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

(Created page with '==Problem== In trapezoid <math> ABCD </math>, <math> \overline{AB} </math> and <math> \overline{CD} </math> are perpendicular to <math> \overline{AD} </math>, with <math> AB+CD=…')
 
(→‎Solution)
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If <math> AB=x </math> and <math> CD=y </math>, we have <math> BC=x+y </math>.
 
If <math> AB=x </math> and <math> CD=y </math>, we have <math> BC=x+y </math>.
  
−
by the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </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>.
+
Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>.
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2001|num-b=23|num-a=25}}
 
{{AMC10 box|year=2001|num-b=23|num-a=25}}

Revision as of 17:55, 1 January 2013

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 $CD=y$, we have $BC=x+y$.

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 (Problems • Answer Key • Resources)
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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