2018 AMC 8 Problems/Problem 24
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
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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