2001 AMC 10 Problems/Problem 24

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 parallel to $\overline{AD}$ 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 $\boxed{\textbf{(B)}\ 12.25}$ .

Solution by IronicNinja

Solution 3

We know it is a trapezoid and that $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ . If they are perpendicular to $\overline{AD}$ that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know $\overline{AD}$ is $7$ . We can then set the length of $\overline{AB}$ to be $x$ and the length of $\overline{DC}$ to be $y$ . $\overline{BC}$ would then be $x+y$ . Let's draw a straight line down from point $B$ which is perpendicular to $\overline{DC}$ and parallel to $\overline{AD}$ . Let's name this line $M$ . Then let's name the point at which line $M$ intersects $\overline{DC}$ point $E$ . Line $M$ partitions the trapezoid into rectangle $ADEB$ and triangle $BEC$ . We will use the triangle to solve for $x*y$ using the Pythagorean theorem. The line segment $\overline{EC}$ would be $y-x$ because $\overline{DC}$ is $y$ and $\overline{DE}$ is $x$ . $\overline{DE}$ is $x$ because it is parallel to $\overline{AB}$ and both are of equal length. Because of the Pythagorean theorem, we know that $(EC)^2+(BE)^2=(BC)^2$ . Substituting the values we have we get $(y-x)^2+(7)^2=(x+y)^2$ . Simplifying this we get $(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)$ . Now we get rid of the $x^2$ and $y^2$ terms from both sides to get $(-2xy)+(49)=(2xy)$ . Combining like terms we get $(49)=(4xy)$ . Then we divide by $4$ to get $(12.25)=(xy)$ . Now we know that $x*y$ (same thing as $xy$ ) is equal to $12.25$ which is answer choice $\boxed{\textbf{(B)}\ 12.25}$ . Solution By: MATHCOUNTSCMS25

P.S. I Don't Know How To Format It Properly Using Latex So Could Someone Please Fix It

EDIT: Fixed! (As much as my ability can)-Mliu630XYZ

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

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Problem

Solution

Solution 2

Solution 3

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