 # ABCD is a field in the shape of a trapezium, `
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:

(i) total area of the four sectors,

(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m. Solution:

(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°

$=\frac{60^{\circ}}{360^{\circ}} \pi(14)^{2}+\frac{90^{\circ}}{360^{\circ}} \pi(14)^{2}+\frac{90^{\circ}}{360^{\circ}} \pi(14)^{2}+\frac{120^{\circ}}{360^{\circ}} \pi(14)^{2}$

$=\left(\frac{60^{\circ}}{360^{\circ}}+\frac{90^{\circ}}{360^{\circ}}+\frac{90^{\circ}}{360^{\circ}}+\frac{120^{\circ}}{360^{\circ}}\right) \pi(14)^{2}$

$=\left(\frac{360^{\circ}}{360^{\circ}}\right) \pi(14)^{2}$

$=616 \mathrm{~m}^{2}$

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

$=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AB}-616$

$=\frac{1}{2}(55+45) \times 30-616$

$=1500-616$

$=884 \mathrm{~m}^{2}$