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 | ||
Line 7: | Line 7: | ||
\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 18:13, 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 | |
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All AHSME Problems and Solutions |
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