Question

Tetrahedron $ABCD$ has $AB = 5,$ $AC = 3,$ $BC = 4,$ $BD = 4,$ $AD = 3,$ and $CD = \dfrac{12}{5}\sqrt{2}.$ What is the volume of the tetrahedron?

$3\sqrt{2}$

$2\sqrt{5}$

$\dfrac{24}{5}$

$3\sqrt{3}$

$\dfrac{24}{5}\sqrt{2}$

Solution:

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.

2015 AMC 12A Problem 16

Solution:

Problem 16 in Other Years