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Difference between revisions of "1999 AMC 8 Problems/Problem 5"
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==Problem==
 
==Problem==
  
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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?
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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?
  
 
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500</math>
 
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500</math>
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==Solution==
 
==Solution==
  
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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>.
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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>.
  
 
==See Also==
 
==See Also==

Latest revision as of 13:17, 13 March 2021

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 $1200$. $1600-1200=400$ which is $\boxed{D}$.

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

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