Difference between revisions of "2001 AMC 10 Problems/Problem 24"

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==Solution==
 
==Solution==
  
If <math> AB=x </math> and <math> CD=y </math>, we have <math> BC=x+y </math>.
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If <math> AB=x </math> and <math> CD=y </math>,then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>.
 
 
By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math>
 
 
 
Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>.
 
  
 
==See Also==
 
==See Also==

Revision as of 12:05, 2 August 2015

Problem

In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB+CD=BC$, $AB<CD$, and $AD=7$. What is $AB\cdot CD$?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$

Solution

If $AB=x$ and $CD=y$,then $BC=x+y$. By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49$ Solving the equation, we get $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$.

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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