Difference between revisions of "1973 AHSME Problems/Problem 1"

(Solution to Problem 1)
 
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<math> \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} </math>
 
<math> \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} </math>
  
==Solutions==
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==Solution==
  
 
<asy>
 
<asy>
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Draw a diagram as shown.  Using the [[Pythagorean Theorem]] (or by using 30-60-90 triangles), half of the chord length is <math>6\sqrt{3}</math>, so the chord’s length is <math>\boxed{\textbf{(D) } 12\sqrt{3}}</math>.  One can also find the length directly by using the [[Law of Cosines]].
 
Draw a diagram as shown.  Using the [[Pythagorean Theorem]] (or by using 30-60-90 triangles), half of the chord length is <math>6\sqrt{3}</math>, so the chord’s length is <math>\boxed{\textbf{(D) } 12\sqrt{3}}</math>.  One can also find the length directly by using the [[Law of Cosines]].
 
 
  
 
==See Also==
 
==See Also==

Revision as of 13:50, 4 July 2018

Problem

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

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

Solution

[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.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AHSME Problems and Solutions