2015 AMC 10A Problem 21

For volume we need a base and a height that is truly perpendicular to the base. None of the given edges is that height, so we build it.

1. Spot the right triangles. $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. So face $ABC$ (sides $3, 4, 5$) has a right angle at $C$, and face $ABD$ (sides $3, 4, 5$) has a right angle at $D$. The two faces are the same triangle, joined along the long side $AB$.
2. Height of one face. In the $3$-$4$-$5$ triangle $ABC$, drop a perpendicular from $C$ to $AB$ and call its foot $F$. The area can be worked out two ways: $\frac{1}{2} \cdot 3 \cdot 4 = 6$, and $\frac{1}{2} \cdot 5 \cdot CF$. So $CF = \frac{12}{5}$.
3. Where $F$ lands. In right triangle $ACF$, $AF^2 = 3^2 - \left(\frac{12}{5}\right)^2 = 9 - \frac{144}{25} = \frac{81}{25}$, so $AF = \frac{9}{5}$.
4. The other face. Triangle $ABD$ has $AD = 3$ and $BD = 4$, exactly like $AC$ and $BC$. So the same calculation gives a perpendicular from $D$ of length $\frac{12}{5}$, landing $\frac{9}{5}$ from $A$: at the same point $F$.
5. The angle between the two heights. Look at triangle $CFD$: $CF = DF = \frac{12}{5}$ and $CD = \frac{12}{5}\sqrt{2}$. Since
$$\left(\tfrac{12}{5}\right)^2 + \left(\tfrac{12}{5}\right)^2 = 2\left(\tfrac{12}{5}\right)^2 = \left(\tfrac{12}{5}\sqrt{2}\right)^2,$$
the Pythagorean theorem works backwards and the angle at $F$ is $90^\circ$.
6. So $DF$ is the height. $DF$ is perpendicular to $AB$ (step 4) and to $CF$ (step 5). Both of those lines lie flat in the base $ABC$ and cross each other at $F$, so $DF$ stands straight up from the base.
7. Volume. A pyramid's volume is $\frac{1}{3} \times \text{base area} \times \text{height}$:
$$V = \frac{1}{3} \times 6 \times \frac{12}{5} = \frac{24}{5}.$$

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

The strange length $\frac{12}{5}\sqrt{2}$ is a clue. $\frac{12}{5}$ is the height of a $3$-$4$-$5$ triangle onto its long side, and a factor $\sqrt{2}$ is what the long side of an isosceles right triangle has. So the two faces meet at a right angle, like a book opened to $90^\circ$. When a problem gives an odd length, ask which right triangle would produce it.

Cut the solid with the flat triangle $CFD$. That triangle is perpendicular to $AB$ and has area $\frac{1}{2} \cdot \frac{12}{5} \cdot \frac{12}{5} = \frac{72}{25}$. It splits the solid into two pyramids, one with tip $A$ and one with tip $B$, whose heights $AF$ and $FB$ add up to $AB = 5$. So
$$V = \frac{1}{3} \cdot \frac{72}{25} \cdot AF + \frac{1}{3} \cdot \frac{72}{25} \cdot FB = \frac{1}{3} \cdot \frac{72}{25} \cdot 5 = \frac{24}{5}.$$

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