2010 AMC 8 Problems/Problem 17
Problem
The diagram shows an octagon consisting of unit squares. The portion below
is a unit square and a triangle with base
. If
bisects the area of the octagon, what is the ratio
?
Solution 1
We see that half the area of the octagon is . We see that the triangle area is
. That means that
.
Meaning,
Solution 2
Like stated in solution 1, we know that half the area of the octagon is .
That means that the area of the trapezoid is .
. Solving for
, we get
.
Subtracting from
, we get
.
Therefore, the answer comes out to
~Hithere22702
Solution 3
We can move the cube on the bottom right to the top left, thus creating a rectangle. We thus know that now this has been separated to 2 triangle, with equal area, although, obviously, one is not a perfect triangle. From this we can subtract 1 from one of the 5's (the area of one of the triangle-shapes-polygons) and add it to the other, since we have just moved the block, original in the triangle-shaped-polygon on the bottom, and thus, we get 4/6, or
-RealityWrites
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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