AMC 8 · 2025 · #8

Grade 8 geometry-3d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

surface-areavolume-rectangular-prismpolyhedron-netsperfect-squares identify-subproblemsdimensional-analysis ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic

📏 Short solution 💡 3 insights 📊 Diagram

Problem

Isaiah cuts open a cardboard cube along some of its edges to form the flat shape shown on the right, which has an area of $18$ square centimeters. What is the volume of the cube in cubic centimeters?

Pick an answer.

(A)

$3\sqrt{3}$

(B)

(C)

(D)

$6\sqrt{3}$

(E)

$9\sqrt{3}$

AMC 8 2025 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.

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Toolkit + CCSS Solution

Understand

Restated: Isaiah cuts a cardboard cube open along some of its edges and flattens it into a single connected shape (a net). That flat shape has total area $18$ square centimeters. Find the volume of the original cube in cubic centimeters.

Givens: The flat shape is the net of a cube — every face of the cube appears exactly once; A cube has $6$ congruent square faces; Total area of the net $= 18 \text{ cm}^2$; Answer choices: (A) $3\sqrt{3}$, (B) $6$, (C) $9$, (D) $6\sqrt{3}$, (E) $9\sqrt{3}$ (cm$^3$)

Unknowns: The volume $V$ of the cube, in cubic centimeters

Understand

Plan

Primary tool: #17 Visualize Spatial Relationships

Secondary: #7 Identify Subproblems, #8 Analyze the Units

The picture is a flat net, but the question asks about a 3D volume — Tool #17 (Visualize Spatial Relationships) is the bridge: mentally re-fold the net into the cube to see that the flat area is exactly the cube's surface area (no overlap, no gaps). From there Tool #7 (Identify Subproblems) splits the work into a clean chain: (a) total surface area $\to$ area of one face, (b) area of one face $\to$ side length $s$, (c) side length $\to$ volume $V = s^3$. Tool #8 (Analyze the Units) keeps us honest as the units march from cm$^2$ (area) down to cm (length) and up to cm$^3$ (volume).

Execute — Answer: A

#17 Visualize Spatial Relationships 6.G.A.4 Step 1

Re-fold the net mentally.
The flat shape was cut along some edges and laid flat without overlapping, so the outside of the cube is exactly covered once by the $6$ square faces of the net.
This means the area of the flat shape equals the cube's surface area.

$$\text{Surface Area} = \text{Area of net} = 18 \text{ cm}^2$$

💡 Folding a flat net into a 3D figure preserves area — that is exactly the Grade 6 "nets and surface area" idea.

#7 Identify Subproblems 3.OA.A.3 Step 2

Subproblem 1: find the area of a single face.
The cube has $6$ congruent square faces, so divide the total surface area by $6$.

$$\text{Area of one face} = \dfrac{18 \text{ cm}^2}{6} = 3 \text{ cm}^2$$

💡 Splitting one big equal-shares quantity into $6$ identical pieces is a Grade 3 division word problem.

#7 Identify Subproblems 8.EE.A.2 Step 3

Subproblem 2: turn the face area into a side length.
Each face is a square, so its area equals (side)$^2$.
To go from area back to side, take the square root.

$$s^2 = 3 \;\Rightarrow\; s = \sqrt{3} \text{ cm}$$

💡 Recovering the side of a square from its area uses the Grade 8 square-root symbol — and $\sqrt{3}$ is an irrational number we keep in exact form.

#8 Analyze the Units 8.EE.A.2 Step 4

Subproblem 3: turn the side length into the cube's volume using $V = s^3$.
Multiply $\sqrt{3}$ by itself three times; pair two copies to get $3$, then multiply the leftover $\sqrt{3}$.

$$V = s^3 = (\sqrt{3})^3 = \sqrt{3}\cdot\sqrt{3}\cdot\sqrt{3} = 3 \cdot \sqrt{3} = 3\sqrt{3} \text{ cm}^3 \;\Rightarrow\; \textbf{(A)}$$

💡 Cubing a length cm gives cm$^3$, and using $\sqrt{3}\cdot\sqrt{3}=3$ is the Grade 8 square-root rule.

[1] #17 6.G.A.4 Re-fold the net mentally. The flat shape was cut along some edges and laid flat

[2] #7 3.OA.A.3 Subproblem 1: find the area of a single face. The cube has $6$ congruent square

[3] #7 8.EE.A.2 Subproblem 2: turn the face area into a side length. Each face is a square, so i

[4] #8 8.EE.A.2 Subproblem 3: turn the side length into the cube's volume using $V = s^3$. Multi

Review

Reasonableness: Quick estimate: $\sqrt{3} \approx 1.73$, so $s \approx 1.73$ cm. Then $s^2 \approx 3$ cm$^2$ (matches one face) and $s^3 \approx 1.73 \times 3 = 5.2$ cm$^3$. Compare to $3\sqrt{3} \approx 3 \times 1.73 = 5.2$ — same value, so the algebra and the estimate agree. The unit chain (cm$^2$ for face area, cm for side, cm$^3$ for volume) is also consistent with what the question asks for.

Alternative: Tool #3 (Eliminate Possibilities): plug each choice back as a volume and read off the side. Choice (A) $3\sqrt{3} = (\sqrt{3})^3$ gives $s=\sqrt{3}$, face area $3$, surface area $18$ ✓. Choice (B) $6$ gives $s=\sqrt[3]{6}\approx 1.82$, face area $\approx 3.30$, surface $\approx 19.8$ ✗. Choices (C), (D), (E) all yield $s>\sqrt{3}$ and thus surface area $>18$. Only (A) survives.

CCSS standards used (min grade 8)

3.OA.A.3 Solve multiplication and division word problems within 100 (Dividing the total surface area $18$ cm$^2$ equally among the $6$ congruent faces to get $3$ cm$^2$ per face.)
6.G.A.4 Represent three-dimensional figures using nets and find surface area (Recognizing that the flat shape is a net of the cube and therefore its area equals the cube's surface area.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Taking $s = \sqrt{3}$ from $s^2 = 3$, and computing $(\sqrt{3})^3 = 3\sqrt{3}$ using the rule $\sqrt{3}\cdot\sqrt{3}=3$.)

⭐ This AMC 8 problem only needs the Grade 8 square-root idea ($\sqrt{3}\cdot\sqrt{3}=3$) you already know — the rest is just "net area $=$ surface area" and a little division!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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