Difference between revisions of "2009 AMC 10A Problems/Problem 17"
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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>. | 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>. | ||
+ | |||
+ | Solution 3 | ||
+ | There is a better solution where we find EF directly instead of in parts (use similarity). The strategy is similar. | ||
== See Also == | == See Also == |
Revision as of 12:36, 4 January 2017
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 |
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