Difference between revisions of "2008 AMC 10A Problems/Problem 21"
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currentprojection=obliqueX; | 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((0.5,0,0)--(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((0.5,0,1)--(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)); | ||
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label("$A$",A,WSW); | label("$A$",A,WSW); | ||
dot((0,1,1)); | dot((0,1,1)); | ||
− | label("$C$",C, | + | label("$C$",C,NNE); |
dot((0.5,1,0.5)); | dot((0.5,1,0.5)); | ||
label("$D$",D,ESE); | label("$D$",D,ESE); | ||
dot((0,0,0.5)); | dot((0,0,0.5)); | ||
− | label("$B$",B, | + | label("$B$",B,NNW); |
draw(B--C--A--B--D,linetype("4 4")); | draw(B--C--A--B--D,linetype("4 4")); | ||
draw(A--D--C); | draw(A--D--C); | ||
</asy></center> | </asy></center> | ||
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{.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>. | ||
+ | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=20|num-a=22}} | {{AMC10 box|year=2008|ab=A|num-b=20|num-a=22}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 00:33, 25 January 2011
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
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 |