Difference between revisions of "2006 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
− | In [[quadrilateral]] <math> ABCD , \angle B </math> is a right angle, diagonal <math> \overline{AC} </math> is [[perpendicular]] to <math> \overline{CD}, | + | In [[quadrilateral]] <math> ABCD</math>, <math>\angle B </math> is a [[right angle]], [[diagonal]] <math>\overline{AC}</math> is [[perpendicular]] to <math>\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]] of <math>ABCD</math>. |
== Solution == | == Solution == |
Revision as of 19:52, 25 April 2008
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:
![[asy] pointpen = black; pathpen = black + linewidth(0.65); pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); [/asy]](http://latex.artofproblemsolving.com/f/e/5/fe5ba5b6cc93f6a188115c7f4281a190091a9844.png)
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 |