Difference between revisions of "2009 AMC 10B Problems/Problem 4"
VelaDabant (talk | contribs) (New page: {{duplicate|2009 AMC 10B #4 and 2009 AMC 12B #4}} == Problem == A rectangular yard contains two flower beds in the shape of congruent i...) |
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− | <math>\mathrm{(A)}\frac | + | <math>\mathrm{(A)}\frac {1}{8}\qquad |
− | \mathrm{(B)}\frac | + | \mathrm{(B)}\frac {1}{6}\qquad |
− | \mathrm{(C)}\frac | + | \mathrm{(C)}\frac {1}{5}\qquad |
− | \mathrm{(D)}\frac | + | \mathrm{(D)}\frac {1}{4}\qquad |
− | \mathrm{(E)}\frac | + | \mathrm{(E)}\frac {1}{3}</math> |
== Solution == | == Solution == |
Revision as of 10:29, 12 April 2009
- The following problem is from both the 2009 AMC 10B #4 and 2009 AMC 12B #4, so both problems redirect to this page.
Problem
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths and meters. What fraction of the yard is occupied by the flower beds?
Solution
Each triangle has leg length meters and area square meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 m and width 5 m, so its area is 125 square meters. The fraction of the yard occupied by the flower beds is . The answer is .
See also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 AMC 10 Problems and Solutions |
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 AMC 12 Problems and Solutions |