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# A sprinkler system on a farm is set to spray water over a distance of 35 meters and to rotate through an angle of 140degrees. find the area

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A sprinkler system on a farm is set to spray water over a distance of 35 meters and to rotate through an angle of 140degrees. find the area of the region that can be watered.

Guest Sep 23, 2014

#3
+20598
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A sprinkler system on a farm is set to spray water over a distance of 35 meters and to rotate through an angle of 140degrees. find the area of the region that can be watered

$$\\ \small{\text{ A=\pi r^2  is the area of the circle.  A=\pi r^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right)  is the area of the Region can be watered. }} \\ \small{\text{  r=35\;m.\quad A=\pi *35^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) = \pi *35^2 *\frac{7}{18} = 1496.62\;m^2 }}\\ \small{\text{ The area of the Region can be watered is  {1496.62\;m^2 }  }}$$

heureka  Sep 24, 2014
#1
+92596
+5

A = (1/2)r2Θ  and Θ is in rads, so converting 140° to rads, we have

A = (1/2)(35m)2(2.44) = 1494.5m2

CPhill  Sep 23, 2014
#2
+5

Thanks and just for laughs.. what does the butcher weigh? meat.

Guest Sep 23, 2014
#3
+20598
+8
$$\\ \small{\text{ A=\pi r^2  is the area of the circle.  A=\pi r^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right)  is the area of the Region can be watered. }} \\ \small{\text{  r=35\;m.\quad A=\pi *35^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) = \pi *35^2 *\frac{7}{18} = 1496.62\;m^2 }}\\ \small{\text{ The area of the Region can be watered is  {1496.62\;m^2 }  }}$$