Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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The area of the trapezoid containing the shaded region and the two isosceles triangles is <math>\frac{1+3}2\cdot 4=8</math>. Next we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio, so the height of the bigger one is 3, while the height of the smaller one is 1. 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>. The answer is (C). | The area of the trapezoid containing the shaded region and the two isosceles triangles is <math>\frac{1+3}2\cdot 4=8</math>. Next we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio, so the height of the bigger one is 3, while the height of the smaller one is 1. 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>. The answer is (C). | ||
Revision as of 11:35, 23 November 2016
Rectangle below is a rectangle with . What is the area of the "bat wings" (shaded area)?
Solution
The area of the trapezoid containing the shaded region and the two isosceles triangles 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 3, while the height of the smaller one is 1. Thus, their areas are and . Subtracting these areas from the trapezoid, we get . The answer is (C).
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 |
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