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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

Best Answer 

 #3
avatar+20598 
+8

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

$$\\ \small{\text{
$A=\pi r^2 $ is the area of the circle.
$ A=\pi r^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) $ is the area of the Region can be watered.
}}
\\
\small{\text{
$ r=35\;m.\quad A=\pi *35^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) = \pi *35^2 *\frac{7}{18} = 1496.62\;m^2$
}}\\
\small{\text{
The area of the Region can be watered is $ \textcolor[rgb]{1,0,0}{1496.62\;m^2 } $
}}$$

heureka  Sep 24, 2014
 #1
avatar+92596 
+5

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

140° x (pi/180°) = about 2.44 rads....so we have..

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

 

CPhill  Sep 23, 2014
 #2
avatar
+5

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

Guest Sep 23, 2014
 #3
avatar+20598 
+8
Best Answer

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

$$\\ \small{\text{
$A=\pi r^2 $ is the area of the circle.
$ A=\pi r^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) $ is the area of the Region can be watered.
}}
\\
\small{\text{
$ r=35\;m.\quad A=\pi *35^2 *\left( \frac{140\ensurement{^{\circ}}}{360\ensurement{^{\circ}}}\right) = \pi *35^2 *\frac{7}{18} = 1496.62\;m^2$
}}\\
\small{\text{
The area of the Region can be watered is $ \textcolor[rgb]{1,0,0}{1496.62\;m^2 } $
}}$$

heureka  Sep 24, 2014

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