Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math> | Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math> | ||
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Revision as of 21:26, 8 January 2018
Rectangle below is a rectangle with . What is the area of the "bat wings" (shaded area)?
Solution
The area of trapezoid is . Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a ratio[SOMEBODY PROVE THIS], so the height of the larger one is while the height of the smaller one is Thus, their areas are and . Subtracting these areas from the trapezoid, we get . Therefore, the answer to this problem is
Solution 2
Setting coordinates!
Let ,
Now, we easily discover that line has lattice coordinates at and . Hence, the slope of line
Plugging in the rest of the coordinate points, we find that line
Doing the same process to line , we find that line .
Hence, setting them equal to find the intersection point...
.
Hence, we find that the intersection point is . Call it Z.
Now, we can see that
.
Shoelace!
Using the well known Shoelace Formula(https://en.m.wikipedia.org/wiki/Shoelace_formula), we find that the area of one of those small shaded triangles is .
Now because there are two of them, we multiple that area by to get
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
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