2015 AMC 10A Problem 21

Since $3^2 + 4^2 = 5^2$, face $ABC$ is a right triangle with the right angle at $C$, and face $ABD$ is a right triangle with the right angle at $D$. The two faces are congruent and share the hypotenuse $AB$.

Let $F$ be the foot of the perpendicular from $C$ to $AB$. In a $3$-$4$-$5$ triangle the altitude to the hypotenuse is $\frac{3 \cdot 4}{5} = \frac{12}{5}$, and it lands at distance $\frac{AC^2}{AB} = \frac{9}{5}$ from $A$. Triangle $ABD$ has the same side lengths in the same positions ($AD = AC$, $BD = BC$), so the perpendicular from $D$ to $AB$ has the same length $\frac{12}{5}$ and lands at the same point $F$.

Now $CF = DF = \frac{12}{5}$ and $CD = \frac{12}{5}\sqrt2$, so $CF^2 + DF^2 = CD^2$ and $\angle CFD = 90^\circ$. The segment $DF$ is therefore perpendicular to both $AB$ and $CF$, two intersecting lines of the plane $ABC$, so $DF$ is perpendicular to that plane: it is the height of the tetrahedron over base $ABC$.

The base has area $\frac12 \cdot 3 \cdot 4 = 6$, so
$$V = \frac13 \cdot 6 \cdot \frac{12}{5} = \frac{24}{5}.$$

The answer is $\boxed{\textbf{(C)}\ \dfrac{24}{5}}$.

A tetrahedron's volume needs a base and a genuine perpendicular height, and the height is rarely one of the given edges. Two congruent right triangles glued along their hypotenuse have altitudes with a common foot, and the odd edge $\frac{12}{5}\sqrt2$ is a signal: $\frac{12}{5}$ is the familiar $3$-$4$-$5$ altitude and the $\sqrt2$ announces an isosceles right triangle $CFD$. Decode unusual lengths by asking what right triangle would produce them.

Slice the solid along the plane through $F$ perpendicular to $AB$. The cross-section is the isosceles right triangle $CFD$ with legs $\frac{12}{5}$, area $\frac{72}{25}$, and it splits the tetrahedron into two pyramids with apexes $A$ and $B$ whose heights add to $AB = 5$. So $V = \frac13 \cdot \frac{72}{25} \cdot 5 = \frac{24}{5}$.

• Taking $AD = 3$ or $BD = 4$ as the height above face $ABC$, or assuming the two right-triangle faces are automatically perpendicular without using $CD$.
• Believing the right angle at $D$ puts $D$ directly above $C$, which would give $CD$ as the height and the wrong volume $\frac{24}{5}\sqrt2$ (choice E).
• Placing the right angles at $A$ or $B$ instead of at $C$ and $D$, which is where the $5$ is the hypotenuse.