Question

A water sprinkler is a device used to irrigate agricultural crops, lawns, landscapes, golf courses, and other areas. Water sprinklers can be used for residential, industrial, and agricultural usage. A water sprinkler is set to shoot a stream of water a distance of 21 m and rotate through an angle which is equal to the complementary angle of 10°.

1. What is the area of sector in terms of arc length?  (1)
2. What is the area of the watered region (in terms of π)?   (1)
3. 1. If the radius (r) changes to 28 m, find the angle θ so that the area of the watered region remains the same.  (2)
      OR
   2. If the radius (r) is increased from 21 m to 28 m and the angle remains the same, what is the increase in the area of the watered region?

Answer 1

i. Area of sector = `(("Arc length" xx "radius"))/2`

ii. Area of sector = `80/360 π xx 441` = 98π m²

iii. (A) `80/360π xx 441 = θ/360π xx 28^2`

`80/360π xx 441 = θ/360π xx 784`

Cancel π from both sides:

`80/360 xx 441 = θ/360 xx 784`

To remove the denominator 360 from both sides, multiply both sides of the equation by 360:

`360 xx (80/360 xx 441) = 360 xx (θ/360 xx 784)`

Now the 360 cancels out on both sides:

80 × 441 = θ × 784

θ = `(80 xx 441)/784`

θ = `(80 xx 9)/16`

θ = `(10 xx 9)/2`

θ = `90/2`

θ = 45°

OR

iii. (B) Increase in the area of the lawn watered

= `80/360π xx (784-441)`

= 239.56 m²