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Difference between revisions of "2009 AMC 10B Problems/Problem 4"

(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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</asy></center>
 
</asy></center>
  
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<math>\mathrm{(A)}\frac 18\qquad
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<math>\mathrm{(A)}\frac {1}{8}\qquad
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\mathrm{(B)}\frac 16\qquad
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\mathrm{(B)}\frac {1}{6}\qquad
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\mathrm{(C)}\frac 15\qquad
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\mathrm{(C)}\frac {1}{5}\qquad
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\mathrm{(D)}\frac 14\qquad
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\mathrm{(D)}\frac {1}{4}\qquad
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\mathrm{(E)}\frac 13</math>
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\mathrm{(E)}\frac {1}{3}</math>
  
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== Solution ==
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== Solution 1==
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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 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 <math>\frac {25}{125} = \boxed{\frac15}</math>.  The answer is <math>\mathrm{(C)}</math>.
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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>.
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== 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>.
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~hastapasta
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2009|ab=B|num-b=3|num-a=5}}
 
{{AMC10 box|year=2009|ab=B|num-b=3|num-a=5}}
 
{{AMC12 box|year=2009|ab=B|num-b=3|num-a=5}}
 
{{AMC12 box|year=2009|ab=B|num-b=3|num-a=5}}
 +
{{MAA Notice}}

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.

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 $15$ and $25$ 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]

$\mathrm{(A)}\frac {1}{8}\qquad \mathrm{(B)}\frac {1}{6}\qquad \mathrm{(C)}\frac {1}{5}\qquad \mathrm{(D)}\frac {1}{4}\qquad \mathrm{(E)}\frac {1}{3}$

Solution 1

Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ 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 $\frac {25}{125} = \boxed{\frac15}$. The answer is $\mathrm{(C)}$.

Solution 2

The length of each triangle is $\frac{25-15}{2}=5$ 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 $5*5=25$ and the total area, which is $5*25=125$. Alternatively we can use side length ratios, but each way we get $\frac{25}{125}=\frac{1}{5}$. Select $\mathrm{(C)}$.

~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

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