Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
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The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math> | The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math> | ||
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==Video Solutions== | ==Video Solutions== | ||
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*https://youtu.be/q3MAXwNBkcg ~savannahsolver | *https://youtu.be/q3MAXwNBkcg ~savannahsolver | ||
Latest revision as of 16:22, 31 August 2024
Contents
Problem
Rectangle below is a
rectangle with
. The area of the "bat wings" (shaded area) is
Solution 1
The area of trapezoid is
. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a
ratio by AA similarity (alternate interior and vertical angles) so the height of the larger 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
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solutions
- https://youtu.be/q3MAXwNBkcg ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14
See Also
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.