Difference between revisions of "1959 AHSME Problems/Problem 22"

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== Solution ==
 
== Solution ==
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<asy>
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import geometry;
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point B = (0,0);
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point A = (3,5);
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point D = (13,5);
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point C = (15,0);
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point M,N;
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// Trapezoid
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draw(A--B--C--D--A);
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dot(A);
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label("A",A,NW);
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dot(B);
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label("B",B,SW);
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dot(C);
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label("C",C,SE);
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dot(D);
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label("D",D,NE);
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// Diagonals and their midpoints
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draw(A--C);
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draw(B--D);
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M = midpoint(A--C);
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dot(M);
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label("M",M,ENE);
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N = midpoint(B--D);
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dot(N);
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label("N",N,WNW);
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draw(M--N);
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// Length Labels
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label("$x$",midpoint(A--D),(0,1));
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label("$97$",midpoint(B--C),S);
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label("$3$",midpoint(M--N),S);
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</asy>
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Let <math>x</math> be the length of the shorter base. Then:
 
Let <math>x</math> be the length of the shorter base. Then:
  

Revision as of 12:00, 21 July 2024

Problem

The line joining the midpoints of the diagonals of a trapezoid has length $3$. If the longer base is $97,$ then the shorter base is: $\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89$

Solution

[asy]  import geometry;  point B = (0,0); point A = (3,5); point D = (13,5); point C = (15,0); point M,N;  // Trapezoid draw(A--B--C--D--A); dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(D); label("D",D,NE);  // Diagonals and their midpoints draw(A--C); draw(B--D);  M = midpoint(A--C); dot(M); label("M",M,ENE);  N = midpoint(B--D); dot(N); label("N",N,WNW); draw(M--N);  // Length Labels label("$x$",midpoint(A--D),(0,1)); label("$97$",midpoint(B--C),S); label("$3$",midpoint(M--N),S);  [/asy]

Let $x$ be the length of the shorter base. Then:

$3 = \frac{97-x}{2}$

$6 = 97-x$

$x = 91$

Thus, our answer is $\boxed{\textbf{(C) }91}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AHSME Problems and Solutions

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