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

Revision as of 13:48, 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

$\fbox{D}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == 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>

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

Revision as of 13:48, 19 July 2024

Problem

Solution

See Also