Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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<asy> | <asy> | ||
− | draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); | + | draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); |
− | draw((3,0)--(1,4)--(0,0)); | + | draw((3,0)--(1,4)--(0,0)); |
− | fill((0,0)--(1,4)--(1.5,3)--cycle, black); | + | fill((0,0)--(1,4)--(1.5,3)--cycle, black); |
− | fill((3,0)--(2,4)--(1.5,3)--cycle, black); | + | fill((3,0)--(2,4)--(1.5,3)--cycle, black); |
− | label("$A(3,4)$",(3.25,4.2)); | + | label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); |
− | label("$B(2,4)$",(2.1,4.2)); | + | label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); |
− | label("$C(1,4)$",(0.9,4.2)); | + | label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); |
− | label("$D(0,4)$",(-0.3,4.2)); | + | label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); |
− | label("$E(0,0)$", (0,-0.2)); | + | label(scale(0.7)*"$E(0,0)$", (0,-0.2)); |
− | label("$Z(\frac{3}{2},3)$", (1.5,1.8)); | + | label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); |
− | label("$F(3,0)$", (3,-0.2)); | + | label(scale(0.7)*"$F(3,0)$", (3,-0.2)); |
− | label("$1$", (0.3, 4), N); | + | label(scale(0.7)*"$1$", (0.3, 4), N); |
− | label("$1$", (1.5, 4), N); | + | label(scale(0.7)*"$1$", (1.5, 4), N); |
− | label("$1$", (2.7, 4), N); | + | label(scale(0.7)*"$1$", (2.7, 4), N); |
− | label("$4$", (3.2, 2), E); | + | label(scale(0.7)*"$4$", (3.2, 2), E); |
</asy> | </asy> | ||
Revision as of 10:25, 15 September 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 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.