2021 AMC 10A Problem 13

1. The lengths $\sqrt{13}$, $2\sqrt5 = \sqrt{20}$ and $5 = \sqrt{25}$ look like hypotenuses of right triangles. Their squares are $13$, $20$ and $25$.
2. Test the three faces that meet at $A$ with the Pythagorean theorem:
- face $ABC$: $AB^2 + AC^2 = 4 + 9 = 13 = BC^2$, so the angle at $A$ is a right angle;
- face $ABD$: $AB^2 + AD^2 = 4 + 16 = 20 = BD^2$, right angle at $A$;
- face $ACD$: $AC^2 + AD^2 = 9 + 16 = 25 = CD^2$, right angle at $A$.
3. So the edges $AB$, $AC$ and $AD$ are all at right angles to each other, like three edges meeting at the corner of a $2 \times 3 \times 4$ box.
4. Lay triangle $ABC$ flat as the base. It is a right triangle with legs $2$ and $3$, so its area is $\frac12 \times 2 \times 3 = 3$.
5. Edge $AD$ is at right angles to both $AB$ and $AC$, so it points straight up from the base. The height is $AD = 4$.
6. A pyramid's volume is $\frac13 \times \text{base area} \times \text{height} = \frac13 \times 3 \times 4 = 4$.

The answer is $\boxed{\textbf{(C)}\ 4}$.

When all six edges of a tetrahedron are given, look first for right angles. A tetrahedron cut from the corner of a box with edges $a$, $b$, $c$ has volume $\frac{abc}{6}$, one sixth of the box. Here that is $\frac{2 \times 3 \times 4}{6} = 4$.

Put the shape on coordinates: $A = (0,0,0)$, $B = (2,0,0)$, $C = (0,3,0)$, $D = (0,0,4)$. Then $BC = \sqrt{4 + 9} = \sqrt{13}$, $BD = \sqrt{4 + 16} = 2\sqrt5$ and $CD = \sqrt{9 + 16} = 5$, all matching. The volume is $\frac13 \times 3 \times 4 = 4$ as before.