question_answer

ABC is a triangle with \[\angle A={{30}^{o}},\] \[BC=10\text{ }cm\]. The area of the circum circle of the triangle is

A)  \[100\pi \,sq\,m\]

B)  \[5\,sq\,cm\]

C)  \[25\,sq\,cm\]

D)  \[\frac{100\pi }{3}\,sq\,cm\]

Correct Answer: A

Solution :

In \[\Delta ABC,\]\[\angle A={{30}^{o}}\] \[BC=10cm\]      O is the centre of circle                 \[\therefore \]  \[\angle BOC={{60}^{o}}\]and OB and OC are the radius \[\therefore \]  \[\angle OBC=\angle OCB={{60}^{o}}\] \[\Rightarrow \]\[\Delta OBC\] OBC is an equilateral triangle \[\therefore \] S Radius of circle is \[OB=OC=BC=10\text{ }cm\] Now, area of the circum circle is \[\pi {{r}^{2}}\]                 \[=\pi {{(10)}^{2}}=100\pi \,\,sq\,cm\]