2015 AMC 12A Problem 16

Triangles $ABC$ and $ABD$ are $3$-$4$-$5$ right triangles with area $6$ and common hypotenuse $AB.$ Let $E$ be the foot of the altitude from $C$ to $AB;$ then $CE = \dfrac{3\cdot 4}{5} = \dfrac{12}{5}.$ Likewise the altitude from $D$ meets $AB$ at the same point $E$ with $DE = \dfrac{12}{5}.$ Triangle $CDE$ has sides $\dfrac{12}{5},$ $\dfrac{12}{5},$ and $CD = \dfrac{12}{5}\sqrt{2},$ so it is an isosceles right triangle with the right angle at $E.$ Thus $DE \perp CE$ and $DE \perp AB,$ making $DE$ perpendicular to the plane of $ABC.$ The tetrahedron’s volume is $\dfrac{1}{3}\cdot [ABC]\cdot DE = \dfrac{1}{3}\cdot 6\cdot \dfrac{12}{5}$ $= \dfrac{24}{5}.$ Thus, the correct answer is C.