Difference between revisions of "1960 AHSME Problems/Problem 13"
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==Solution== | ==Solution== | ||
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Two of the side lengths are <math>\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}</math> while one of the side lengths is <math>4</math>. That makes the triangle isosceles, so the answer is <math>\boxed{\textbf{(B)}}</math>. | Two of the side lengths are <math>\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}</math> while one of the side lengths is <math>4</math>. That makes the triangle isosceles, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
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==See Also== | ==See Also== | ||
{{AHSME box|year=1960|num-b=12|num-a=14}} | {{AHSME box|year=1960|num-b=12|num-a=14}} |
Revision as of 19:52, 8 May 2018
Problem
The polygon(s) formed by , and , is (are):
Solution
The points of intersection of two of the lines are and , so use the Distance Formula to find the sidelengths.
Two of the side lengths are while one of the side lengths is . That makes the triangle isosceles, so the answer is .
See Also
1960 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AHSME Problems and Solutions |