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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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Though counterintuitive, the answer is not E!!!
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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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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 (3,4,5)*5=(15,20,25) triple, The foot of the ladder is now 15 feet from the building. Thus, it slides 15-7=8 ft. <math>\text{(D)}</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 22:10, 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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