Difference between revisions of "2000 AMC 8 Problems/Problem 16"
(Created page with "The length of the rectangle is 1000/25=40 meters. The perimeter equals 1000/10=100 meters. Perimeter is 2L+2W. Solving for W, we get 10 inches. The area is 40*10=400 square meter...") |
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− | The length of the rectangle is 1000 | + | ==Problem== |
+ | |||
+ | In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters? | ||
+ | |||
+ | <math>\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | The length <math>L</math> of the rectangle is <math>\frac{1000}{25}=40</math> meters. The perimeter <math>P</math> is <math>\frac{1000}{10}=100</math> meters. Since <math>P_{rect} = 2L + 2W</math>, we plug values in to get: | ||
+ | |||
+ | <math>100 = 2\cdot 40 + 2W</math> | ||
+ | |||
+ | <math>100 = 80 + 2W</math> | ||
+ | |||
+ | <math>2W = 20</math> | ||
+ | |||
+ | <math>W = 10</math> meters | ||
+ | |||
+ | Since <math>A_{rect} = LW</math>, the area is <math>40\cdot 10=400</math> square meters or <math>\boxed{C}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=ib2x_b3LNj8 ~David | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:05, 15 April 2023
Contents
Problem
In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
Solution
The length of the rectangle is meters. The perimeter is meters. Since , we plug values in to get:
meters
Since , the area is square meters or .
Video Solution
https://www.youtube.com/watch?v=ib2x_b3LNj8 ~David
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.