2015 AMC 10A Problems/Problem 21

The following problem is from both the 2015 AMC 12A #16 and 2015 AMC 10A #21, so both problems redirect to this page.

Problem

Tetrahedron $ABCD$ has $AB=5$ , $AC=3$ , $BC=4$ , $BD=4$ , $AD=3$ , and $CD=\tfrac{12}5\sqrt2$ . What is the volume of the tetrahedron?

$\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2$

Solutions

Solution 1

Drop altitudes of triangle $ABC$ and triangle $ABD$ down from $C$ and $D$ , respectively. Both will hit the same point; let this point be $T$ . Because both triangle $ABC$ and triangle $ABD$ are 3-4-5 triangles, $CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}$ . Because $CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}$ , it follows that the $CTD$ is a right triangle, meaning that $\angle CTD = 90^\circ$ , and it follows that planes $ABC$ and $ABD$ are perpendicular to each other. Now, we can treat $ABC$ as the base of the tetrahedron and $TD$ as the height. Thus, the desired volume is $V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}$ which is answer $\boxed{\textbf{(C) } \dfrac{24}{5}}$

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 10 Problems and Solutions

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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2015 AMC 10A Problems/Problem 21

Contents

Problem

Solutions

Solution 1

See Also