Difference between revisions of "2016 AMC 8 Problems/Problem 22"
(→Solution) |
|||
Line 23: | Line 23: | ||
{{AMC8 box|year=2016|num-b=21|num-a=23}} | {{AMC8 box|year=2016|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>E=(0,0)</math>, <math>F=(3,0)</math> | ||
+ | |||
+ | Now, we easily discover that line <math>CF</math> has lattice coordinates at <math>(1,4)</math> and <math>(3,0)</math>. Hence, the slope of line <math>CF=-2x</math> | ||
+ | |||
+ | Plugging in the rest of the coordinate points, we find that <math>CF=-2x+6</math> | ||
+ | |||
+ | Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>. | ||
+ | |||
+ | Hence, setting them equal to find the intersection point... | ||
+ | |||
+ | <math>2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}</math>. | ||
+ | |||
+ | Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z. | ||
+ | |||
+ | Now, we can see that | ||
+ | |||
+ | <math>E=(0,0)</math> | ||
+ | |||
+ | <math>Z=(\frac{3}{2},3)</math> | ||
+ | |||
+ | <math>C=(1,4)</math>. | ||
+ | |||
+ | Shoelace! | ||
+ | |||
+ | 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> |
Revision as of 11:13, 13 August 2017
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