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: [CBSE 2013C]
(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.

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°
$$\begin{gathered} =\frac{60°}{360°}\mathrm{\pi }{\left(14\right)}^2+\frac{90°}{360°}\mathrm{\pi }{\left(14\right)}^2+\frac{90°}{360°}\mathrm{\pi }{\left(14\right)}^2+\frac{120°}{360°}\mathrm{\pi }{\left(14\right)}^2 \\ =\left(\frac{60°}{360°}+\frac{90°}{360°}+\frac{90°}{360°}+\frac{120°}{360°}\right)\mathrm{\pi }{\left(14\right)}^2 \\ =\left(\frac{360°}{360°}\right)\mathrm{\pi }{\left(14\right)}^2 \\ =616{\mathrm{m}}^2 \end{gathered}$$

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants
$$\begin{gathered} =\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)\times \mathrm{AB}-616 \\ =\frac{1}{2}\left(55+45\right)\times 30-616 \\ =1500-616 \\ =884{\mathrm{m}}^2 \end{gathered}$$