Difference between revisions of "1999 AMC 8 Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | The perimeter of the rectangular garden is <math>2(50+10)=120</math> feet. A square with this perimeter has sidelength <math>120/4=30</math> feet. The area of the rectangular garden is <math>(50)(10=500</math> and the area of the square garden is <math>(30)(30)=900</math>, so the area increases by <math>900-500=\boxed{\text{(D)}\ 400}</math>. | + | The perimeter of the rectangular garden is <math>2(50+10)=120</math> feet. A square with this perimeter has sidelength <math>120/4=30</math> feet. The area of the rectangular garden is <math>(50)(10)=500</math> and the area of the square garden is <math>(30)(30)=900</math>, so the area increases by <math>900-500=\boxed{\text{(D)}\ 400}</math>. |
==See Also== | ==See Also== | ||
Revision as of 12:52, 21 October 2016
Problem
A rectangular garden 50 feet long and 10 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
The perimeter of the rectangular garden is 
feet. A square with this perimeter has sidelength 
feet. The area of the rectangular garden is 
and the area of the square garden is 
, so the area increases by 
.
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. 
