Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
 
(3 intermediate revisions by the same user not shown)
Line 14: Line 14:
 
<cmath>x^2 = 576</cmath>
 
<cmath>x^2 = 576</cmath>
 
<cmath>x = 24</cmath>
 
<cmath>x = 24</cmath>
 +
 +
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,
 +
 +
<cmath>20^2 + (7+y)^2 = 25^2</cmath>
 +
<cmath>400 + 49 + y^2 + 14y = 625</cmath>
 +
<cmath>y^2 + 14y - 176 = 0</cmath>
 +
<cmath>y^2 + 22y - 8y - 176</cmath>
 +
<cmath>x(y+22) - 8(y+22)</cmath>
 +
<cmath>(y-8)(y+22)</cmath>
 +
 +
Disregarding the negative solution to equation the solution to the problem is <math>\boxed{\textbf{(D)}\ 8\text{ ft}}</math>.
  
 
== See Also ==
 
== See Also ==

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_32&oldid=208328"