In the diagram, ABCDEO is two- thirds of a circle centre O. The radius AO is 7cm and /AB/ = /BC/ = /CD/ = /DE/. Calculate, correct to the nearest whole number, the area of the shaded portion. [Take \(\pi = \frac{22}{7}\)].
\(\stackrel\frown{AOB} = \stackrel\frown{BOC} = \stackrel\frown{COD} = \stackrel\frown{DOE} = 60°\)
In \(\Delta AOB\), Area of sector = \(\frac{60}{360} \times \frac{22}{7} \times 7^{2}\)
= \(25.67 cm^{2}\)
Area of triangle AOB = \(\frac{1}{2} \times \7 \times 7 \times \sin 60\)
= \(24.5 \times 0.866\)
= \(21.22 cm^{2}\)
Area of the 4 sectors = \(4 \times 25.67 = 102.68 cm^{2}\)
Area of the 4 triangles = \(4 \times 21.22 = 84.88 cm^{2}\)
Area of the shaded portion = \((102.68 - 84.88)cm^{2}\)
= \(17.80 cm^{2}\)
\(\approxeq 18 cm^{2}\)
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