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Difference between revisions of "2008 AMC 10A Problems/Problem 21"
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draw(A--D--C);
 
draw(A--D--C);
 
</asy></center>
 
</asy></center>
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Since <math>AB = AD = CB = CD = \sqrt{.5^2+1^2}</math>, it follows that <math>ABCD</math> is a [[rhombus]]. The area of the rhombus can be computed by the formula <math>A = \frac 12 d_1d_2</math>, where <math>d_1,\,d_2</math> are the diagonals of the rhombus (or of a [[kite]] in general). <math>BD</math> has the same length as a face diagonal, or <math>\sqrt{1^2 + 1^2} = \sqrt{2}</math>. <math>AC</math> is a space diagonal, with length <math>\sqrt{1^2+1^2+1^2} = \sqrt{3}</math>. Thus <math>A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}</math>.
+
Since <math>AB = AD = CB = CD = \sqrt{\frac{1}{2}^2+1^2}</math>, it follows that <math>ABCD</math> is a [[rhombus]]. The area of the rhombus can be computed by the formula <math>A = \frac 12 d_1d_2</math>, where <math>d_1,\,d_2</math> are the diagonals of the rhombus (or of a [[kite]] in general). <math>BD</math> has the same length as a face diagonal, or <math>\sqrt{1^2 + 1^2} = \sqrt{2}</math>. <math>AC</math> is a space diagonal, with length <math>\sqrt{1^2+1^2+1^2} = \sqrt{3}</math>. Thus <math>A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}</math>.
  
 
==See also==
 
==See also==

Revision as of 18:48, 25 January 2015

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 $ABCD$?

[asy] import three; unitsize(3cm); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; 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("$A$",(0.5,0,0),WSW); dot((0,1,1)); label("$C$",(0,1,1),NE); dot((0.5,1,0.5)); label("$D$",(0.5,1,0.5),ESE); dot((0,0,0.5)); label("$B$",(0,0,0.5),NW);[/asy]

$\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}$

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]

Since $AB = AD = CB = CD = \sqrt{\frac{1}{2}^2+1^2}$, 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 $A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}$.

See also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AMC 10 Problems and Solutions

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