Here are further examples of area of a sector. These are advanced problems that involve other mathematical concepts as well.

**Example 1** : A cow is tethered by a 6 m rope to a corner post in a paddock. What is the area of grass (shaded in the diagram) on which the cow is able to graze. Give your answer correct to 1 decimal place.

Here the cow can go through an angle of (360° – 90°) = 270° across the farm. This represents a sector of 270°.

Area of the sector = r^{2}

= x 3.14 x 6^{2}

= 84.8 m^{2}

So the area of grass available to the cow is 84.8 m^{2}

**Example 2** : A searchlight lights up the ground to a distance of 105 m. What area does the searchlight illuminate if it can swing through an angle of 135°. Give your answer correct to 1 decimal place.

Area of a sector = r^{2}

= x 3.14 x 105^{2 }(since r = 105 m)

= 12,981.9 m^{2}

Hence the area illuminated by the searchlight is 12,981.9 m^{2}.

**Example 3** : A goat is tethered to a corner post in a square field with length 70 m. What should be the length of the rope so that the goat can eat only half the grass in the field.

Here we need to find the area of the sector that is half the area of the square field. Then we need to find the length of the rope (radius r) given the area of the sector.

Area of the square field = a^{2}

= 70 x 70

= 4900 m^{2}

Area available to the goat =

= 2450 m^{2}

Angle formed by the sector is 90°

Area of the sector = r^{2}

2450 = x 3.14 x r^{2}

r =

= 55.9 m (to 1 decimal place)

The length of the rope should be 55.9 m for the goat to be able to eat half the grass of the square field.

**Example 4** : A beam of light is projected onto a theatre stage as shown in the diagram.

a) Calculate the illuminated area (correct to 1 decimal place)

b) What percentage of the total stage area is illuminated by the light beam

We need to find the area of the illuminated area by finding the area of the sector, then find the area of the stage, and finally, the proportion of the illuminated area to the total stage area.

Area of the sector = r^{2}

= x 3.14 x 6^{2}

= 23.6 m^{2} (to 1 decimal place)

So the illuminated area = 23.6 m^{2}

Area of the stage = lb

= 20 x 14

= 280 m^{2}

So percentage of stage area illuminated = x 100

= 8.4%