Difference between revisions of "2006 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
− | In | + | 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 == | ||
+ | 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]]: | ||
− | + | <div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div> | |
− | |||
− | + | <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>: | |
− | + | <div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div> | |
− | |||
− | + | Plugging in the given information: | |
− | <math> | + | <div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div> |
− | <math> | + | <div style="text-align:center"><math> (AD)^2 = 961 </math></div> |
− | <math> | + | <div style="text-align:center"><math> (AD)= 31 </math></div> |
− | <math> | + | So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{084}</math>. |
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== See also == | == See also == | ||
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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:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.