1965 AHSME Problems/Problem 35

Revision as of 13:23, 19 July 2024 by Thepowerful456 (talk | contribs) (Solution: with diagram)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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