Difference between revisions of "2009 AMC 10A Problems/Problem 17"
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Obviously, from the [[Pythagorean theorem]] we have <math>BD=5</math>. | Obviously, from the [[Pythagorean theorem]] we have <math>BD=5</math>. | ||
− | Triangle <math>EAB</math> is similar to <math> | + | Triangle <math>EAB</math> is similar to <math>DAB</math>, as they have the same angles. Segment <math>BA</math> is perpendicular to <math>DA</math>, meaning that angle <math>DAB</math> and <math>BAE</math> are right angles and congruent. Also, angle <math>DBE</math> is a right angle. Because it is a rectangle, angle <math>BDC</math> is congruent to <math>DBA</math> and angle <math>ADC</math> is also a right angle. By the transitive property: |
+ | <math>mADB + mBDC = mDBA + mABE</math> | ||
+ | <math>mBDC = mDBA</math> | ||
+ | <math>mADB + mBDC = mBDC + mABE</math> | ||
+ | <math>mADB = mABE</math> | ||
+ | |||
+ | Next, because every triangle has a degree measure of 180, angle <math>E</math> and angle <math>DBA</math> are congruent. | ||
+ | |||
+ | Hence <math>BE/AB = DB/AD</math>, and therefore <math>BE = AB\cdot DB/AD = 20/3</math>. | ||
Also triangle <math>CBF</math> is similar to <math>ABD</math>. Hence <math>BF/BC = DB/AB</math>, and therefore <math>BF=BC\cdot DB / AB = 15/4</math>. | Also triangle <math>CBF</math> is similar to <math>ABD</math>. Hence <math>BF/BC = DB/AB</math>, and therefore <math>BF=BC\cdot DB / AB = 15/4</math>. |
Revision as of 16:12, 3 January 2013
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. 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 . .
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 |