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