2016 AMC 8 Problems/Problem 22
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, 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 is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution 2
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
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!
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