question_answer

A chord of a circle subtends an angle of \[60{}^\circ \] at the centre of the circle. If the length of the chord is \[\text{1}0\text{ cm},\] then the area of the major segment is: (Take \[\pi =3.14\] and \[\sqrt{3}=1.732\])

A) \[304.97\,c{{m}^{2}}\]

B) \[295\,c{{m}^{2}}\]

C) \[310\,c{{m}^{2}}\]

D) \[335\,c{{m}^{2}}\] 

Correct Answer: A

Solution :

[a] We have, \[OA=OB\]   (Radii of circle)

\[\Rightarrow \,\,\,\angle A=\angle B=60{}^\circ \,\,\,\,\,\,\,\,\,\,\,\,\,\,(\angle AOB=60{}^\circ )\]

\[\Rightarrow \,\,\,\Delta AOB\]is equilateral triangle.

\[\therefore \,\,\,\,\,\,OA=OB=AB=10cm\]

and \[\theta =60{}^\circ \]

Area of the minor segment

\[=\text{Area of minor sector}-\text{Area of  }\Delta \text{AOB}\]

\[=\left( \frac{\pi {{r}^{2}}\theta }{360{}^\circ }-\frac{\sqrt{3}}{4}{{r}^{2}} \right)\]

\[=\left\{ \left( 3.14\times 10\times 10\times \frac{60{}^\circ }{360{}^\circ } \right)-\left( \frac{\sqrt{3}}{4}\times 10\times 10 \right) \right\}\]

\[=\left( \frac{157}{3}-25\times \sqrt{3} \right)=\left( \frac{157}{3}\,\,\frac{433}{10} \right)\]

\[=\frac{(1570-1299)}{30}=\frac{271}{30}=9.03c{{m}^{2}}\]

Area of major segment

\[=\pi {{r}^{2}}-\] Area of minor segment

\[=[(\pi \times 10\times 10)-9.03]\,c{{m}^{2}}\]

\[=\{(3.14\times 100)-9.03\}c{{m}^{2}}\]

\[=(314-9.03)c{{m}^{2}}=304.97c{{m}^{2}}\]