Difference between revisions of "1973 AHSME Problems/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> | ||
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<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]]. | ||
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==See Also== | ==See Also== | ||
Revision as of 14:50, 4 July 2018
Problem
A chord which is the perpendicular bisector of a radius of length 12 in a circle, has length
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]](http://latex.artofproblemsolving.com/5/8/1/5811cf5242e0d777fab9f039e8a3153791f6adc4.png)
Draw a diagram as shown. Using the Pythagorean Theorem (or by using 30-60-90 triangles), half of the chord length is 
, so the chord’s length is 
. One can also find the length directly by using the Law of Cosines.
See Also
| 1973 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
