Question

Square pyramid $ABCDE$ has base $ABCD,$ which measures $3$ cm on a side, and altitude $\overline{AE}$ perpendicular to the base, which measures $6$ cm. Point $P$ lies on $\overline{BE},$ one third of the way from $B$ to $E;$ point $Q$ lies on $\overline{DE},$ one third of the way from $D$ to $E;$ and point $R$ lies on $\overline{CE},$ two thirds of the way from $C$ to $E.$ What is the area, in square centimeters, of $\triangle PQR?$

$\dfrac{3\sqrt2}{2}$

$\dfrac{3\sqrt3}{2}$

$2\sqrt2$

$2\sqrt3$

$3\sqrt2$

Answer: C

Concepts:3D geometry vector triangle area

Difficulty rating: 1620

Solution:

Place $A=(0,0,0),$ $\ B=(3,0,0),$ $\ C=(3,3,0),$ $\ D=(0,3,0),$ $\ E=(0,0,6).$ Then $\begin{gathered} P=(2,0,2), \\ Q=(0,2,2), \\ R=(1,1,4). \end{gathered}$

So $\vec{PQ}=(-2,2,0)$ and $\vec{PR}=(-1,1,2),$ giving $\vec{PQ}\times\vec{PR}=(4,4,0).$

The area is $\dfrac12\bigl|(4,4,0)\bigr|=\dfrac12\cdot4\sqrt2=2\sqrt2.$

Thus, C is the correct answer.

2019 AMC 12B Problem 18

Answer: C

Difficulty rating: 1620

Solution:

Problem 18 in Other Years