2015 AMC 10A Problem 21

We claim that the planes $ABC$ and $ABD$ are perpendicular to each other. We can show this by dropping the perpendiculars from $C$ and $D$ to $AB$. Since $AC = AD$ and $BC = BD,$ we have that the feet of these altitudes will coincide at point $P.$ Then we have that $$CP = DP = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5}.$$ We also have $CD = CP\sqrt{2},$ so $\triangle CPD$ is an isosceles right triangle and $CP\perp DP.$ Since $CP\perp AB$ and $CP\perp DP$, the segment $CP$ is perpendicular to the plane $ABD$. Finally, the volume of the tetrahedron is $$\dfrac{1}{3}[ABD] \cdot CP = \dfrac{6}{3} \cdot \dfrac{12}{5} = \dfrac{24}{5}.$$ Thus, C is the correct answer.