Difference between revisions of "2000 AMC 10 Problems/Problem 18"
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<math>\mathrm{(A)} 24 \qquad\mathrm{(B)}\ 27 \qquad\mathrm{(C)}\ 39 \qquad\mathrm{(D)}\ 40 \qquad\mathrm{(E)}\ 42</math> | <math>\mathrm{(A)} 24 \qquad\mathrm{(B)}\ 27 \qquad\mathrm{(C)}\ 39 \qquad\mathrm{(D)}\ 40 \qquad\mathrm{(E)}\ 42</math> | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/j7Hi5I8INII - Happytwin | ||
==Solution== | ==Solution== |
Revision as of 15:31, 21 January 2021
Contents
Problem
Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?
Video Solution
https://youtu.be/j7Hi5I8INII - Happytwin
Solution
The area she sees looks at follows:
The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius .
Therefore the total area she can see is , which rounded to the nearest integer is .
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
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