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2006 AIME I Problems/Problem 1: Difference between revisions

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


== Solution ==
== Solution ==
From the problem statement, we construct the following diagram: <div style="text-align:center">[[Image:Aime06i.1.PNG]]</div>
From the problem statement, we construct the following diagram:  
 
<center><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></center><!-- Asymptote replacement for Image:Aime06i.1.PNG by joml88 -->
Using the [[Pythagorean Theorem]]:
Using the [[Pythagorean Theorem]]:


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<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>084</math>.
So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{084}</math>.


== See also ==
== See also ==

Revision as of 18:43, 25 April 2008

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:

[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]

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 $\boxed{084}$.

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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