaway

When a solid has no formula, slice it where its shape stops changing: the steady middle becomes a prism and the leftover ends become pyramids you can measure and add.

  • Find the ridge height
  • Cut into a prism and two pyramids
  • Volume of the middle prism
  • Volume of the two end pyramids
  • Add the pieces

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