Difference between revisions of "2002 AMC 10A Problems/Problem 25"
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== Problem == | == Problem == | ||
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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 | 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> | <math>\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260</math> | ||
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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> | 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 === | === Solution 2 but quicker === | ||
Line 144: | Line 169: | ||
First note how the answer choices are all integers. | First note how the answer choices are all integers. | ||
− | The area of the trapezoid is <math>\frac{39+52}{2} | + | 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 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. | + | 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 | + | 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 | + | 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 h is 4 which we ruled out. | + | 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 | So the answer is <math>\boxed{\textbf{(C)} 210}</math>. - megateleportingrobots | ||
== See also == | == See also == | ||
− | {{AMC10 box|year=2002|num-b=24|after=Last | + | {{AMC10 box|year=2002|num-b=24|after=-(Last question)|ab=A}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
[[Category:Area Problems]] | [[Category:Area Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:05, 19 December 2022
Contents
Problem
In trapezoid with bases
and
, we have
,
,
, and
. The area of
is
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 and
to meet at point
:
![[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]](http://latex.artofproblemsolving.com/0/8/f/08fd9a3001f04eb9624f966ea35af9f5984fd7a8.png)
Since we have
, with the ratio of proportionality being
. Thus
So the sides of
are
, which we recognize to be a
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),
Solution 2
Draw altitudes from points and
:
![[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]](http://latex.artofproblemsolving.com/2/5/5/255b3ee5b7acb3f4baca3b52720811aa66cba79e.png)
Translate the triangle so that
coincides with
. 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]](http://latex.artofproblemsolving.com/a/e/9/ae9e06d7e09ef1c45cb1eb1354f20938e46480b6.png)
The length of in this triangle is equal to the length of the original
, minus the length of
.
Thus
.
Therefore is a well-known
right triangle. Its area is
, and therefore its altitude
is
.
Now the area of the original trapezoid is
Solution 3
Draw altitudes from points and
:
![[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]](http://latex.artofproblemsolving.com/2/5/5/255b3ee5b7acb3f4baca3b52720811aa66cba79e.png)
Call the length of to be
, the length of
to be
, and the height of the trapezoid to be
.
By the Pythagorean Theorem, we have:
Subtracting these two equation yields:
We also have: .
We can substitute the value of into the equation we just obtained, so we now have:
.
We can add the and the
equation to find the value of
, which simplifies down to be
. Finally, we can plug in
and use the Pythagorean theorem to find the height of the trapezoid.
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 , and multiplying this and the height of the trapezoid gets us:
Solution 4
We construct a line segment parallel to from point
to line
and label the intersection of this segment with line
as point
Then quadrilateral
is a parallelogram, so
and
Triangle
is therefore a right triangle, with area
By continuing to split and
into segments of length
we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides
and
and each with area
The total area is therefore
Alternative: Instead of creating seven congruent right triangles, we can find the height of parallelogram by drawing an altitude from
to side
, creating the new point
. By recognizing that triangle
is similar to triangle
, we can use properties of similar triangles and find that
. Thus, the area of parallelogram
is
. Finally, we add the areas of the parallelogram
and the right triangle
together and we get
. ~scarletsyc
Solution 2 but quicker
From Solution we know that the the altitude of the trapezoid is
and the triangle's area is
.
Note that once we remove the triangle we get a rectangle with length
and height
.
The numbers multiply nicely to get
-harsha12345
Quick Time Trouble Solution 5
First note how the answer choices are all integers.
The area of the trapezoid is . So h divides 2. Let
be
. The area is now
.
Trying
and
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
.
Seeing how
and
assume that the denominator divides 13. Letting
the area is now
.
Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean
is
which we ruled out.
So the answer is
. - megateleportingrobots
See also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by -(Last question) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.