Difference between revisions of "2016 AMC 8 Problems/Problem 22"
(/* Solution *1/) |
m (→Solution 2) |
||
Line 68: | Line 68: | ||
Shoelace! | Shoelace! | ||
− | Using the well known | + | Using the well known [url=https://en.m.wikipedia.org/wiki/Shoelace_formula]Shoelace Formula[/url], we find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>. |
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> | ||
{{AMC8 box|year=2016|num-b=21|num-a=23}} | {{AMC8 box|year=2016|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:32, 10 November 2019
Rectangle below is a rectangle with . What is the area of the "bat wings" (shaded area)?
Solution 1
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 by AA similarity (alternate interior and vertical angles) 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 [url=https://en.m.wikipedia.org/wiki/Shoelace_formula]Shoelace Formula[/url], 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.