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

Revision as of 12:53, 25 September 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

Let the side length be called $x$ .

Then $AB=BC=CD=DE=EF=AF=x$ .

The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$ .

Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$ , and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$

Then we have to solve the equation

$2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2$ .

$2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)$

$2116=x^2$

$x=46$

Therefore, AB is 46.

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 == Problem ==
-In convex hexagon <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are right angles, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are congruent. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</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 ==

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

Revision as of 12:53, 25 September 2007

Problem

Solution

See also