Difference between revisions of "2016 AMC 8 Problems/Problem 22"
m (→Solution) |
Reaganchoi (talk | contribs) m |
||
Line 11: | Line 11: | ||
label("$E$", (0,-0.2)); | label("$E$", (0,-0.2)); | ||
label("$F$", (3,-0.2)); | label("$F$", (3,-0.2)); | ||
+ | label("$1$", (0.5, 4), N); | ||
+ | label("$1$", (1.5, 4), N); | ||
+ | label("$1$", (2.5, 4), N); | ||
+ | label("$4$", (3.2, 2), E); | ||
</asy> | </asy> | ||
Revision as of 14:50, 28 November 2016
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 bigger 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.