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 numb...
Question 1 Report
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 \(\pi = \frac{22}{7}\)].
Answer Details
Setting up. The figure \(ABCDEO\) is two-thirds of a circle, centre \(O\), radius \(AO=7\text{ cm}\). The sector angle is
\[\tfrac23\times360^\circ=240^\circ.\]The equal chords \(|AB|=|BC|=|CD|=|DE|\) divide the \(240^\circ\) arc into \(4\) equal arcs, each
\[\frac{240^\circ}{4}=60^\circ.\]So each of \(\triangle OAB,\ \triangle OBC,\ \triangle OCD,\ \triangle ODE\) has two sides equal to the radius \(7\) with a \(60^\circ\) angle between them, making each an equilateral triangle of side \(7\text{ cm}\).
Shaded portion = area of the \(240^\circ\) sector \(-\) area of the four triangles (it is the four segments between the chords and the outer arc).
Area of the \(240^\circ\) sector with \(\pi=\tfrac{22}{7}\):
\[\frac{240}{360}\times\frac{22}{7}\times7^2=\frac{2}{3}\times\frac{22}{7}\times49=\frac{2}{3}\times154=102.67\text{ cm}^2.\]Area of one equilateral triangle (side \(7\)):
\[\frac{\sqrt3}{4}\times7^2=\frac{\sqrt3}{4}\times49=21.22\text{ cm}^2.\]Four of them: \(4\times21.22=84.87\text{ cm}^2.\)
Shaded area:
\[102.67-84.87=17.8\approx \boxed{18\text{ cm}^2}.\](Equivalently, one \(60^\circ\) segment \(=\tfrac16(154)-21.22=25.67-21.22=4.45\text{ cm}^2\), and \(4\times4.45=17.8\text{ cm}^2.\))
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