AMC 8 · 2022 · #24

Grade 7 geometry-3d

Archetype Net Folding / Face Adjacency └ Sub-archetype Net-Edge Matching for Length, Area, or Volume

volume-rectangular-prismpolyhedron-netsspatial-visualization physical-representationidentify-subproblems ↑ Prerequisites: area-trianglespolyhedron-nets

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A flat polygon ABCDEFGH (made of rectangles and right triangles) is the net of a triangular prism: folding along the dotted lines glues the right triangles △ GJB and △ FIC together as the two triangular bases, while the rectangles become the three lateral faces. We are told AH = EF = 8 and GH = 14, and we need to compute the volume of the resulting prism.

Pick an answer.

(A)

~112

(B)

~128

(C)

~192

(D)

~240

(E)

~288

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Visualize Spatial Relationships

This is a classic "net to 3D" problem, which is exactly the trigger for Tool #17 (Visualize Spatially): we must mentally fold the flat figure along the dotted lines and identify which edges glue together. Once we see that the two right triangles △ GJB and △ FIC become the two bases and the three rectangles become the lateral faces, Tool #1 (Draw a Diagram) lets us re-label the net with the prism's measurements (height of the prism, legs of the base triangle). Finally, Tool #7 (Identify Subproblems) splits the volume question into two independent pieces — "What is the area of the triangular base?" and "What is the height of the prism?" — that combine via the prism volume formula. We deliberately avoid Tool #13 (Algebra); the problem is purely a geometric bookkeeping exercise once the folding is understood.

1STEP 1

Fold the net: creases CF and BG lift the two right triangles into the bases, and the three rectangles wrap around as a triangular prism.

2STEP 2

Folding lands edge EF onto FG, so the two bases are separated by the lateral height h = 8.

h = FG = EF = 8

3STEP 3

Rectangle face ABJH has AH and BJ as opposite sides, so the first base leg is BJ = 8.

BJ = AH = 8

4STEP 4

The bottom edge GH = 14 splits into lateral edge JH = 8 and the second base leg, so GJ = 6.

GJ = GH - JH = 14 - 8 = 6

5STEP 5

Right triangle GJB has legs BJ = 8 and GJ = 6, so the base area is half their product, 24.

Base area =

\frac{1}{2}

× 8 × 6 = 24

6STEP 6

Multiply base area by height for the volume: V = 24 × 8 = 192, matching choice (C).

V = Base area × h = 24 × 8 = 192 → (C)

Answer

~192

Sanity-check the dimensions against the answer choices. The base triangle 6-8-10 has area 24, and the height is 8, so the volume 24 × 8 = 192 lands cleanly in the middle of the choices (between 128 and 240). If we had wrongly used the full bottom edge GH = 14 as a leg instead of GJ = 6, we would get

\frac{1}{2}

× 14 × 8 × 8 = 448, which is not even a choice — confirming we correctly separated the lateral edge JH = 8 from the base leg GJ = 6. Also, the units "length cubed" make sense for a volume.

💡Key takeaway

This AMC 8 problem only needs Grade 7 volume reasoning — base area times height — and a little net-folding visualization you already know!

More like this