Difference between revisions of "1999 AMC 8 Problems/Problem 14"

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There is a rectangle present, with both horizontal bases being <math>8</math> units in length. The excess units on the bottom base must then be 8. The fact that <math>AB</math> and <math>CD</math> are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of <math>8</math> units, so each is <math>4</math> units. The triangle has a hypotenuse of 5, because the triangles are <math>3-4-5</math> right triangles. So, the sides of the trapezoid are <math>8</math>, <math>5</math>, <math>16</math>, and <math>5</math>. Adding those up gives us the perimeter, <math>8 + 5 + 16 + 5 = 13 + 21 = 34</math> units.
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==Problem==
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In trapezoid <math>ABCD</math>, the sides <math>AB</math> and <math>CD</math> are equal. The perimeter of <math>ABCD</math> is
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<asy>
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draw((0,0)--(4,3)--(12,3)--(16,0)--cycle);
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draw((4,3)--(4,0),dashed);
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draw((3.2,0)--(3.2,.8)--(4,.8));
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label("$A$",(0,0),SW);
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label("$B$",(4,3),NW);
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label("$C$",(12,3),NE);
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label("$D$",(16,0),SE);
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label("$8$",(8,3),N);
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label("$16$",(8,0),S);
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label("$3$",(4,1.5),E);
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</asy>
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<math>\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48</math>
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==Solution==
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<asy>
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draw((0,0)--(4,3)--(12,3)--(16,0)--cycle);
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draw((4,3)--(4,0),dashed);
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draw((12,3)--(12,0),dashed);
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draw((3.2,0)--(3.2,.8)--(4,.8));
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label("$A$",(0,0),SW);
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label("$B$",(4,3),NW);
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label("$C$",(12,3),NE);
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label("$D$",(16,0),SE);
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label("$8$",(8,3),N);
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label("$8$",(8,0),S);
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label("$3$",(4,1.5),E);
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label("$4$",(2,0),S);
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label("$4$",(14,0),S);
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label("$5$",(0,0)--(4,3),NW);
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label("$5$",(12,3)--(16,0),NE);
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</asy>
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 +
There is a rectangle present, with both horizontal bases being <math>8</math> units in length. The excess units on the bottom base must then be <math>16-8=8</math>. The fact that <math>AB</math> and <math>CD</math> are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of <math>8</math> units, so each is <math>4</math> units. The triangle has a hypotenuse of <math>5</math>, because the triangles are <math>3-4-5</math> right triangles. So, the sides of the trapezoid are <math>8</math>, <math>5</math>, <math>16</math>, and <math>5</math>. Adding those up gives us the perimeter, <math>8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}</math> units.
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==See Also==
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{{AMC8 box|year=1999|num-b=13|num-a=15}}

Revision as of 12:53, 23 December 2012

Problem

In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8));  label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$16$",(8,0),S); label("$3$",(4,1.5),E); [/asy]

$\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48$

Solution

[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((12,3)--(12,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8));  label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$8$",(8,0),S); label("$3$",(4,1.5),E); label("$4$",(2,0),S); label("$4$",(14,0),S); label("$5$",(0,0)--(4,3),NW); label("$5$",(12,3)--(16,0),NE); [/asy]

There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$. The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$, because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$, $5$, $16$, and $5$. Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}$ units.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions
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