Indian Institute of Foreign Trade 2011 Solved Paper

Let ABC be the arc as shown in the figure. It is known that the diameter of the circle = 84 cm. Hence, the radius of the circle = 42 cm.We can see that both CO and BO are equivalent to the radius of the circle. Hence,$\mathit{C}\mathit{O}=42$ cm and $\mathit{B}\mathit{D}=\mathit{B}\mathit{O}-\mathit{O}\mathit{D}=42-21=21\mathit{c}\mathit{m}.$In triangle ODC $\mathrm{cosDOC}=\frac{21}{42}$
Hence $\angle \mathit{D}\mathit{O}\mathit{C}=60°$
In triangle BOC, OB = OC. Hence, we can say that $\angle \mathit{O}\mathit{C}\mathit{B}=\angle \mathit{O}\mathit{B}\mathit{C}=60°.$
Therefore, OBC is an equilateral triangle. Hence, $\mathit{O}\mathit{B}=\mathit{O}\mathit{C}=\mathit{B}\mathit{C}=42\mathit{c}\mathit{m}.$
Hence, we can say that the length of the chord of ‘half of the arc’ = 42 cm. Therefore, option C is the correct answer.