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.
(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
$=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AB}-616$
$=\frac{1}{2}(55+45) \times 30-616$
$=1500-616$
$=884 \mathrm{~m}^{2}$
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