2015 AMC 12A Problem 16

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}

Answer: C

Concepts:3D geometry volume right triangle

Difficulty rating: 1840

Small hint:

Triangles

ABC

and

ABD

are both

3

-

4

-

5

right triangles sharing the edge

AB

Big hint:

Drop altitudes from

C

and

D

to

AB;

their common foot shows

CD

is perpendicular to base

ABC

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.