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Difference between revisions of "1999 AMC 8 Problems/Problem 5"

(Created page with "The perimeter of the rectangle is <math>50(2) + 10(2) = 100 + 20 = 120</math> feet. This fence is used for a square, which has four sides, so each side is <math>30</math> feet lo...")
 
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The perimeter of the rectangle is <math>50(2) + 10(2) = 100 + 20 = 120</math> feet. This fence is used for a square, which has four sides, so each side is <math>30</math> feet long.
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==Problem==
  
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Now to compare areas. The area of the rectangle is <math>(50)(10)</math>, or <math>500</math> square feet. The square's area is <math>30^2</math> square feet, or <math>900</math> square feet. Subtracting the rectangular area from the square area, we have an increase of <math>400</math> square feet.
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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?
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<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==
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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>.
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==See Also==
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{{AMC8 box|year=1999|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 14:17, 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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