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>(60)(10)=600</math> and the area of the square garden is <math>(30)(30)=900</math>, so the area increases by <math>900-600=\boxed{\text{(C)}\ 300}</math>.
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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?

$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

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 (ProblemsAnswer KeyResources)
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

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