Difference between revisions of "1953 AHSME Problems/Problem 26"
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==Problem 26== | ==Problem 26== | ||
| − | The base of a triangle is <math> | + | The base of a triangle is <math>144</math> 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: |
<math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad | <math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad | ||
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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== | ||
Revision as of 19:13, 26 January 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
