Difference between revisions of "1961 AHSME Problems/Problem 10"
Rockmanex3 (talk | contribs) (Solution to Problem 10) |
Rockmanex3 (talk | contribs) (→Solution) |
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label("$12$",(40,25)); | label("$12$",(40,25)); | ||
draw((25,3)--(28,3)--(28,0)); | draw((25,3)--(28,3)--(28,0)); | ||
− | label("$3\sqrt{3}$",( | + | label("$3\sqrt{3}$",(30.5,11)); |
</asy> | </asy> | ||
From the [[Pythagorean Theorem]] (or by using 30-60-90 triangles), <math>AD = 6\sqrt{3}</math>. That means <math>DE = 3\sqrt{3}</math>. Using the Pythagorean Theorem again, <math>BE = \sqrt{63}</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>. | From the [[Pythagorean Theorem]] (or by using 30-60-90 triangles), <math>AD = 6\sqrt{3}</math>. That means <math>DE = 3\sqrt{3}</math>. Using the Pythagorean Theorem again, <math>BE = \sqrt{63}</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>. |
Revision as of 13:52, 19 May 2018
Problem 10
Each side of is units. is the foot of the perpendicular dropped from on , and is the midpoint of . The length of , in the same unit, is:
Solution
From the Pythagorean Theorem (or by using 30-60-90 triangles), . That means . Using the Pythagorean Theorem again, , which is answer choice .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.