1973 AHSME Problems/Problem 1

Problem

A chord which is the perpendicular bisector of a radius of length 12 in a circle, has length

$\textbf{(A)}\ 3\sqrt3\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 6\sqrt3\qquad\textbf{(D)}\ 12\sqrt3\qquad\textbf{(E)}\ \text{ none of these}$

Solutions

$[asy] draw(circle((0,0),12)); draw((0,0)--(0,12)); draw((0,0)--(10.392,6)--(-10.392,6)--(0,0)); label("$12$",(5.196,3),SE); label("$12$",(-5.196,3),SW); label("$6$",(0,3),E); draw((1,6)--(1,5)--(0,5)); [/asy]$

Draw a diagram as shown. Using the Pythagorean Theorem (or by using 30-60-90 triangles), half of the chord length is $6\sqrt{3}$ , so the chord’s length is $\boxed{\textbf{(D) } 12\sqrt{3}}$ . One can also find the length directly by using the Law of Cosines.

1973 AHSC (Problems • Answer Key • Resources)
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1973 AHSME Problems/Problem 1

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