Difference between revisions of "2010 AMC 8 Problems/Problem 17"
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Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
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==Solution 3== | ==Solution 3== |
Latest revision as of 21:04, 23 December 2023
Contents
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
~kempwood
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
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=937
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
Video Solution by WhyMath
~savannahsolver
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.