question_answer

A chord of a circle subtends an angle of \[{{60}^{o}}\] at the centre. If the length of the chord is 100 cm, find the area of the major segment.

A)  \[30720.5\,c{{m}^{2}}\]    

B)  \[31021.42\,c{{m}^{2}}\]

C)        \[30391.7\,c{{m}^{2}}\]   

D)         \[30520.61\,c{{m}^{2}}\]

Correct Answer: D

Solution :

In \[\Delta \,\,OAB,\] by angle sum property           \[={{60}^{o}}+\angle OAB+\angle OBA={{180}^{o}}\] \[\Rightarrow \] \[{{60}^{o}}+\angle OAB+\angle OBA={{180}^{o}}\] \[\Rightarrow \]\[2\angle OAB={{120}^{o}}\Rightarrow \angle OAB={{60}^{o}}\] \[\Rightarrow \] \[\Delta OAB\] is equilaterai triangle. \[\Rightarrow \]  \[r=100\text{ }cm\] Area of major segment = Area of major sector + Area of \[\Delta \,OAB\] \[=\frac{\left( {{360}^{o}}-{{60}^{o}} \right)}{{{360}^{o}}}\times \pi {{r}^{2}}+\frac{\sqrt{3}}{4}{{r}^{2}}\] \[=\frac{300}{360}\times \frac{22}{7}\times {{(100)}^{2}}+\frac{\sqrt{3}}{4}\times {{100}^{2}}\] \[=\frac{5}{6}\times \frac{22}{7}\times {{(100)}^{2}}+\sqrt{3}\times 2500\] \[=26190.48+4330.13=30520.61\text{ }c{{m}^{2}}\]