Difference between revisions of "2009 AMC 10A Problems/Problem 17"
(→Solution 1) |
(→Solution 1) |
||
Line 16: | Line 16: | ||
== Solution == | == Solution == | ||
− | === Solution 1 === | + | === [[Solution 1]] === |
The situation is shown in the picture below. | The situation is shown in the picture below. |
Revision as of 11:21, 18 February 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.
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 |