question_answer

A chord of a circle of radius 10 cmsubtends a right angle at the centre. Then, the area of the minor segment is

A) \[78.5\,c{{m}^{2}}\]

B) \[28.5c{{m}^{2}}\]

C) \[50c{{m}^{2}}\]

D) \[128.5c{{m}^{2}}\]

Correct Answer: B

Solution :

Given, radius of a circle, AO =10 cm and \[\angle AOC=90{}^\circ \]   Area of \[\Delta AOC=\frac{1}{2}\times OA\times OC\] \[=\frac{1}{2}\times 10\times 10=50c{{m}^{2}}\] Area of sector OAECO \[=\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}\] \[=\frac{90{}^\circ }{360{}^\circ }\times 3.14\times {{\left( 10 \right)}^{2}}=\frac{314}{4}=78.5c{{m}^{2}}\] Area of minor segment AECDA = Area of sector OAECO - Area of \[\Delta AOC\] \[=\text{ }78.5\text{ }-\text{ }50\text{ }=\text{ }28.5\text{ }c{{m}^{2}}\]