Difference between revisions of "1950 AHSME Problems/Problem 32"
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==Solution== | ==Solution== | ||
| − | + | By the Pythagorean triple <math>(7,24,25)</math>, the point where the ladder meets the wall is <math>24</math> feet above the ground. When the ladder slides, it becomes <math>20</math> feet above the ground. By the <math>(15,20,25)</math> Pythagorean triple, The foot of the ladder is now <math>15</math> feet from the building. Thus, it slides <math>15-7=\textbf{(D)}8 \text{ ft}</math>. | |
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==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 19:21, 10 April 2013
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
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 
.
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 | ||
