2008 AMC 10A Problems/Problem 21
Problem
A cube with side length is sliced by a plane that passes through two diagonally opposite vertices
and
and the midpoints
and
of two opposite edges not containing
or
, as shown. What is the area of quadrilateral
?
Solution
![[asy] import three; unitsize(3cm); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; triple A=(0.5,0,0),C=(0,1,1),D=(0.5,1,0.5),B=(0,0,0.5); draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4")); draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1)); draw((0.5,1,0)--(0,1,0)--(0,1,1)); dot((0.5,0,0)); label("$A$",A,WSW); dot((0,1,1)); label("$C$",C,NNE); dot((0.5,1,0.5)); label("$D$",D,ESE); dot((0,0,0.5)); label("$B$",B,NNW); draw(B--C--A--B--D,linetype("4 4")); draw(A--D--C); [/asy]](http://latex.artofproblemsolving.com/f/3/4/f34f547fc1686f55e873a8878bf4656915e53829.png)
Since , it follows that
is a rhombus. The area of the rhombus can be computed by the formula
, where
are the diagonals of the rhombus (or of a kite in general).
has the same length as a face diagonal, or
.
is a space diagonal, with length
. Thus
.
See also
2008 AMC 10A (Problems • Answer Key • 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 | ||
All AMC 10 Problems and Solutions |
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