Difference between revisions of "1950 AHSME Problems/Problem 32"
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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. | 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. | ||
| − | <math>x^2 + 7^2 = 25^2</math> | + | <math>x^2 + 7^2 = 25^2</math>\\ |
| − | <math>x^2 = 625 - 49</math> | + | <math>x^2 = 625 - 49</math>\\ |
| − | <math>x^2 = 576</math> | + | <math>x^2 = 576</math>\\ |
| − | <math>x = 24</math> | + | <math>x = 24</math>\\ |
== See Also == | == See Also == | ||
Revision as of 21:02, 24 December 2023
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.
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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. 
