Question

ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m. Find the following:

1. total area of the four sectors,
2. area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.

Answer 1

(i) Area of fours sector = Area of sector having central angle 60° + Area of sector having central angle 90° + Area of sector having central angle 90° + Area of sector having central angle 120°

`= 60^circ/360^circ pi(14)^2+90^circ/360^circpi(14)^2+(90°)/360^circpi(14)^2+120^circ/360^circ(14)^2`

`=(60^circ/360^circ + "90^circ/360^circ +(90^circ)/360^circ+120^circ/360^circ)pi(14)^2`

`=((360^circ)/(360^circ))pi(14)^2`

= 616 m² 

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants

`=1/2 ("AB"+"BC")xx"AB"- 616`

= `1/2(55+45)×30-616`

= 1500 − 616

= 884 m²