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

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==Problem==
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== Problem ==
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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:
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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>
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== Solution 1==
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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>.
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== 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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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:
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<cmath>x^2 + 7^2 = 25^2</cmath>
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<cmath>x^2 = 625 - 49</cmath>
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<cmath>x^2 = 576</cmath>
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<cmath>x = 24</cmath>
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Since the top of the ladder slipped by 4 ft the new height is <math>24 - 4 = 20 ft</math>. The base of the ladder has moved so the new base is say <math>(7+y)</math>. The hypotenuse remains the same at 25ft. So,
  
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<math>\textbf{(A)}\ 9\text{ ft} \qquad
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<cmath>20^2 + (7+y)^2 = 25^2</cmath>
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\textbf{(B)}\ 15\text{ ft} \qquad
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<cmath>400 + 49 + y^2 + 14y = 625</cmath>
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\textbf{(C)}\ 5\text{ ft} \qquad
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<cmath>y^2 + 14y - 176 = 0</cmath>
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\textbf{(D)}\ 8\text{ ft} \qquad
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<cmath>y^2 + 22y - 8y - 176</cmath>
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\textbf{(E)}\ 4\text{ ft}</math>
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<cmath>x(y+22) - 8(y+22)</cmath>
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==Solution==
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<cmath>(y-8)(y+22)</cmath>
  
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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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Disregarding the negative solution to equation the solution to the problem is <math>\boxed{\textbf{(D)}\ 8\text{ ft}}</math>.
  
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==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]]
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{{MAA Notice}}

Latest revision as of 21:10, 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.

\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]

Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$. The base of the ladder has moved so the new base is say $(7+y)$. The hypotenuse remains the same at 25ft. So,

\[20^2 + (7+y)^2 = 25^2\] \[400 + 49 + y^2 + 14y = 625\] \[y^2 + 14y - 176 = 0\] \[y^2 + 22y - 8y - 176\] \[x(y+22) - 8(y+22)\] \[(y-8)(y+22)\]

Disregarding the negative solution to equation the solution to the problem is $\boxed{\textbf{(D)}\ 8\text{ ft}}$.

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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