Difference between revisions of "2005 PMWC Problems/Problem T7"
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The big sector has a radius of 2 and is 2/3 of the total circle that it used to be a part of, and thus has an area of <math>4\pi*\frac{2}{3}=\frac{8}{3}\pi</math>. The two little sectors each have radius 1 and are 1/6 the total area of their circle, so the area of each of those is <math>\frac{\pi}{6}</math>. Adding, we get <math>\frac{8}{3}\pi+\frac{\pi}{6}+\frac{\pi}{6}=3\pi</math>. Then approximate. | The big sector has a radius of 2 and is 2/3 of the total circle that it used to be a part of, and thus has an area of <math>4\pi*\frac{2}{3}=\frac{8}{3}\pi</math>. The two little sectors each have radius 1 and are 1/6 the total area of their circle, so the area of each of those is <math>\frac{\pi}{6}</math>. Adding, we get <math>\frac{8}{3}\pi+\frac{\pi}{6}+\frac{\pi}{6}=3\pi</math>. Then approximate. | ||
Revision as of 10:30, 18 September 2008
Problem
Skipper’s doghouse has a regular hexagonal base that measures one metre on each side. Skipper is tethered to a 2-metre rope which is fixed to a vertex. What is the area of the region outside the doghouse that Skipper can reach? Calculate an approximate answer by using 
or 
.
Solution
<geogebra>8adbc5cf46e8c80136a021d43648b8653cfbf839</geogebra>
The big sector has a radius of 2 and is 2/3 of the total circle that it used to be a part of, and thus has an area of 
. The two little sectors each have radius 1 and are 1/6 the total area of their circle, so the area of each of those is 
. Adding, we get 
. Then approximate.
See also
| 2005 PMWC (Problems) | ||
| Preceded by Problem T6 |
Followed by Problem T8 | |
| I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
