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Difference between revisions of "1966 AHSME Problems/Problem 8"

(New page: ==Problem== The length of the common chord of two intersecting circles is <math>16</math> feet. If the radii are <math>10</math> feet and <math>17</math> feet, a possible value for the dis...)
 
(→‎Solution)
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[[Image:1966_AHSME-8.jpg]]
 
[[Image:1966_AHSME-8.jpg]]
 
Let <math>O</math> be the center of the circle of radius <math>10</math> and <math>P</math> be the center of the circle of radius <math>17</math>. Chord <math>\overline{AB} = 16</math> feet.
 
Let <math>O</math> be the center of the circle of radius <math>10</math> and <math>P</math> be the center of the circle of radius <math>17</math>. Chord <math>\overline{AB} = 16</math> feet.
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<math>\overline{OA} = \overline{OB} = 10</math> feet, since they are radii of a circle. Hence, <math>\triangle OAB</math> is isoceles with base <math>AB</math>. The height of <math>\trianlge OAB</math> from <math>O</math> to <math>AB</math> is <math>\sqrt {\overline{OB}^2 - \frac{\overline{AB}}{2}^2}</math>
+
<math>\overline{OA} = \overline{OB} = 10</math> feet, since they are radii of a circle. Hence, <math>\triangle OAB</math> is isoceles with base <math>AB</math>. The height of <math>\triangle OAB</math> from <math>O</math> to <math>AB</math> is <math>\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2}</math>

Revision as of 18:38, 1 July 2008

Problem

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of the circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$

Solution

File:1966 AHSME-8.jpg Let $O$ be the center of the circle of radius $10$ and $P$ be the center of the circle of radius $17$. Chord $\overline{AB} = 16$ feet. $\overline{OA} = \overline{OB} = 10$ feet, since they are radii of a circle. Hence, $\triangle OAB$ is isoceles with base $AB$. The height of $\triangle OAB$ from $O$ to $AB$ is $\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2}$

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