$A B C D E F G H$ 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_e85ef69091c57cb633f3g-135.jpg?height=391&width=396&top_left_y=352&top_left_x=1336)
$A B C D E F G H$ 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) 6
(D) $\sqrt{2}$
(E) $\sqrt{32}$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-244.jpg?height=434&width=458&top_left_y=1621&top_left_x=1430)

## 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_688a3b42f69739f2e244g-245.jpg?height=361&width=309&top_left_y=313&top_left_x=1491)

ANSWER: (A)