Question

A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C,$ as shown. What is the area of quadrilateral $ABCD?$

$\dfrac{\sqrt{6}}{2}$

$\dfrac{5}{4}$

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{3}$

Solution:

Each side of $ABCD$ joins a vertex of the cube to the midpoint of an edge, so all four sides are equal and $ABCD$ is a rhombus.

Its diagonals are the space diagonal $AC = \sqrt{3}$ and the face diagonal $BD = \sqrt{2}.$

The area of a rhombus is half the product of its diagonals: $\dfrac{1}{2}\cdot\sqrt{3}\cdot\sqrt{2} = \dfrac{\sqrt{6}}{2}.$

Thus, the correct answer is A.

2008 AMC 10A Problem 21

Solution:

Problem 21 in Other Years