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>. | ||
+ | |||
+ | Thus, the answer is <math>E</math>. | ||
== See also == | == See also == |
Revision as of 13:19, 15 January 2018
Problem
In a circle of radius units,
and
are perpendicular diameters. A chord
cutting
at
is
units long. The diameter
is divided into two segments whose dimensions are:
Solution
Let the intersection of and
be
,
be the center of the circle,
and
. By power of a point on
, we have
is a right triangle, so we also know that
, thus
It follows that
.
Thus, the answer is .
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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