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

Latest revision as of 21:28, 12 January 2025

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 ▭ $ADEB$ and $\triangle$ $BEC$ . We will use the triangle to solve for $xy$ 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 $xy\ =\ 12.25$ which is answer choice $\boxed{\textbf{(B)}\ 12.25}$ .

Solution By: MATHCOUNTSCMS25

Fixed $\text{\LaTeX}$ - Mliu630XYZ, palaashgang, anshulb, JoyfulSapling

Solution 4 (EZ Cheez)

Choose any value for $BC$ , and then use Pythagorean theorem to get $CD - AB$ , and $AB = (BC - (CD - AB))/2$ ). Then multiply $AB \cdot CD$ .

For example:

$BC=25$ . $CD - AB = \sqrt{25^2 - 7^2} = 24$ . $AB = (25 - 24)/2=0.5$ . $CD = 0.5 + 24 = 24.5$ . $AB \cdot CD = (0.5)(24.5)= \boxed{\textbf{(B)}\ 12.25}$ .

Solution 5 (Quick Solve)

Denote $AB$ to $x$ , and $CD$ to $y$ . This means that $BC = x+y$ . This means that if we draw an altitude from $B$ $CD$ and we call the foot of that $P$ , we have that $BP$ is $7$ due to symmetry. This means $7^2+(y-x)^2 = (x+y)^2$ . Simplifying gets you: $49 +y ^2-2xy+x^2 = x^2+2xy+y^2 \Longrightarrow xy = \dfrac{49}{4} \Longrightarrow xy = \boxed{\textbf{(B)}\ 12.25}$ .

-jb2015007

@@ Line 62: / Line 62: @@
 <math>BC=25</math>. <math>CD - AB = \sqrt{25^2 - 7^2} = 24</math>. <math>AB = (25 - 24)/2=0.5</math>. <math>CD = 0.5 + 24 = 24.5</math>.
 <math>AB \cdot CD = (0.5)(24.5)= \boxed{\textbf{(B)}\ 12.25}</math>.
+==Solution 5 (Quick Solve)==
+Denote <math>AB</math> to <math>x</math>, and <math>CD</math> to <math>y</math>. This means that <math>BC = x+y</math>. This means that if we draw an altitude from <math>B</math> <math>CD</math> and we call the foot of that <math>P</math>, we have that <math>BP</math> is <math>7</math> due to symmetry. This means <cmath>7^2+(y-x)^2 = (x+y)^2</cmath>. Simplifying gets you: <cmath>49 +y ^2-2xy+x^2 = x^2+2xy+y^2 \Longrightarrow xy = \dfrac{49}{4} \Longrightarrow xy = \boxed{\textbf{(B)}\ 12.25}</cmath>.
+-jb2015007
 ==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
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