Difference between revisions of "1950 AHSME Problems/Problem 32"
| Line 9: | Line 9: | ||
\textbf{(E)}\ 4\text{ ft}</math> | \textbf{(E)}\ 4\text{ ft}</math> | ||
==Solution== | ==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. <math>\text{(D)}</math> | ||
==See Also== | ==See Also== | ||
{{AHSME 50p box|year=1950|num-b=31|num-a=33}} | {{AHSME 50p box|year=1950|num-b=31|num-a=33}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 13:58, 5 July 2012
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
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. 
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 | ||
