Difference between revisions of "2006 AIME I Problems/Problem 1"
(→Solution) |
m (→Solution) |
||
(One intermediate revision by one other user not shown) | |||
Line 27: | Line 27: | ||
<div style="text-align:center"><math> (AD)= 31 </math></div> | <div style="text-align:center"><math> (AD)= 31 </math></div> | ||
− | So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{ | + | So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{084}</math>. |
== See also == | == See also == | ||
Line 33: | Line 33: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:43, 2 February 2014
Problem
In quadrilateral , is a right angle, diagonal is perpendicular to , , , and . Find the perimeter of .
Solution
From the problem statement, we construct the following diagram:
Using the Pythagorean Theorem:
Substituting for :
Plugging in the given information:
So the perimeter is , and the answer is .
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.