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

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==Problem==
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== Problem ==
 
 
 
A <math>25</math> foot ladder is placed against a vertical wall of a building. The foot of the ladder is <math>7</math> feet from the base of the building. If the top of the ladder slips <math>4</math> feet, then the foot of the ladder will slide:
 
A <math>25</math> foot ladder is placed against a vertical wall of a building. The foot of the ladder is <math>7</math> feet from the base of the building. If the top of the ladder slips <math>4</math> feet, then the foot of the ladder will slide:
  
<math>\textbf{(A)}\ 9\text{ ft} \qquad
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<math>\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}</math>
\textbf{(B)}\ 15\text{ ft} \qquad
 
\textbf{(C)}\ 5\text{ ft} \qquad
 
\textbf{(D)}\ 8\text{ ft} \qquad
 
\textbf{(E)}\ 4\text{ ft}</math>
 
==Solution==
 
  
By the Pythagorean triple <math>(7,24,25)</math>, the point where the ladder meets the wall is <math>24</math> feet above the ground. When the ladder slides, it becomes <math>20</math> feet above the ground. By the <math>(15,20,25)</math> Pythagorean triple, The foot of the ladder is now <math>15</math> feet from the building. Thus, it slides <math>15-7=\textbf{(D)} 8 \text{ ft}</math>.
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== Solution ==
 +
By the Pythagorean triple <math>(7,24,25)</math>, the point where the ladder meets the wall is <math>24</math> feet above the ground. When the ladder slides, it becomes <math>20</math> feet above the ground. By the <math>(15,20,25)</math> Pythagorean triple, The foot of the ladder is now <math>15</math> feet from the building. Thus, it slides <math>15-7 = \boxed{\textbf{(D)}\ 8\text{ ft}}</math>.
  
==See Also==
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== See Also ==
 
{{AHSME 50p box|year=1950|num-b=31|num-a=33}}
 
{{AHSME 50p box|year=1950|num-b=31|num-a=33}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:02, 12 October 2020

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

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 $15-7 = \boxed{\textbf{(D)}\ 8\text{ ft}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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All AHSME Problems and Solutions

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