Difference between revisions of "1951 AHSME Problems/Problem 30"
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<math> \textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these} </math> | <math> \textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these} </math> | ||
− | ==Solution== | + | ==Solution 1== |
The two pole formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>. | The two pole formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
The two lines can be represented as <math>y=\frac{-x}{5}+20</math> and <math>y=\frac{4x}{5}</math>. | The two lines can be represented as <math>y=\frac{-x}{5}+20</math> and <math>y=\frac{4x}{5}</math>. | ||
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<math>y=\frac{4\cdot 20}{5}</math> | <math>y=\frac{4\cdot 20}{5}</math> | ||
− | + | <math>y=16</math> <math>\boxed{16 \textbf{ (C)}}</math> | |
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== See Also == | == See Also == |
Revision as of 20:26, 19 March 2017
Contents
Problem
If two poles and high are apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
Solution 1
The two pole formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is , or .
Solution 2
The two lines can be represented as and . Solving the system,
So the lines meet at an of 20.
Solving for the height they meet,
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Problem 31 | |
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All AHSME Problems and Solutions |
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