1950 AHSME Problems/Problem 32
Contents
Problem
A foot ladder is placed against a vertical wall of a building. The foot of the ladder is feet from the base of the building. If the top of the ladder slips feet, then the foot of the ladder will slide:
Solution 1
By the Pythagorean triple , the point where the ladder meets the wall is feet above the ground. When the ladder slides, it becomes feet above the ground. By the Pythagorean triple, The foot of the ladder is now feet from the building. Thus, it slides .
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.
Since the top of the ladder slipped by 4 ft the new height is . The base of the ladder has moved so the new base is say . The hypotenuse remains the same at 25ft. So,
Disregarding the negative solution to equation the solution to the problem is .
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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