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

m (Solution: added in asy diagram)
(Solution)
Line 8: Line 8:
  
 
==Solution==
 
==Solution==
[asy]
 
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
 
import graph; size(11.6cm);
 
real labelscalefactor = 0.5; /* changes label-to-point distance */
 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
 
pen dotstyle = black; /* point style */
 
real xmin = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3;  /* image dimensions */
 
  
/* draw figures */
 
draw(circle((0.2,4.92), 1.3));
 
draw(circle((1.04,1.58), 2.14));
 
draw((-1.1,4.92)--(0.2,4.92));
 
draw((0.2,4.92)--(1.04,1.58));
 
draw((1.04,1.58)--(-1.1,1.58));
 
draw((-1.1,1.58)--(-1.1,4.92));
 
/* dots and labels */
 
dot((-1.1,4.92),dotstyle);
 
label("<math>A</math>", (-1.02,5.12), NE * labelscalefactor);
 
dot((0.2,4.92),dotstyle);
 
label("<math>B</math>", (0.28,5.12), NE * labelscalefactor);
 
dot((-1.1,1.58),dotstyle);
 
label("<math>D</math>", (-1.02,1.78), NE * labelscalefactor);
 
dot((1.04,1.58),dotstyle);
 
label("<math>C</math>", (1.12,1.78), NE * labelscalefactor);
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 
/* end of picture */
 
[\asy]
 
 
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>.
 
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>.
  

Revision as of 11:09, 1 December 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png