Difference between revisions of "2009 AMC 10B Problems/Problem 4"
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\mathrm{(E)}\frac {1}{3}</math> | \mathrm{(E)}\frac {1}{3}</math> | ||
| − | == Solution == | + | == Solution 1== |
Each triangle has leg length <math>\frac 12 \cdot (25 - 15) = 5</math> meters and area <math>\frac 12 \cdot 5^2 = \frac {25}{2}</math> square meters. Thus the flower beds have a total area of <math>25</math> square meters. The entire yard has length <math>25</math> m and width <math>5</math> m, so its area is <math>125</math> square meters. The fraction of the yard occupied by the flower beds is <math>\frac {25}{125} = \boxed{\frac15}</math>. The answer is <math>\mathrm{(C)}</math>. | Each triangle has leg length <math>\frac 12 \cdot (25 - 15) = 5</math> meters and area <math>\frac 12 \cdot 5^2 = \frac {25}{2}</math> square meters. Thus the flower beds have a total area of <math>25</math> square meters. The entire yard has length <math>25</math> m and width <math>5</math> m, so its area is <math>125</math> square meters. The fraction of the yard occupied by the flower beds is <math>\frac {25}{125} = \boxed{\frac15}</math>. The answer is <math>\mathrm{(C)}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | The length of each triangle is <math>\frac{25-15}{2}=5</math> meters. By translating then rotating the top right triangle so that it forms a square with the top left triangle, the ratio we desire is literally the ratio of the <math>5*5=25</math> and the total area, which is <math>5*25=125</math>. Alternatively we can use side length ratios, but each way we get <math>\frac{25}{125}=\frac{1}{5}</math>. Select <math>\mathrm{(C)}</math>. | ||
| + | |||
| + | ~hastapasta | ||
== See also == | == See also == | ||
Latest revision as of 17:07, 30 November 2022
- The following problem is from both the 2009 AMC 10B #4 and 2009 AMC 12B #4, so both problems redirect to this page.
Contents
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?
![[asy] unitsize(2mm); defaultpen(linewidth(.8pt)); fill((0,0)--(0,5)--(5,5)--cycle,gray); fill((25,0)--(25,5)--(20,5)--cycle,gray); draw((0,0)--(0,5)--(25,5)--(25,0)--cycle); draw((0,0)--(5,5)); draw((20,5)--(25,0)); [/asy]](http://latex.artofproblemsolving.com/3/e/e/3ee29f78e4a8eecdb8faec4092f2f1c361a869e4.png)
Solution 1
Each triangle has leg length 
meters and area 
square meters. Thus the flower beds have a total area of square meters. The entire yard has length m and width 
m, so its area is 
square meters. The fraction of the yard occupied by the flower beds is 
. The answer is 
.
Solution 2
The length of each triangle is 
meters. By translating then rotating the top right triangle so that it forms a square with the top left triangle, the ratio we desire is literally the ratio of the 
and the total area, which is 
. Alternatively we can use side length ratios, but each way we get 
. Select .
~hastapasta
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
