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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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| − | Let the side length be called <math>x</math>.
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| − | [[Image:Diagram1.png]]
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| − | Then <math>AB=BC=CD=DE=EF=AF=x</math>.
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| − | The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>.
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| − | Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,
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| − | and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
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| − | Then we have to solve the equation
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| − | <math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
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| − | <math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
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| − | <math>2116=x^2</math>
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| − | <math>x=46</math>
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| − | Therefore, AB is 46.
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| − | == See also ==
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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Geometry Problems]]
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