1950 AHSME Problems/Problem 32

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

1950 AHSC (Problems • Answer Key • Resources)
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1950 AHSME Problems/Problem 32

Contents

Problem

Solution 1

Solution 2

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