Difference between revisions of "1953 AHSME Problems/Problem 26"
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\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad | \textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad | ||
\textbf{(D)}\ 7.5\text{ inches}\\ | \textbf{(D)}\ 7.5\text{ inches}\\ | ||
| − | \textbf{( | + | \textbf{(4)}\ \text{none of these} </math> |
==Solution== | ==Solution== | ||
Latest revision as of 21:51, 24 February 2020
Problem 26
The base of a triangle is 
inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:
Solution
Let the triangle be 
where 
is the base. Then let the parallels be 
and 
, where is closer to our base .
It's obvious that 
, where 
, so 
. Since we know 
, 
Hence our answer is 
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
