2009 AMC 10B Problem 4

Below is the professionally curated solution for Problem 4 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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Concepts:triangle areatrapezoidarea ratio

Difficulty rating: 960

4.

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 1515 and 2525 meters. What fraction of the yard is occupied by the flower beds?

18\dfrac18

16\dfrac16

15\dfrac15

14\dfrac14

13\dfrac13

Solution:

The two parallel sides differ by 2515=10,25-15=10, split evenly between the two triangles, so each isosceles right triangle has legs of length 55 and area 1252=252.\dfrac12\cdot5^2=\dfrac{25}{2}.

Together the beds cover 2525 square meters. The rectangle has length 2525 and width 5,5, so area 125.125. The fraction is 25125=15.\dfrac{25}{125}=\dfrac15.

Thus, the correct answer is C.

Problem 4 in Other Years