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

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==Solution 3==
 
==Solution 3==
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We know it is a trapezoid and that AB and CD are perpendicular to AD. If they are perpendicular to 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 AD is 7. We can then set the length of AB to be x and the length of DC to be y. BC would then be x+y. Let's draw a straight line down from point B which is perpendicular to DC and parallel to AD. Let's name this line M. Then let's name the point at which line M intersects 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 EC would be y-x because DC is y and DE is x. DE is x because it is parallel to 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 B.
+
We know it is a trapezoid and that AB and CD are perpendicular to AD. If they are perpendicular to 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 AD is 7. We can then set the length of AB to be x and the length of DC to be y. BC would then be x+y. Let's draw a straight line down from point B which is perpendicular to DC and parallel to AD. Let's name this line M. Then let's name the point at which line M intersects 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 EC would be y-x because DC is y and DE is x. DE is x because it is parallel to 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 <math>\boxed{\textbf{(B)}\ 12.25} </math>.
  
 
==See Also==
 
==See Also==

Revision as of 10:01, 5 January 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 AB and CD are perpendicular to AD. If they are perpendicular to 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 AD is 7. We can then set the length of AB to be x and the length of DC to be y. BC would then be x+y. Let's draw a straight line down from point B which is perpendicular to DC and parallel to AD. Let's name this line M. Then let's name the point at which line M intersects 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 EC would be y-x because DC is y and DE is x. DE is x because it is parallel to 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 $\boxed{\textbf{(B)}\ 12.25}$.

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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