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**+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**+5 **

A = (1/2)r^{2}Θ 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) = 1494.5m^{2}

CPhill
Sep 23, 2014

#3**+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