AMC 10 · 2021 · #13
Grade 8 geometry-3dProblem
What is the volume of tetrahedron ABCD with edge lengths AB=2, AC=3, AD=4, BC=13, BD=25, and CD=5 ?
Pick an answer.
AMC 10 2021 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 tetrahedron $ABCD$ has six edge lengths: $AB = 2$, $AC = 3$, $AD = 4$, $BC = \sqrt{13}$, $BD = 2\sqrt{5}$, $CD = 5$. Find its volume.
Givens: Three edges meeting at $A$: $AB = 2, AC = 3, AD = 4$; The opposite three edges: $BC = \sqrt{13}, BD = 2\sqrt{5}, CD = 5$; Answer choices: $3,\;\; 2\sqrt{3},\;\; 4,\;\; 3\sqrt{3},\;\; 6$
Unknowns: Volume of tetrahedron $ABCD$
Understand
Restated: A tetrahedron $ABCD$ has six edge lengths: $AB = 2$, $AC = 3$, $AD = 4$, $BC = \sqrt{13}$, $BD = 2\sqrt{5}$, $CD = 5$. Find its volume.
Givens: Three edges meeting at $A$: $AB = 2, AC = 3, AD = 4$; The opposite three edges: $BC = \sqrt{13}, BD = 2\sqrt{5}, CD = 5$; Answer choices: $3,\;\; 2\sqrt{3},\;\; 4,\;\; 3\sqrt{3},\;\; 6$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities
Tool #1 (Draw): sketch vertex $A$ with the three edges $AB, AC, AD$ going to $B, C, D$. The trick is to test whether each triangular face containing $A$ is right-angled at $A$ — use the Pythagorean converse (Tool #6, Guess and Check on the equation $a^2 + b^2 = c^2$). Indeed: $2^2 + 3^2 = 13 = BC^2$, $2^2 + 4^2 = 20 = BD^2$, $3^2 + 4^2 = 25 = CD^2$. All three! So the three edges from $A$ are pairwise perpendicular — the tetrahedron is a 'corner of a box' and its volume is $\tfrac{1}{6} \cdot AB \cdot AC \cdot AD$. Tool #7 (Subproblems) splits the work into: detect right angles, then plug into the corner-of-box formula. Tool #3 (Eliminate) confirms the final number against the choices.
Execute — Answer: C
6.EE.A.1 Step 1 - List the squares of the three edges at $A$: $AB^2 = 4$, $AC^2 = 9$, $AD^2 = 16$.
- Then list the squares of the three opposite edges: $BC^2 = 13$, $BD^2 = 20$, $CD^2 = 25$.
💡 Grade 6 expressions: squaring each length surfaces sum-of-squares patterns.
8.G.B.6 Step 2 - Check the Pythagorean relation in face $ABC$.
- Does $AB^2 + AC^2 = BC^2$?
- $4 + 9 = 13$ ✓.
- So $\triangle ABC$ has a right angle at $A$ (converse of the Pythagorean theorem).
- That means edge $AB$ is perpendicular to edge $AC$.
💡 Grade 8 Pythagorean converse: if sides satisfy $a^2 + b^2 = c^2$, the triangle is right-angled.
8.G.B.6 Step 3 - Repeat for face $ABD$ and face $ACD$.
- $4 + 16 = 20 = BD^2$ ✓, so $\angle BAD = 90^\circ$.
- $9 + 16 = 25 = CD^2$ ✓, so $\angle CAD = 90^\circ$.
💡 Same Pythagorean converse applied to the other two faces meeting at $A$.
8.G.B.8 Step 4 - Translate the three right angles into a 3D picture.
- All three edges from $A$ are pairwise perpendicular — exactly like the three edges meeting at a corner of a rectangular box.
- Pin $A$ at the origin with $B$ on the $x$-axis, $C$ on the $y$-axis, $D$ on the $z$-axis.
- The tetrahedron is the 'corner piece' of a $2 \times 3 \times 4$ box.
💡 Grade 8 coordinates in space: place perpendicular edges on the axes for a clean picture.
6.G.A.1 Step 5 - Compute the volume using face $ABC$ (a right triangle, legs $2$ and $3$) as the base.
- Its area is $\tfrac{1}{2} \cdot 2 \cdot 3 = 3$.
- The height from $D$ to the plane of $ABC$ (the $xy$-plane) is simply $AD = 4$, since $AD$ is perpendicular to that plane.
💡 Grade 6 base-times-height area of a right triangle, plus 'perpendicular edge = height'.
7.G.B.6 Step 6 - Apply the pyramid volume formula.
- $V = \tfrac{1}{3} \cdot \text{base} \cdot \text{height} = \tfrac{1}{3} \cdot 3 \cdot 4 = 4$.
- This is also the well-known 'corner of a box' formula $V = \tfrac{1}{6} \cdot AB \cdot AC \cdot AD = \tfrac{1}{6} \cdot 2 \cdot 3 \cdot 4 = 4$.
💡 Grade 7 volume formula: any pyramid's volume is one-third of base times height.
6.EE.B.5 Step 7 - Match $V = 4$ to the choices.
- It's (C).
- The other values ($3, 2\sqrt{3}, 3\sqrt{3}, 6$) don't agree with the box-corner calculation.
💡 Grade 6 multiple-choice match: only one option equals $4$.
6.EE.A.1 List the squares of the three edges at $A$: $AB^2 = 4$, $AC^2 = 9$, $AD^2 = 16$. 8.G.B.6 Check the Pythagorean relation in face $ABC$. Does $AB^2 + AC^2 = BC^2$? $4 + 9 8.G.B.6 Repeat for face $ABD$ and face $ACD$. $4 + 16 = 20 = BD^2$ ✓, so $\angle BAD = 9 8.G.B.8 Translate the three right angles into a 3D picture. All three edges from $A$ are 6.G.A.1 Compute the volume using face $ABC$ (a right triangle, legs $2$ and $3$) as the 7.G.B.6 Apply the pyramid volume formula. $V = \tfrac{1}{3} \cdot \text{base} \cdot \tex 6.EE.B.5 Match $V = 4$ to the choices. It's (C). The other values ($3, 2\sqrt{3}, 3\sqrt{ Review
Reasonableness: The 'box-corner' shape is a sanity check by itself: the full $2 \times 3 \times 4$ box has volume $24$, and a corner tetrahedron is exactly $1/6$ of the box (because slicing a box along the three diagonal planes through one corner produces six congruent tetrahedra). So $V = 24/6 = 4$ ✓. The magnitude is reasonable for edge lengths of order $2$–$5$: a 'typical' tetrahedron with those edges should have a volume around a handful of cubic units, and $4$ fits.
Alternative: Tool #6 (Guess and Check) using the scalar triple product. With the coordinates $A = (0,0,0)$, $B = (2,0,0)$, $C = (0,3,0)$, $D = (0,0,4)$, the volume of the tetrahedron is $\tfrac{1}{6} |{\vec{AB} \cdot (\vec{AC} \times \vec{AD})}| = \tfrac{1}{6} |\det\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix}| = \tfrac{1}{6} \cdot 24 = 4$. Same answer (C). (This is the high-school version of the same calculation we did with the pyramid formula.)
CCSS standards used (min grade 8)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Computing $AB^2, AC^2, AD^2, BC^2, BD^2, CD^2$ to surface Pythagorean patterns.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Matching the computed $V = 4$ against the five answer choices to select (C).)6.G.A.1Find the area of right triangles, other triangles, special quadrilaterals, and polygons (Computing the area of right triangle $ABC$ as $\tfrac{1}{2} \cdot 2 \cdot 3 = 3$.)7.G.B.6Solve real-world and mathematical problems involving area, volume and surface area (Applying $V = \tfrac{1}{3} \cdot \text{base} \cdot \text{height}$ to the tetrahedron.)8.G.B.6Explain a proof of the Pythagorean Theorem and its converse (Using the converse to detect right angles at $A$ in faces $ABC, ABD, ACD$.)8.G.B.8Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Placing $A, B, C, D$ on the coordinate axes once $AB, AC, AD$ are pairwise perpendicular.)
⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know! Square every edge: $2^2+3^2 = 13$, $2^2+4^2 = 20$, $3^2+4^2 = 25$ — these are exactly $BC^2, BD^2, CD^2$. So $AB, AC, AD$ are all perpendicular to each other, like the three edges at the corner of a $2 \times 3 \times 4$ box. The tetrahedron is $1/6$ of that box, so its volume is $24/6 = 4$, answer (C).
⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know! Square every edge: $2^2+3^2 = 13$, $2^2+4^2 = 20$, $3^2+4^2 = 25$ — these are exactly $BC^2, BD^2, CD^2$. So $AB, AC, AD$ are all perpendicular to each other, like the three edges at the corner of a $2 \times 3 \times 4$ box. The tetrahedron is $1/6$ of that box, so its volume is $24/6 = 4$, answer (C).
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