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Difference between revisions of "2006 AIME I Problems/Problem 1"
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== Solution ==
 
== Solution ==
−
Using the Pythagorean Theorem:
+
From the problem statement, we construct the following diagram: <center>[[Image:Aime06i.1.PNG]]</center>
  
−
<math> (AD)^2 = (AC)^2 + (CD)^2 </math>
+
Using the [[Pythagorean Theorem]]:
  
−
<math> (AC)^2 = (AB)^2 + (BC)^2 </math>
+
<center><math> (AD)^2 = (AC)^2 + (CD)^2 </math></center>
 +
 
 +
<center><math> (AC)^2 = (AB)^2 + (BC)^2 </math></center>
  
 
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:  
 
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:  
  
−
<math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math>
+
<center><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></center>
  
 
Plugging in the given information:  
 
Plugging in the given information:  
  
−
<math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math>
+
<center><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></center>
  
−
<math> (AD)^2 = 961 </math>
+
<center><math> (AD)^2 = 961 </math></center>
  
−
<math> (AD)= 31 </math>
+
<center><math> (AD)= 31 </math></center>
  
 
So the perimeter is:  
 
So the perimeter is:  
Line 26: Line 28:
 
== See also ==
 
== See also ==
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 16:58, 18 July 2006

Problem

In quadrilateral $ABCD , \angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}, AB=18, BC=21,$ and $CD=14.$ Find the perimeter of $ABCD.$

Solution

From the problem statement, we construct the following diagram:

Aime06i.1.PNG

Using the Pythagorean Theorem:

$(AD)^2 = (AC)^2 + (CD)^2$
$(AC)^2 = (AB)^2 + (BC)^2$

Substituting $(AB)^2 + (BC)^2$ for $(AC)^2$:

$(AD)^2 = (AB)^2 + (BC)^2 + (CD)^2$

Plugging in the given information:

$(AD)^2 = (18)^2 + (21)^2 + (14)^2$
$(AD)^2 = 961$
$(AD)= 31$

So the perimeter is: $18+21+14+31=084$

See also

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