Question

How do I complete this problem on arc length?

Answer 1

a) Length of the arc:
`(25\pi )/(3)`
`m` or ≈ 26.18
`m`

b) Area of the sector:
`(125\pi )/(6)`
`m^2` or ≈ 65.45
`m^2`

c) 16
`\pi` rad or 2880°

Explanation:

If a is the length of the arc, r is the radius of the sector and 0 the central angle in radians.

Use the formula:

[set r = 5 m, 0 =
`(5\pi )/(3)` (given)]

`a=0` ·
`r`

⇒
`a=(5\pi )/(3) .`
`5`

⇒
`a=(25\pi )/(3)` ≈
`26.18`
`m`

The arc length of the watered sector is
`(25\pi )/(3)`
`m` or ≈ 26.18.

----------------------------------------------------------------------------------------------

Recall, the area A of a circle (obviously
`0=2\pi`
`rad`) with a radius
`r` is given by the formula:

`A=\pi r^2=(2\pi )/(2) r^2`

So the area of the circular sector with radius r and central 0 (in radians) will be:

[set r = 5m , 0 =
`(5\pi )/(3)` (given)]

`Asector=(0)/(2) r^2`

⇒
`Asector=(5\pi/3 )/(2) (5)^2`

⇒
`Asector` =
`(5\pi )/(6)(25)`

⇒
`Asector` =
`(125\pi )/(6)` ≈ 65.45
`m^2`

The area of the watered sector is
`(125\pi )/(6)`
`m^2` or ≈ 65.45
`m^2`.

-----------------------------------------------------------------------------------------------

From the given information we know that the sprinkler perform a full revolution (2
`\pi` rad) every 15 sec so we will make the following proportion:

`(15 sec)/( 2 min)` =
`(2\pi )/(0)`

⇒
`(15 sec)/(120 sec)` =
`(2\pi )/(0)`

⇒ 0 =
`(2\pi . 120 sec)/(15 sec)`

⇒ 0 =
`(240\pi )/(15)`

⇒ 0 =
`16\pi`

We can convert the angle of 0 =
`16\pi` rad to degrees as shown:

`0deg` = 16
`\pi` ×
`(180°)/(\pi )`

⇒
`0deg` = 2880°

Finally the sprinkler rotates 16
`\pi` rad or 2880° in 2 minutes.