Difference between revisions of "1960 AHSME Problems/Problem 32"
Rockmanex3 (talk | contribs) (Solution to Problem 32) |
Rockmanex3 (talk | contribs) m (→Solution) |
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For option B, substitution results in | For option B, substitution results in | ||
<cmath>(r \sqrt{5} - r)r = (3r - r\sqrt{5})(r \sqrt{5} - r)</cmath> | <cmath>(r \sqrt{5} - r)r = (3r - r\sqrt{5})(r \sqrt{5} - r)</cmath> | ||
− | |||
<cmath>r^2 \sqrt{5} - r^2 = 4r^2 \sqrt{5} - 8r^2</cmath> | <cmath>r^2 \sqrt{5} - r^2 = 4r^2 \sqrt{5} - 8r^2</cmath> | ||
Revision as of 22:53, 12 May 2018
Problem
In this figure the center of the circle is . , is a straight line, , and has a length twice the radius. Then:
Solution
Let be the radius of the circle, so . By the Pythagorean Theorem, . That means, , so .
Substitute values for each answer choice to determine which one is correct for all .
For option A, substitution results in
For option B, substitution results in
For option C, substitution results in
For option D, substitution results in
From each option, only option A has both sides equaling each other, so the answer is .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |