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

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(Solution 3)
 
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==Solution==
 
==Solution==
  
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>.
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<asy>
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/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
 +
import graph; size(7cm);
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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));
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draw((-1.1,4.92)--(0.2,4.92));
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draw((0.2,4.92)--(1.04,1.58));
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draw((1.04,1.58)--(-1.1,1.58));
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draw((-1.1,1.58)--(-1.1,4.92));
 +
/* dots and labels */
 +
dot((-1.1,4.92),dotstyle);
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label("$A$", (-1.02,5.12), NE * labelscalefactor);
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dot((0.2,4.92),dotstyle);
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label("$B$", (0.28,5.12), NE * labelscalefactor);
 +
dot((-1.1,1.58),dotstyle);
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label("$D$", (-1.02,1.78), NE * labelscalefactor);
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dot((1.04,1.58),dotstyle);
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label("$C$", (1.12,1.78), NE * labelscalefactor);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 +
/* end of picture */
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</asy>
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 +
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>.
 +
 
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==Solution 2==
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Simpler is just drawing the trapezoid and then using what is given to solve.
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Draw a line parallel to <math>\overline{AD}</math> that connects the longer side to the corner of the shorter side. Name the bottom part <math>x</math> and top part <math>a</math>.
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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{\textbf{(B)}\ 12.25} </math>.
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Solution by IronicNinja
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==Solution 3==
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We know it is a trapezoid and that <math>\overline{AB}</math> and <math>\overline{CD}</math> are perpendicular to <math>\overline{AD}</math>. If they are perpendicular to <math>\overline{AD}</math> 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 <math>\overline{AD}</math> is <math>7</math>. We can then set the length of <math>\overline{AB}</math> to be <math>x</math> and the length of <math>\overline{DC}</math> to be <math>y</math>. <math>\overline{BC}</math> would then be <math>x+y</math>. Let's draw a straight line down from point <math>B</math> which is perpendicular to <math>\overline{DC}</math> and parallel to <math>\overline{AD}</math>. Let's name this line <math>M</math>. Then let's name the point at which line <math>M</math> intersects <math>\overline{DC}</math> point <math>E</math>. Line <math>M</math> partitions the trapezoid into rectangle <math>ADEB</math> and triangle <math>BEC</math>. We will use the triangle to solve for <math>x \times y</math> using the Pythagorean theorem. The line segment <math>\overline{EC}</math> would be <math>y-x</math> because <math>\overline{DC}</math> is <math>y</math> and <math>\overline{DE}</math> is <math>x</math>. <math>\overline{DE}</math> is <math>x</math> because it is parallel to <math>\overline{AB}</math> and both are of equal length. Because of the Pythagorean theorem, we know that <math>(EC)^2+(BE)^2=(BC)^2</math>. Substituting the values we have we get <math>(y-x)^2+(7)^2=(x+y)^2</math>. Simplifying this we get <math>(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)</math>. Now we get rid of the <math>x^2</math> and <math>y^2</math> terms from both sides to get <math>(-2xy)+(49)=(2xy)</math>. Combining like terms we get <math>(49)=(4xy)</math>. Then we divide by <math>4</math> to get <math>(12.25)=(xy)</math>. Now we know that <math>x \times y</math> (same thing as <math>xy</math>) is equal to <math>12.25</math> which is answer choice <math>\boxed{\textbf{(B)}\ 12.25} </math>.
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Solution By: MATHCOUNTSCMS25
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P.S. I Don't Know How to Format It Properly Using <math>\LaTeX</math> So Could Someone Please Fix It
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EDIT: Fixed! (As much as my ability can) - Mliu630XYZ
 +
 
 +
EDIT: Fixed Completely! - palaashgang
  
 
==See Also==
 
==See Also==

Latest revision as of 00:56, 1 June 2021

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 $x \times y$ 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 $x \times y$ (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

EDIT: Fixed Completely! - palaashgang

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

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