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 == | ||
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[[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:
![[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:
![$(AD)^2 = (AC)^2 + (CD)^2$](http://latex.artofproblemsolving.com/a/7/d/a7d216f23824f0248399bfcf94327646da38392c.png)
![$(AC)^2 = (AB)^2 + (BC)^2$](http://latex.artofproblemsolving.com/0/1/9/0195b55e6d09d7d778973a0c221c3ab651e6db7b.png)
Substituting for
:
![$(AD)^2 = (AB)^2 + (BC)^2 + (CD)^2$](http://latex.artofproblemsolving.com/8/8/7/88701f8a91bf212b3086977771394589ed016aed.png)
Plugging in the given information:
![$(AD)^2 = (18)^2 + (21)^2 + (14)^2$](http://latex.artofproblemsolving.com/f/1/c/f1cdddae8be3cb267c3926faed6cb19a2ee5a96b.png)
![$(AD)^2 = 961$](http://latex.artofproblemsolving.com/0/7/7/0772ff8d7eb763ef832decbf901937b9b64b3181.png)
![$(AD)= 31$](http://latex.artofproblemsolving.com/c/8/7/c87f29dea59845f942599cc4924dc064152c154b.png)
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.