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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 1]] |
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| − | 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>
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| − | == Solution ==
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| − | From the problem statement, we construct the following diagram: <div style="text-align:center">[[Image:Aime06i.1.PNG]]</div>
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| − | Using the [[Pythagorean Theorem]]:
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| − | <div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div>
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| − | <div style="text-align:center"><math> (AC)^2 = (AB)^2 + (BC)^2 </math></div>
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| − | Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:
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| − | <div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div>
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| − | Plugging in the given information:
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| − | <div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div>
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| − | <div style="text-align:center"><math> (AD)^2 = 961 </math></div>
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| − | <div style="text-align:center"><math> (AD)= 31 </math></div>
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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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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|before=First Problem|num-a=2}}
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| − | [[Category:Intermediate Geometry Problems]]
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