https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Megateleportingrobots&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T09:25:57ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=1284912002 AMC 10A Problems/Problem 252020-07-17T22:55:49Z<p>Megateleportingrobots: /* Quick Time Trouble Solution 5 */</p>
<hr />
<div>== Problem ==<br />
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<br />
<br />
<math>\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:<br />
<br />
<center><asy><br />
size(250);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br />
draw(A--B--C--D--cycle);<br />
draw(D--F--C,dashed);<br />
label("\(A\)",A,S);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,W);<br />
label("\(E\)",F,N);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,E);<br />
label("12",(B+C)/2,WSW);<br />
</asy></center><br />
<br />
Since <math>\overline{AB} || \overline{CD}</math> we have <math>\triangle AEB \sim \triangle DEC</math>, with the ratio of [[proportion]]ality being <math>\frac {39}{52} = \frac {3}{4}</math>. Thus<br />
<cmath><br />
\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\<br />
\frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}<br />
</cmath><br />
So the sides of <math>\triangle CDE</math> are <math>15,36,39</math>, which we recognize to be a <math>5 - 12 - 13</math> [[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),<br />
<cmath><br />
[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 2 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
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);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Translate the triangle <math>ADD'</math> so that <math>DD'</math> coincides with <math>CC'</math>. We get the following triangle:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br />
draw(A--B--C--cycle);<br />
draw(C--F,dashed);<br />
label("\(A'\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,N);<br />
label("\(C'\)",F,SE);<br />
label("5",(A+C)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
The length of <math>A'B</math> in this triangle is equal to the length of the original <math>AB</math>, minus the length of <math>CD</math>.<br />
Thus <math>A'B = 52 - 39 = 13</math>.<br />
<br />
Therefore <math>A'BC</math> is a well-known <math>(5,12,13)</math> right triangle. Its area is <math>[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30</math>, and therefore its altitude <math>CC'</math> is <math>\frac{[A'BC]}{A'B} = \frac{60}{13}</math>.<br />
<br />
Now the area of the original trapezoid is <math>\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}</math><br />
<br />
=== Solution 3 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
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);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Call the length of <math>AD'</math> to be <math>y</math>, the length of <math>BC'</math> to be <math>z</math>, and the height of the trapezoid to be <math>x</math>.<br />
By the Pythagorean Theorem, we have:<br />
<cmath>z^2 + x^2 = 144</cmath><br />
<cmath>y^2 + x^2 = 25</cmath><br />
<br />
Subtracting these two equation yields:<br />
<cmath>z^2-y^2=119 \implies (z+y)(z-y)=119</cmath><br />
<br />
We also have: <math>z+y=52-39=13</math>.<br />
<br />
We can substitute the value of <math>z+y</math> into the equation we just obtained, so we now have:<br />
<br />
<cmath>(13) (z-y)=119 \implies z-y=\frac{119}{13}</cmath>.<br />
<br />
We can add the <math>z+y</math> and the <math>z-y</math> equation to find the value of <math>z</math>, which simplifies down to be <math>\frac{144}{13}</math>. Finally, we can plug in <math>z</math> and use the Pythagorean theorem to find the height of the trapezoid.<br />
<br />
<cmath>\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}</cmath><br />
<br />
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br />
<br />
The median of the trapezoid is <math>\frac{39+52}{2} = \frac{91}{2}</math>, and multiplying this and the height of the trapezoid gets us:<br />
<br />
<cmath>\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 4 ===<br />
<br />
We construct a line segment parallel to <math>\overline{AD}</math> from point <math>C</math> to line <math>\overline{AB},</math> and label the intersection of this segment with line <math>\overline{AB}</math> as point <math>E.</math> Then quadrilateral <math>AECD</math> is a parallelogram, so <math>CE=5, AE=39,</math> and <math>EB=13.</math> Triangle <math>EBC</math> is therefore a right triangle, with area <math>\frac12 \cdot 5 \cdot 12 = 30.</math><br />
<br />
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><br />
<br />
=== Solution 2 but quicker ===<br />
From Solution <math>2</math> we know that the the altitude of the trapezoid is <math>\frac{60}{13}</math> and the triangle's area is <math>30</math>.<br />
Note that once we remove the triangle we get a rectangle with length <math>39</math> and height <math>\frac{60}{13}</math>.<br />
The numbers multiply nicely to get <math>180+30=\boxed{(C) 210}</math><br />
-harsha12345<br />
<br />
=== Quick Time Trouble Solution 5 ===<br />
<br />
First note how the answer choices are all integers.<br />
The area of the trapezoid is <math>\frac{39+52}{2} * 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>. <br />
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. <br />
Since the area is an integer the denominator of x must divide either 13 or 7 since <math>91 = 13*7</math>.<br />
Seeing how <math>39 = 3*13</math> and <math>52 = 4*13</math> assume that the denominator divides 13. Letting <math>y = \frac{x}{13}</math> the area is now <math>7y</math>.<br />
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. <br />
So the answer is <math>\boxed{\textbf{(C)} 210}</math>. - megateleportingrobots<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|num-b=24|after=Last Problem|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Megateleportingrobotshttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=1284902002 AMC 10A Problems/Problem 252020-07-17T22:55:28Z<p>Megateleportingrobots: /* Quick Time Trouble Solution 5 */</p>
<hr />
<div>== Problem ==<br />
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<br />
<br />
<math>\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:<br />
<br />
<center><asy><br />
size(250);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br />
draw(A--B--C--D--cycle);<br />
draw(D--F--C,dashed);<br />
label("\(A\)",A,S);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,W);<br />
label("\(E\)",F,N);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,E);<br />
label("12",(B+C)/2,WSW);<br />
</asy></center><br />
<br />
Since <math>\overline{AB} || \overline{CD}</math> we have <math>\triangle AEB \sim \triangle DEC</math>, with the ratio of [[proportion]]ality being <math>\frac {39}{52} = \frac {3}{4}</math>. Thus<br />
<cmath><br />
\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\<br />
\frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}<br />
</cmath><br />
So the sides of <math>\triangle CDE</math> are <math>15,36,39</math>, which we recognize to be a <math>5 - 12 - 13</math> [[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),<br />
<cmath><br />
[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 2 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
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);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Translate the triangle <math>ADD'</math> so that <math>DD'</math> coincides with <math>CC'</math>. We get the following triangle:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br />
draw(A--B--C--cycle);<br />
draw(C--F,dashed);<br />
label("\(A'\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,N);<br />
label("\(C'\)",F,SE);<br />
label("5",(A+C)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
The length of <math>A'B</math> in this triangle is equal to the length of the original <math>AB</math>, minus the length of <math>CD</math>.<br />
Thus <math>A'B = 52 - 39 = 13</math>.<br />
<br />
Therefore <math>A'BC</math> is a well-known <math>(5,12,13)</math> right triangle. Its area is <math>[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30</math>, and therefore its altitude <math>CC'</math> is <math>\frac{[A'BC]}{A'B} = \frac{60}{13}</math>.<br />
<br />
Now the area of the original trapezoid is <math>\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}</math><br />
<br />
=== Solution 3 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
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);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Call the length of <math>AD'</math> to be <math>y</math>, the length of <math>BC'</math> to be <math>z</math>, and the height of the trapezoid to be <math>x</math>.<br />
By the Pythagorean Theorem, we have:<br />
<cmath>z^2 + x^2 = 144</cmath><br />
<cmath>y^2 + x^2 = 25</cmath><br />
<br />
Subtracting these two equation yields:<br />
<cmath>z^2-y^2=119 \implies (z+y)(z-y)=119</cmath><br />
<br />
We also have: <math>z+y=52-39=13</math>.<br />
<br />
We can substitute the value of <math>z+y</math> into the equation we just obtained, so we now have:<br />
<br />
<cmath>(13) (z-y)=119 \implies z-y=\frac{119}{13}</cmath>.<br />
<br />
We can add the <math>z+y</math> and the <math>z-y</math> equation to find the value of <math>z</math>, which simplifies down to be <math>\frac{144}{13}</math>. Finally, we can plug in <math>z</math> and use the Pythagorean theorem to find the height of the trapezoid.<br />
<br />
<cmath>\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}</cmath><br />
<br />
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br />
<br />
The median of the trapezoid is <math>\frac{39+52}{2} = \frac{91}{2}</math>, and multiplying this and the height of the trapezoid gets us:<br />
<br />
<cmath>\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 4 ===<br />
<br />
We construct a line segment parallel to <math>\overline{AD}</math> from point <math>C</math> to line <math>\overline{AB},</math> and label the intersection of this segment with line <math>\overline{AB}</math> as point <math>E.</math> Then quadrilateral <math>AECD</math> is a parallelogram, so <math>CE=5, AE=39,</math> and <math>EB=13.</math> Triangle <math>EBC</math> is therefore a right triangle, with area <math>\frac12 \cdot 5 \cdot 12 = 30.</math><br />
<br />
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><br />
<br />
=== Solution 2 but quicker ===<br />
From Solution <math>2</math> we know that the the altitude of the trapezoid is <math>\frac{60}{13}</math> and the triangle's area is <math>30</math>.<br />
Note that once we remove the triangle we get a rectangle with length <math>39</math> and height <math>\frac{60}{13}</math>.<br />
The numbers multiply nicely to get <math>180+30=\boxed{(C) 210}</math><br />
-harsha12345<br />
<br />
=== Quick Time Trouble Solution 5 ===<br />
<br />
First note how the answer choices are all integers.<br />
The area of the trapezoid is <math>\frac{39+52}{2} * 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>. <br />
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. <br />
Since the area is an integer the denominator of x must divide either 13 or 7 since <math>91 = 13*7</math>.<br />
Seeing how <math>39 = 3*13</math> and <math>52 = 4*13</math> assume that the denominator divides 13. Letting <math>y = \frac{x}{13}</math> the area is now <math>7y</math>.<br />
Note that only (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. <br />
So the answer is <math>\boxed{\textbf{(C)} 210}</math>. - megateleportingrobots<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|num-b=24|after=Last Problem|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Megateleportingrobotshttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=1284892002 AMC 10A Problems/Problem 252020-07-17T22:54:32Z<p>Megateleportingrobots: /* Solution */</p>
<hr />
<div>== Problem ==<br />
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<br />
<br />
<math>\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:<br />
<br />
<center><asy><br />
size(250);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br />
draw(A--B--C--D--cycle);<br />
draw(D--F--C,dashed);<br />
label("\(A\)",A,S);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,W);<br />
label("\(E\)",F,N);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,E);<br />
label("12",(B+C)/2,WSW);<br />
</asy></center><br />
<br />
Since <math>\overline{AB} || \overline{CD}</math> we have <math>\triangle AEB \sim \triangle DEC</math>, with the ratio of [[proportion]]ality being <math>\frac {39}{52} = \frac {3}{4}</math>. Thus<br />
<cmath><br />
\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\<br />
\frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}<br />
</cmath><br />
So the sides of <math>\triangle CDE</math> are <math>15,36,39</math>, which we recognize to be a <math>5 - 12 - 13</math> [[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),<br />
<cmath><br />
[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 2 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
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);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Translate the triangle <math>ADD'</math> so that <math>DD'</math> coincides with <math>CC'</math>. We get the following triangle:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br />
draw(A--B--C--cycle);<br />
draw(C--F,dashed);<br />
label("\(A'\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,N);<br />
label("\(C'\)",F,SE);<br />
label("5",(A+C)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
The length of <math>A'B</math> in this triangle is equal to the length of the original <math>AB</math>, minus the length of <math>CD</math>.<br />
Thus <math>A'B = 52 - 39 = 13</math>.<br />
<br />
Therefore <math>A'BC</math> is a well-known <math>(5,12,13)</math> right triangle. Its area is <math>[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30</math>, and therefore its altitude <math>CC'</math> is <math>\frac{[A'BC]}{A'B} = \frac{60}{13}</math>.<br />
<br />
Now the area of the original trapezoid is <math>\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}</math><br />
<br />
=== Solution 3 ===<br />
<br />
Draw altitudes from points <math>C</math> and <math>D</math>:<br />
<br />
<center><asy><br />
unitsize(0.2cm);<br />
defaultpen(0.8);<br />
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);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E,dashed);<br />
draw(D--F,dashed);<br />
label("\(A\)",A,SW);<br />
label("\(B\)",B,S);<br />
label("\(C\)",C,NE);<br />
label("\(D\)",D,N);<br />
label("\(D'\)",F,SSE);<br />
label("\(C'\)",E,S);<br />
label("39",(C+D)/2,N);<br />
label("52",(A+B)/2,S);<br />
label("5",(A+D)/2,W);<br />
label("12",(B+C)/2,ENE);<br />
</asy></center><br />
<br />
Call the length of <math>AD'</math> to be <math>y</math>, the length of <math>BC'</math> to be <math>z</math>, and the height of the trapezoid to be <math>x</math>.<br />
By the Pythagorean Theorem, we have:<br />
<cmath>z^2 + x^2 = 144</cmath><br />
<cmath>y^2 + x^2 = 25</cmath><br />
<br />
Subtracting these two equation yields:<br />
<cmath>z^2-y^2=119 \implies (z+y)(z-y)=119</cmath><br />
<br />
We also have: <math>z+y=52-39=13</math>.<br />
<br />
We can substitute the value of <math>z+y</math> into the equation we just obtained, so we now have:<br />
<br />
<cmath>(13) (z-y)=119 \implies z-y=\frac{119}{13}</cmath>.<br />
<br />
We can add the <math>z+y</math> and the <math>z-y</math> equation to find the value of <math>z</math>, which simplifies down to be <math>\frac{144}{13}</math>. Finally, we can plug in <math>z</math> and use the Pythagorean theorem to find the height of the trapezoid.<br />
<br />
<cmath>\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}</cmath><br />
<br />
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br />
<br />
The median of the trapezoid is <math>\frac{39+52}{2} = \frac{91}{2}</math>, and multiplying this and the height of the trapezoid gets us:<br />
<br />
<cmath>\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}</cmath><br />
<br />
=== Solution 4 ===<br />
<br />
We construct a line segment parallel to <math>\overline{AD}</math> from point <math>C</math> to line <math>\overline{AB},</math> and label the intersection of this segment with line <math>\overline{AB}</math> as point <math>E.</math> Then quadrilateral <math>AECD</math> is a parallelogram, so <math>CE=5, AE=39,</math> and <math>EB=13.</math> Triangle <math>EBC</math> is therefore a right triangle, with area <math>\frac12 \cdot 5 \cdot 12 = 30.</math><br />
<br />
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><br />
<br />
=== Solution 2 but quicker ===<br />
From Solution <math>2</math> we know that the the altitude of the trapezoid is <math>\frac{60}{13}</math> and the triangle's area is <math>30</math>.<br />
Note that once we remove the triangle we get a rectangle with length <math>39</math> and height <math>\frac{60}{13}</math>.<br />
The numbers multiply nicely to get <math>180+30=\boxed{(C) 210}</math><br />
-harsha12345<br />
<br />
=== Quick Time Trouble Solution 5 ===<br />
<br />
First note how the answer choices are all integers.<br />
The area of the trapezoid is <math>\frac{39+52}{2} * 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>. <br />
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. <br />
Since the area is an integer the denominator of x must divide either 13 or 7 since <math>91 = 13*7</math>.<br />
Seeing how <math>39 = 3*13</math> and <math>52 = 4*13</math> assume that the denominator divides 13. Letting <math>y = \frac{x}{13}</math> the area is now <math>7y</math>.<br />
Note that only (A) and (C) are multiples of 7. We know that A doesn't work because that would mean h is 4 which we ruled out. <br />
So the answer is <math>\boxed{\textbf{(C)} 210}</math>. - megateleportingrobots<br />
<br />
== See also ==<br />
{{AMC10 box|year=2002|num-b=24|after=Last Problem|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Megateleportingrobots