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

Though counterintuitive, the answer is not E!!!

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. $\text{(D)}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
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1950 AHSME Problems/Problem 32

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