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Difference between revisions of "1952 AHSME Problems/Problem 29"

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<cmath>25 - a^2 = 25-(x^2-25) = 50-x^2 = 8x-x^2 \Rightarrow x = \frac{50}{8} = \frac{25}{4}.</cmath>
 
<cmath>25 - a^2 = 25-(x^2-25) = 50-x^2 = 8x-x^2 \Rightarrow x = \frac{50}{8} = \frac{25}{4}.</cmath>
 
It follows that <math>a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}</math>.
 
It follows that <math>a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}</math>.
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 +
Thus, the answer is <math>E</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:19, 15 January 2018

Problem

In a circle of radius $5$ units, $CD$ and $AB$ are perpendicular diameters. A chord $CH$ cutting $AB$ at $K$ is $8$ units long. The diameter $AB$ is divided into two segments whose dimensions are:

$\textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75,7.25 \qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)}\ 4,6 \qquad \textbf{(E)}\ \text{none of these}$

Solution

$\fbox{}$


Let the intersection of $CH$ and $AB$ be $N$, $O$ be the center of the circle, $ON = a$ and $CN = x$. By power of a point on $N$, we have \[(5+a)(5-a) = 25-a^2 = x(8-x).\] $\triangle CON$ is a right triangle, so we also know that $x^2 = a^2 + 25$, thus \[25 - a^2 = 25-(x^2-25) = 50-x^2 = 8x-x^2 \Rightarrow x = \frac{50}{8} = \frac{25}{4}.\] It follows that $a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}$.

Thus, the answer is $E$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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All AHSME Problems and Solutions

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