AMC 8 · 2022 · #24
Grade 7 geometry-3dProblem
The figure below shows a polygon ABCDEFGH, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that AH=EF=8 and GH=14. What is the volume of the prism?
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $\triangle GJB$ and $\triangle FIC$ 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.
Givens: The 2D polygon $ABCDEFGH$ folds (along the dotted lines $JB$, $BG$, $CF$, $FI$) into a triangular prism; $\triangle GJB$ and $\triangle FIC$ are right triangles (right angles at $J$ and $I$) and become the two congruent triangular bases; $AH = 8$ (one of the rectangular-face edges); $EF = 8$ (an edge that folds onto $FG$, the lateral height of the prism); $GH = 14$ (the total length of the bottom edge of the net, split into $GJ$ and $JH$); Answer choices: (A) $112$, (B) $128$, (C) $192$, (D) $240$, (E) $288$
Unknowns: The volume of the triangular prism in cubic units
Understand
Restated: 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 $\triangle GJB$ and $\triangle FIC$ 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.
Givens: The 2D polygon $ABCDEFGH$ folds (along the dotted lines $JB$, $BG$, $CF$, $FI$) into a triangular prism; $\triangle GJB$ and $\triangle FIC$ are right triangles (right angles at $J$ and $I$) and become the two congruent triangular bases; $AH = 8$ (one of the rectangular-face edges); $EF = 8$ (an edge that folds onto $FG$, the lateral height of the prism); $GH = 14$ (the total length of the bottom edge of the net, split into $GJ$ and $JH$); Answer choices: (A) $112$, (B) $128$, (C) $192$, (D) $240$, (E) $288$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
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 $\triangle GJB$ and $\triangle FIC$ 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.
Execute — Answer: C
6.G.A.4 Step 1 - Mentally fold the net.
- The two dashed segments $CF$ and $BG$ are crease lines: the triangles $\triangle FIC$ and $\triangle GJB$ lift up and become the two parallel triangular bases.
- The three rectangles between them ($BGFC$, $ABJH$, and the strip $CIDE \cup FIDE$ region around $EF$) wrap around as the lateral faces of a triangular prism.
- So the solid is a triangular prism whose base is the right triangle $\triangle GJB$ (congruent to $\triangle FIC$).
💡 Recognizing that a 2D net folds into a specific 3D solid (here a triangular prism) is the Grade 6 "represent 3D figures using nets" standard.
6.G.A.4 Step 2 - Find the lateral height of the prism.
- The lateral edges of the prism are the segments of the net that connect the two triangular bases — namely $FG$ and $JH$.
- When folded, edge $EF$ on the net is glued onto $FG$ (they share point $F$ and become the same lateral edge), so $FG = EF = 8$.
- Therefore the prism's height is $h = 8$.
💡 Edges that meet when a net is folded must have the same length — a direct net-and-solid relationship from Grade 6.
4.MD.A.3 Step 3 - Find the first leg of the triangular base.
- The rectangular face $ABJH$ (on the net, between the lines $AB$ and $HJ$) has $AH$ and $BJ$ as opposite sides, so they are equal.
- Since $AH = 8$, we get $BJ = 8$.
- After folding, $BJ$ becomes one of the two legs of the right-triangular base $\triangle GJB$.
💡 Opposite sides of a rectangle are equal — a Grade 4 perimeter/area rectangle fact, drawn directly from the labelled diagram.
4.OA.A.3 Step 4 - Find the second leg of the triangular base.
- The bottom edge $GH = 14$ is made up of $GJ$ (the second leg of $\triangle GJB$) and $JH$.
- But $JH$ is also a lateral edge of the prism, so $JH = h = 8$.
- Subtracting, $GJ = GH - JH = 14 - 8 = 6$.
💡 Splitting a known total ($GH = 14$) into a known piece ($JH = 8$) and an unknown piece ($GJ$) is a Grade 4 multi-step word-problem move.
6.G.A.1 Step 5 - Compute the area of the triangular base.
- $\triangle GJB$ is a right triangle with legs $BJ = 8$ and $GJ = 6$, so its area is half the product of the legs.
💡 Area of a right triangle as half the product of its legs is a Grade 6 "area of triangles" standard.
7.G.B.6 Step 6 Multiply base area by prism height to get the volume, then match to the answer choices.
💡 Volume of a (non-rectangular) prism as base-area times height is exactly the Grade 7 "area, surface area, and volume" standard.
6.G.A.4 Mentally fold the net. The two dashed segments $CF$ and $BG$ are crease lines: t 6.G.A.4 Find the lateral height of the prism. The lateral edges of the prism are the seg 4.MD.A.3 Find the first leg of the triangular base. The rectangular face $ABJH$ (on the n 4.OA.A.3 Find the second leg of the triangular base. The bottom edge $GH = 14$ is made up 6.G.A.1 Compute the area of the triangular base. $\triangle GJB$ is a right triangle wit 7.G.B.6 Multiply base area by prism height to get the volume, then match to the answer c Review
Reasonableness: 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 $24 \times 8 = 192$ 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 $\tfrac{1}{2} \times 14 \times 8 \times 8 = 448$, 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.
Alternative: Tool #10 (Create a Physical Representation): cut out the net on paper, label $AH = EF = 8$ and $GH = 14$, and fold along the dotted lines $JB$, $BG$, $CF$, $FI$. You will physically see that $EF$ glues onto $FG$ (so the height is $8$), $AH$ becomes opposite to $BJ$ (so $BJ = 8$), and the remaining segment $GJ = 14 - 8 = 6$ is the second leg of the base. From there the formula $V = \tfrac{1}{2} \times 6 \times 8 \times 8 = 192$ gives the same answer (C).
CCSS standards used (min grade 7)
4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Using the fact that opposite sides of the rectangular face $ABJH$ are equal to conclude $BJ = AH = 8$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Subtracting the known lateral edge $JH = 8$ from the total bottom edge $GH = 14$ to get the base leg $GJ = 6$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of the right-triangular base $\triangle GJB$ as $\tfrac{1}{2} \times 8 \times 6 = 24$.)6.G.A.4Represent three-dimensional figures using nets and find surface area (Recognizing that the polygon $ABCDEFGH$ is a net of a triangular prism and matching net-edges that glue together when folded ($EF$ onto $FG$, $JH$ as a lateral edge).)7.G.B.6Solve real-world problems involving area, surface area, and volume (Applying the volume-of-a-prism formula $V = (\text{base area}) \times (\text{height}) = 24 \times 8 = 192$ for a non-rectangular (triangular) prism.)
⭐ This AMC 8 problem only needs Grade 7 volume reasoning — base area times height — and a little net-folding visualization you already know!
⭐ This AMC 8 problem only needs Grade 7 volume reasoning — base area times height — and a little net-folding visualization you already know!
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