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Difference between revisions of "1965 AHSME Problems/Problem 35"

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== Solution ==
 
== Solution ==
 +
 +
<asy>
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 +
import geometry;
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 +
point M;
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segment l;
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 +
draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5)));
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dot((0,sqrt(5)));
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label("A", (0,sqrt(5)), NW);
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dot((0,0));
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label("B", (0,0), SW);
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dot((5,0));
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label("C", (5,0), SE);
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dot((5,sqrt(5)));
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label("D", (5, sqrt(5)), NE);
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M=(2.5,sqrt(5)/2);
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l=line((0,sqrt(5)),(5,0));
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draw(l);
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draw(perpendicular(M,l));
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dot(M);
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label("M",M,W);
 +
 +
</asy>
 +
 +
 
<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
  

Revision as of 12:23, 19 July 2024

Problem

The length of a rectangle is $5$ inches and its width is less than $4$ inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is $\sqrt {6}$, then the width is:

$\textbf{(A)}\ \sqrt {2} \qquad \textbf{(B) }\ \sqrt {3} \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ \sqrt{5}\qquad \textbf{(E) }\ \sqrt{\frac{11}{2}}$

Solution

[asy] import geometry; point M; segment l; draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5))); dot((0,sqrt(5))); label("A", (0,sqrt(5)), NW); dot((0,0)); label("B", (0,0), SW); dot((5,0)); label("C", (5,0), SE); dot((5,sqrt(5))); label("D", (5, sqrt(5)), NE); M=(2.5,sqrt(5)/2); l=line((0,sqrt(5)),(5,0)); draw(l); draw(perpendicular(M,l)); dot(M); label("M",M,W); [/asy]


$\fbox{D}$

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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