Difference between revisions of "1951 AHSME Problems/Problem 30"
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==Solution 1== | ==Solution 1== | ||
− | The two | + | The [[two poles 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>. |
==Solution 2== | ==Solution 2== | ||
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<math>\frac{-x}{5}+20=\frac{4x}{5}</math> | <math>\frac{-x}{5}+20=\frac{4x}{5}</math> | ||
− | <math>20=x</math> | + | <math>20=x.</math> |
So the lines meet at an <math>x</math>-coordinate of 20. | So the lines meet at an <math>x</math>-coordinate of 20. |
Latest revision as of 19:41, 17 August 2021
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 poles 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 -coordinate 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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