Difference between revisions of "2018 AMC 8 Problems/Problem 24"
Mutinykids (talk | contribs) (→Solution) |
(→Note) |
||
Line 39: | Line 39: | ||
==Note== | ==Note== | ||
− | In the 2008 AMC 10A, Question 21 was nearly identical to this question. | + | In the 2008 AMC 10A, Question 21 was nearly identical to this question, its the same but for this question, you have to look for the square not the real thing. |
==See Also== | ==See Also== |
Revision as of 15:10, 23 November 2018
Contents
[hide]Problem 24
In the cube with opposite vertices and and are the midpoints of edges and respectively. Let be the ratio of the area of the cross-section to the area of one of the faces of the cube. What is
Solution
Note that is a rhombus. Let the side length of the cube be . By the Pythagorean theorem, and . Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is . . Thus
Note
In the 2008 AMC 10A, Question 21 was nearly identical to this question, its the same but for this question, you have to look for the square not the real thing.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.