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

m (→‎Solution)
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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>.
  
 +
==Solution 2==
 +
Simpler is just drawing the trapezoid and then using what is given to solve.
 +
Draw a line perpendicular to AD that connects the longer side to the corner of the shorter side. Name the bottom part x and top part a.
 +
By the Pythagorean theorem, it is obvious that <math>a^{2} + 49 = (2x+a)^{2}</math> (the RHS is the fact the two sides added together equals that). Then, we get <math>a^2 + 49 = 4x^2 + 4ax + a^2</math>, cancel out and factor and we get <math>49 = 4x(x+a)</math>. Notice that <math>x(x+a)</math> is what the question asks, so the answer is <math>\boxed{\frac{49}{4}}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:13, 8 January 2019

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(7cm); 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 $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$.

Solution 2

Simpler is just drawing the trapezoid and then using what is given to solve. Draw a line perpendicular to AD that connects the longer side to the corner of the shorter side. Name the bottom part x and top part a. By the Pythagorean theorem, it is obvious that $a^{2} + 49 = (2x+a)^{2}$ (the RHS is the fact the two sides added together equals that). Then, we get $a^2 + 49 = 4x^2 + 4ax + a^2$, cancel out and factor and we get $49 = 4x(x+a)$. Notice that $x(x+a)$ is what the question asks, so the answer is $\boxed{\frac{49}{4}}$

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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