2015 AMC 10A Problem 21

Question

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?

3\sqrt2

2\sqrt5

\dfrac{24}5

3\sqrt{3}

\dfrac{24}5\sqrt2

Put the two

3

-

4

-

5

triangles on a common base

AB

CD=(\frac{12}{5})\sqrt2

makes the two altitudes perpendicular

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.

Problem 21 in other years