$Q$ is the point of intersection of the diagonals of one face of a cube whose edges have length 2 units. The length of $Q R$ is
(A) 2
(B) $\sqrt{8}$
(C) $\sqrt{5}$
(D) $\sqrt{12}$
(E) $\sqrt{6}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-149.jpg?height=415&width=401&top_left_y=123&top_left_x=1339)
$Q$ is the point of intersection of the diagonals of one face of a cube whose edges have length 2 units. The length of $Q R$ is
(A) 2
(B) $\sqrt{8}$
(C) $\sqrt{5}$
(D)
(E) $\sqrt{6}$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-305.jpg?height=461&width=455&top_left_y=2008&top_left_x=1301)

## Solution

Label points $P$ and $S$ as shown. Since each face of the cube is a square with sides of length 2, use the Pythagorean Theorem to find the length of diagonal $P S$.

$$
\begin{aligned}
P S^{2} & =2^{2}+2^{2} \\
& =8 \\
P S & =2 \sqrt{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-306.jpg?height=501&width=482&top_left_y=300&top_left_x=1301)

Then $Q S$ has length $\sqrt{2}$, as $Q$ is the midpoint of diagonal $P S$.

Because we are working with a cube, $\angle Q S R=90^{\circ}$ and $\triangle Q R S$ is a right - angled triangle. Use the Pythagorean Theorem in $\triangle Q R S$ to get

$$
\begin{aligned}
Q R^{2} & =2^{2}+(\sqrt{2})^{2} \\
& =6 \\
Q R & =\sqrt{6}
\end{aligned}
$$

ANSWER: (E)