Difference between revisions of "1999 AMC 8 Problems/Problem 5"
(→Solution) |
(→Solution) |
||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | + | We need the same perimeter as a 60 by 20 rectangle, but the greatest area we can get. right now the perimeter is 160. To get the greatest area while keeping a perimeter of 160, the sides should all be 40. that means an area of 1600. Right now, the area is 20 times 60 which is 1200. 1600-1200=\boxed{\text{(D)}\ 400}$. | |
==See Also== | ==See Also== |
Revision as of 10:04, 23 November 2020
Problem
A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?
Solution
We need the same perimeter as a 60 by 20 rectangle, but the greatest area we can get. right now the perimeter is 160. To get the greatest area while keeping a perimeter of 160, the sides should all be 40. that means an area of 1600. Right now, the area is 20 times 60 which is 1200. 1600-1200=\boxed{\text{(D)}\ 400}$.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.