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})].
Explanation
(stackrelfrown{AOB} = stackrelfrown{BOC} = stackrelfrown{COD} = stackrelfrown{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})