Difference between revisions of "2002 AMC 10A Problems/Problem 25"

Latest revision as of 06:05, 19 December 2022

Problem

In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ , $BC = 12$ , $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is

$[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]$

$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution

Solution 1

It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$ :

$[asy] size(250); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13); draw(A--B--C--D--cycle); draw(D--F--C,dashed); label("$A$",A,S); label("$B$",B,S); label("$C$",C,NE); label("$D$",D,W); label("$E$",F,N); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,E); label("12",(B+C)/2,WSW); [/asy]$

Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$ , with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$ . Thus $\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}$ So the sides of $\triangle CDE$ are $15,36,39$ , which we recognize to be a $5 - 12 - 13$ right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared), $[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}$

Solution 2

Draw altitudes from points $C$ and $D$ :

$[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,NE); label("$D$",D,N); label("$D'$",F,SSE); label("$C'$",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); [/asy]$

Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$ . We get the following triangle:

$[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label("$A'$",A,SW); label("$B$",B,S); label("$C$",C,N); label("$C'$",F,SE); label("5",(A+C)/2,W); label("12",(B+C)/2,ENE); [/asy]$

The length of $A'B$ in this triangle is equal to the length of the original $AB$ , minus the length of $CD$ . Thus $A'B = 52 - 39 = 13$ .

Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$ , and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$ .

Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}$

Solution 3

Draw altitudes from points $C$ and $D$ :

$[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,NE); label("$D$",D,N); label("$D'$",F,SSE); label("$C'$",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); [/asy]$

Call the length of $AD'$ to be $y$ , the length of $BC'$ to be $z$ , and the height of the trapezoid to be $x$ . By the Pythagorean Theorem, we have: $z^2 + x^2 = 144$ $y^2 + x^2 = 25$

Subtracting these two equation yields: $z^2-y^2=119 \implies (z+y)(z-y)=119$

We also have: $z+y=52-39=13$ .

We can substitute the value of $z+y$ into the equation we just obtained, so we now have:

$(13) (z-y)=119 \implies z-y=\frac{119}{13}$ .

We can add the $z+y$ and the $z-y$ equation to find the value of $z$ , which simplifies down to be $\frac{144}{13}$ . Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.

$\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}$

Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.

The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$ , and multiplying this and the height of the trapezoid gets us:

$\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}$

Solution 4

We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$

By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{\textbf{(C)} 210}.$

Alternative: Instead of creating seven congruent right triangles, we can find the height of parallelogram $AECD$ by drawing an altitude from $D$ to side $AE$ , creating the new point $F$ . By recognizing that triangle $DAF$ is similar to triangle $BEC$ , we can use properties of similar triangles and find that $DE = 12 \cdot \frac{5}{13} = \frac{60}{13}$ . Thus, the area of parallelogram $AECD$ is $\frac{60}{13} \cdot 39 = 180$ . Finally, we add the areas of the parallelogram $AECD$ and the right triangle $BEC$ together and we get $180 + 30 = \boxed{\textbf{(C)} 210}$ . ~scarletsyc

Solution 2 but quicker

From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$ . Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$ . The numbers multiply nicely to get $180+30=\boxed{(C) 210}$ -harsha12345

Quick Time Trouble Solution 5

First note how the answer choices are all integers. The area of the trapezoid is $\frac{39+52}{2} \cdot h = \frac{91}{2} h$ . So h divides 2. Let $x$ be $2h$ . The area is now $91x$ . Trying $x=1$ and $x=2$ can easily be seen to not work. Those make the only integers possible so now you know x is a fraction. Since the area is an integer the denominator of x must divide either 13 or 7 since $91 = 13\cdot7$ . Seeing how $39 = 3\cdot13$ and $52 = 4\cdot13$ assume that the denominator divides 13. Letting $y = \frac{x}{13}$ the area is now $7y$ . Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean $h$ is $4$ which we ruled out. So the answer is $\boxed{\textbf{(C)} 210}$ . - megateleportingrobots

@@ Line 1: / Line 1: @@
 == Problem ==
 In [[trapezoid]] <math>ABCD</math> with bases <math>AB</math> and <math>CD</math>, we have <math>AB = 52</math>, <math>BC = 12</math>, <math>CD = 39</math>, and <math>DA = 5</math>. The area of <math>ABCD</math> is
+<asy>
+pair A,B,C,D;
+A=(0,0);
+B=(52,0);
+C=(38,20);
+D=(5,20);
+dot(A);
+dot(B);
+dot(C);
+dot(D);
+draw(A--B--C--D--cycle);
+label("$A$",A,S);
+label("$B$",B,S);
+label("$C$",C,N);
+label("$D$",D,N);
+label("52",(A+B)/2,S);
+label("39",(C+D)/2,N);
+label("12",(B+C)/2,E);
+label("5",(D+A)/2,W);
+</asy>
 <math>\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260</math>
@@ Line 134: / Line 157: @@
 By continuing to split <math>\overline{AB}</math> and <math>\overline{CD}</math> into segments of length <math>13,</math> we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides <math>5,12,</math> and <math>13,</math> and each with area <math>30.</math> The total area is therefore <math>7 \cdot 30 = \boxed{\textbf{(C)} 210}.</math>
+Alternative: Instead of creating seven congruent right triangles, we can find the height of parallelogram <math>AECD</math> by drawing an altitude from <math>D</math> to side <math>AE</math>, creating the new point <math>F</math>. By recognizing that triangle <math>DAF</math> is similar to triangle <math>BEC</math>, we can use properties of similar triangles and find that <math>DE = 12 \cdot \frac{5}{13} = \frac{60}{13}</math>. Thus, the area of parallelogram <math>AECD</math> is <math>\frac{60}{13} \cdot 39 = 180</math>. Finally, we add the areas of the parallelogram <math>AECD</math> and the right triangle <math>BEC</math> together and we get <math>180 + 30 = \boxed{\textbf{(C)} 210}</math>. ~scarletsyc
 === Solution 2 but quicker ===
@@ Line 140: / Line 165: @@
 The numbers multiply nicely to get <math>180+30=\boxed{(C) 210}</math>
 -harsha12345
+=== Quick Time Trouble Solution 5 ===
+First note how the answer choices are all integers.
+The area of the trapezoid is <math>\frac{39+52}{2} \cdot h = \frac{91}{2} h</math>. So h divides 2. Let <math>x</math> be <math>2h</math>. The area is now <math>91x</math>.
+Trying <math>x=1</math> and <math>x=2</math> can easily be seen to not work. Those make the only integers possible so now you know x is a fraction.
+Since the area is an integer the denominator of x must divide either 13 or 7 since <math>91 = 13\cdot7</math>.
+Seeing how <math>39 = 3\cdot13</math> and <math>52 = 4\cdot13</math> assume that the denominator divides 13. Letting <math>y = \frac{x}{13}</math> the area is now <math>7y</math>.
+Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean <math>h</math> is <math>4</math> which we ruled out.
+So the answer is <math>\boxed{\textbf{(C)} 210}</math>. - megateleportingrobots
 == See also ==
-{{AMC10 box|year=2002|num-b=24|after=Last Problem|ab=A}}
+{{AMC10 box|year=2002|num-b=24|after=-(Last question)|ab=A}}
 [[Category:Introductory Geometry Problems]]
 [[Category:Area Problems]]
 {{MAA Notice}}

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by -(Last question)
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