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Area of a Sector – Extension Problems

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.area of a sector extension_1

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  =  theta/360~*~pir2

=  270/360 x  3.14  x  62

=  84.8 m2

So the area of grass available to the cow is 84.8 m2

 

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 extension_2

Area of a sector = theta/360~*~pi r2

=  135/360 x  3.14  x  1052 (since r = 105 m)

=  12,981.9 m2

Hence the area illuminated by the searchlight is 12,981.9 m2.

 

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. area of a sector extension_3

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  =  a2

=  70 x 70

=  4900 m2

Area available to the goat  =  4900/2

=  2450 m2

Angle formed by the sector is 90°

Area of the sector  =  theta/360~*~pir2

2450  =  90/360x 3.14 x r2

r  =  sqrt{2450~*~4/3.14}

=  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.area of a sector extension_4

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  =  theta/360~*~pir2

=  75/360x 3.14 x 62

=  23.6 m2 (to 1 decimal place)

So the illuminated area  =  23.6 m2

Area of the stage  =  lb

=  20 x 14

=  280 m2

So percentage of stage area illuminated  =  23.6/280x 100

=  8.4%