2008 AMC 10A Problems/Problem 21

From AoPSWiki

Problem

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 $ABCD$ ?

$\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qq...$

Solution

import three;unitsize(3cm);defaultpen(fontsize(8)+linewidth(0.7));currentprojection=obliqueX;pair A=(0.5,0,0),C=(0,1,1),D=(0....

Since $AB = AD = CB = CD = \sqrt{.5^2+1^2}$ , it follows that $ABCD$ is a rhombus. The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$ , where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^2 + 1^2} = \sqrt{2}$ . $AC$ is a space diagonal, with length $\sqrt{1^2+1^2+1^2} = \sqrt{3}$ . Thus $A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}$ .

2008 AMC 10A (Problems • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Personal tools

Navigation

Search

Toolbox

2008 AMC 10A Problems/Problem 21

From AoPSWiki

Problem

Solution

See also