2001 AMC 10 Problems/Problem 24: Difference between revisions

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 $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2001 AMC 10 Problems/Problem 24: Difference between revisions

Revision as of 11:08, 1 December 2015

Problem

Solution

See Also

@@ Line 8: / Line 8: @@
 ==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>.