$A B C D E F G H$ is a cube having a side length of 2. $P$ is the midpoint of $E F$, as shown. The area of $\triangle A P B$ is
(A) $\sqrt{8}$
(B) 3
(C) $\sqrt{32}$
(D) $\sqrt{2}$
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-134.jpg?height=387&width=398&top_left_y=1140&top_left_x=1335)
$A B C D E F G H$ is a cube having a side length of 2. $P$ is the midpoint of $E F$, as shown. The area of $\triangle A P B$ is
(A) $\sqrt{8}$
(D) $\sqrt{2}$
(B) 3
(C) $\sqrt{32}$
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-262.jpg?height=420&width=428&top_left_y=516&top_left_x=1445)

## Solution

By symmetry, the lengths of $A P$ and $B P$ will be equal, and $A P=\sqrt{A D^{2}+D E^{2}+E P^{2}}=\sqrt{2^{2}+2^{2}+1^{2}}=3$.

If $M$ is the midpoint of $A B$, then $P M$ is perpendicular to $A B$. By Pythagoras, $M P=\sqrt{3^{2}-1^{2}}=\sqrt{8}$.

So the area of $\triangle A P B$ is

$$
\text { Area }=\frac{1}{2}(2)(\sqrt{8})=\sqrt{8}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-262.jpg?height=361&width=306&top_left_y=1039&top_left_x=1495)

ANSWER: (A)