Difference between revisions of "1999 AMC 8 Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | + | We need the same perimeter as a <math>60</math> by <math>20</math> rectangle, but the greatest area we can get. right now the perimeter is <math>160</math>. To get the greatest area while keeping a perimeter of <math>160</math>, the sides should all be <math>40</math>. that means an area of <math>1600</math>. Right now, the area is <math>20 \times 60</math> which is <math>1200</math>. <math>1600-1200=400</math> which is <math>\boxed{D}</math>. | |
| + | |||
| + | ==Video(By YippieMath)== | ||
| + | https://www.youtube.com/watch?v=glxI3hrkAlA | ||
==See Also== | ==See Also== | ||
Latest revision as of 17:33, 17 May 2025
Problem
A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?
Solution
We need the same perimeter as a 
by 
rectangle, but the greatest area we can get. right now the perimeter is 
. To get the greatest area while keeping a perimeter of , the sides should all be 
. that means an area of 
. Right now, the area is 
which is 
. 
which is 
.
Video(By YippieMath)
https://www.youtube.com/watch?v=glxI3hrkAlA
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
