2009 AMC 10B Problems/Problem 4

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?

$[asy] unitsize(2mm); defaultpen(linewidth(.8pt)); fill((0,0)--(0,5)--(5,5)--cycle,gray); fill((25,0)--(25,5)--(20,5)--cycle,gray); draw((0,0)--(0,5)--(25,5)--(25,0)--cycle); draw((0,0)--(5,5)); draw((20,5)--(25,0)); [/asy]$

$\mathrm{(A)}\frac 18\qquad \mathrm{(B)}\frac 16\qquad \mathrm{(C)}\frac 15\qquad \mathrm{(D)}\frac 14\qquad \mathrm{(E)}\frac 13$

Solution

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 $\mathrm{(C)}$ .

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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2009 AMC 10B Problems/Problem 4

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