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Difference between revisions of "1950 AHSME Problems/Problem 32"

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== Solution 2==
 
== Solution 2==
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We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
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<math>x^2 + 7^2 = 25^2</math>
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<math>x^2 = 625 - 49</math>
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<math>x^2 = 576</math>
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<math>x = 24</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 21:02, 24 December 2023

Problem

A $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:

$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$

Solution 1

By the Pythagorean triple $(7,24,25)$, the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{\textbf{(D)}\ 8\text{ ft}}$.

Solution 2

We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.

$x^2 + 7^2 = 25^2$ $x^2 = 625 - 49$ $x^2 = 576$ $x = 24$

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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All AHSME Problems and Solutions

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