Difference between revisions of "1967 AHSME Problems/Problem 11"
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| + | The least possible value of the diagonal is when the side lengths are as close as possible. In this case, the perimeter is divisible by <math>4</math>, so we can make a square. Since the perimeter is <math>20</math>, <math>20</math> divided by <math>4</math> is <math>5</math>. So when <math>5</math> is the side length of the square, the diagonal is <math>5\sqrt2</math> or <math>\sqrt50</math>. | ||
<math>\fbox{B}</math> | <math>\fbox{B}</math> | ||
Revision as of 17:27, 21 November 2016
Problem
If the perimeter of rectangle 
is 
inches, the least value of diagonal 
, in inches, is:
Solution
The least possible value of the diagonal is when the side lengths are as close as possible. In this case, the perimeter is divisible by 
, so we can make a square. Since the perimeter is , divided by is 
. So when is the side length of the square, the diagonal is 
or 
.
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
