2008 AMC 10A Problem 21

Attempt Problem 21 of the 2008 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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21.

A cube with side length 11 is sliced by a plane that passes through two diagonally opposite vertices AA and CC and the midpoints BB and DD of two opposite edges not containing AA or C,C, as shown. What is the area of quadrilateral ABCD?ABCD?

62\dfrac{\sqrt{6}}{2}

54\dfrac{5}{4}

2\sqrt{2}

32\dfrac{3}{2}

3\sqrt{3}

Answer: A
Concepts:cube geometryrhombusdiagonal
Difficulty rating: 1770
Solution:

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

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

The area of a rhombus is half the product of its diagonals: 1232=62.\dfrac{1}{2}\cdot\sqrt{3}\cdot\sqrt{2} = \dfrac{\sqrt{6}}{2}.

Thus, the correct answer is A.

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