Difference between revisions of "2009 AMC 10A Problems/Problem 17"
(New page: == Problem == Rectangle <math>ABCD</math> has <math>AB=4</math> and <math>BC=3</math>. Segment <math>EF</math> is constructed through <math>B</math> so that <math>EF</math> is perpendicula...) |
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
The situation is shown in the picture below. | The situation is shown in the picture below. | ||
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We then have <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | We then have <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | Since <math>BD</math> is the altitude from <math>B</math> to <math>EF</math>, we can use the equation <math>BD^2 = EB\cdot BF</math>. | ||
+ | |||
+ | Looking at the angles, we see that triangle <math>EAB</math> is similar to <math>DCB</math>. Because of this, <math>\frac{AB}{CB} = \frac{EB}{DB}</math>. From the given information and the [[Pythagorean theorem]], <math>AB=4</math>, <math>CB=3</math>, and <math>DB=5</math>. Solving gives <math>EB=20/3</math>. | ||
+ | |||
+ | We can use the above formula to solve for <math>BF</math>. <math>BD^2 = 20/3\cdot BF</math>. Solve to obtain <math>BF=15/4</math>. | ||
+ | |||
+ | We now know <math>EB</math> and <math>BF</math>. <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}} |
Revision as of 00:25, 1 February 2012
Problem
Rectangle has and . Segment is constructed through so that is perpendicular to , and and lie on and , respectively. What is ?
Solution
Solution 1
The situation is shown in the picture below.
Obviously, from the Pythagorean theorem we have .
Triangle is similar to , as they have the same angles. Hence , and therefore .
Also triangle is similar to . Hence , and therefore .
We then have .
Solution 2
Since is the altitude from to , we can use the equation .
Looking at the angles, we see that triangle is similar to . Because of this, . From the given information and the Pythagorean theorem, , , and . Solving gives .
We can use the above formula to solve for . . Solve to obtain .
We now know and . .
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |