Difference between revisions of "2008 AMC 10A Problems/Problem 21"
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− | Since <math>AB = AD = CB = CD = \sqrt{ | + | 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 19:48, 25 January 2015
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 |
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