Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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<math>\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}</math> | <math>\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}</math> | ||
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==Note== | ==Note== |
Revision as of 14:51, 28 October 2020
Problem 24
In the cube with opposite vertices and and are the midpoints of segments 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
Note
In the 2008 AMC 10A, Question 21 was nearly identical to this question, except that in this question you have to look for the square of the area, not the actual area.
Video Solution
https://www.youtube.com/watch?v=04pV_rZw8bg - Happytwin
Video Solution
https://www.youtube.com/watch?v=ji9_6XNxyIc ~ MathEx
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.