Difference between revisions of "2009 AMC 10A Problems/Problem 17"
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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>. | 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> | + | Looking at the angles, we see that triangle <math>BDE</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 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>. |
Revision as of 20:21, 30 October 2017
Contents
[hide]Problem
Rectangle has
and
. Segment
is constructed through
so that
is perpendicular to
, and
and
lie on
and
, respectively. What is
?
Solutions
Solution 1
The situation is shown in the picture below.
From the Pythagorean theorem we have .
Triangle is similar to
, as they have the same angles. Segment
is perpendicular to
, meaning that angle
and
are right angles and congruent. Also, angle
is a right angle. Because it is a rectangle, angle
is congruent to
and angle
is also a right angle. By the transitive property:
Next, because every triangle has a degree measure of 180, angle and angle
are congruent.
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
.
.
Solution 3 There is a better solution where we find EF directly instead of in parts (use similarity). The strategy is similar.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.