Difference between revisions of "2005 PMWC Problems/Problem T7"

(Solution)
(Solution)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
<geogebra>8adbc5cf46e8c80136a021d43648b8653cfbf839</geogebra>
+
<math><geogebra>8adbc5cf46e8c80136a021d43648b8653cfbf839</geogebra></math>
  
 
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 14:31, 23 October 2021

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 $\pi=3.14$ or $\pi=22/7$.

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 $4\pi*\frac{2}{3}=\frac{8}{3}\pi$. 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 $\frac{\pi}{6}$. Adding, we get $\frac{8}{3}\pi+\frac{\pi}{6}+\frac{\pi}{6}=3\pi$. 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