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

(→‎Solution)
m (→‎Solution: added in asy diagram)
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==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 11:08, 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

[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("$A$", (-1.02,5.12), NE * labelscalefactor); dot((0.2,4.92),dotstyle); label("$B$", (0.28,5.12), NE * labelscalefactor); dot((-1.1,1.58),dotstyle); label("$D$", (-1.02,1.78), NE * labelscalefactor); dot((1.04,1.58),dotstyle); label("$C$", (1.12,1.78), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 

/* end of picture */

[\asy] 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 (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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