Difference between revisions of "2017 AMC 10A Problems/Problem 3"
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Find the area of the entire rectangle, and subtract the beds. | Find the area of the entire rectangle, and subtract the beds. | ||
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To do this, the length of the garden is <math> 2+2+2+1+1+1+1=10 </math> | To do this, the length of the garden is <math> 2+2+2+1+1+1+1=10 </math> | ||
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The width is <math> 6+6+1+1+1=15 </math> | The width is <math> 6+6+1+1+1=15 </math> | ||
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Therefore, the area of the entire garden is <math> 15*10=150 </math> | Therefore, the area of the entire garden is <math> 15*10=150 </math> | ||
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Each flower bed is <math> 6*2=12 </math>, so the combined area of the beds is <math> 12*6=72 </math> | Each flower bed is <math> 6*2=12 </math>, so the combined area of the beds is <math> 12*6=72 </math> | ||
− | So, the total area of the walkways is <math> 150-72= </math> \boxed{\textbf{(B)}\ 78} | + | |
+ | So, the total area of the walkways is <math> 150-72= </math> <math> \boxed{\textbf{(B)}\ 78}</math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 16:13, 31 May 2021
Problem
Tamara has three rows of two -feet by -feet flower beds in her garden. The beds are separated and also surrounded by -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?
Solution
Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.
Since the width of Tamara's garden contains three margins, the total width is feet.
Similarly, the height of Tamara's garden is feet.
Therefore, the total area of the garden is square feet.
Finally, since the six flower beds each have an area of square feet, the area we seek is , and our answer is
Alternate Solution
Find the area of the entire rectangle, and subtract the beds.
To do this, the length of the garden is
The width is
Therefore, the area of the entire garden is
Each flower bed is , so the combined area of the beds is
So, the total area of the walkways is
Video Solution
~savannahsolver
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.