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Difference between revisions of "2006 AIME I Problems/Problem 1"

m (→‎See also: box)
m (→‎Solution: div align=center)
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== Solution ==
 
== Solution ==
−
From the problem statement, we construct the following diagram: <center>[[Image:Aime06i.1.PNG]]</center>
+
From the problem statement, we construct the following diagram: <div style="text-align:center">[[Image:Aime06i.1.PNG]]</div>
  
 
Using the [[Pythagorean Theorem]]:
 
Using the [[Pythagorean Theorem]]:
  
−
<center><math> (AD)^2 = (AC)^2 + (CD)^2 </math></center>
+
<div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div>
  
−
<center><math> (AC)^2 = (AB)^2 + (BC)^2 </math></center>
+
<div style="text-align:center"><math> (AC)^2 = (AB)^2 + (BC)^2 </math></div>
  
 
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:  
 
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:  
  
−
<center><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></center>
+
<div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div>
  
 
Plugging in the given information:  
 
Plugging in the given information:  
  
−
<center><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></center>
+
<div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div>
  
−
<center><math> (AD)^2 = 961 </math></center>
+
<div style="text-align:center"><math> (AD)^2 = 961 </math></div>
  
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<center><math> (AD)= 31 </math></center>
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<div style="text-align:center"><math> (AD)= 31 </math></div>
  
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So the perimeter is:
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So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>084</math>.
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<math> 18+21+14+31=84 </math>
 
−
 
 
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The answer is 084.
 
  
 
== See also ==
 
== See also ==

Revision as of 20:24, 11 March 2007

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 $18+21+14+31=84$, and the answer is $084$.

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
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