Plz help me with this .
Asked by | 1st Jan, 2013, 11:15: PM
Expert Answer:
ABC is an equilateral triangle.
AB = BC = CA = 9 cm
O is the circumcentre of ABC.
OD is the perpendicular bisector of the side BC.
(O is the point of intersection of the perpendicular bisectors of the sides of the triangle)
In OBD and OCD,
OB = OC (Radius of the circle)
BD = DC (D is the mid point of BC)
OD = OD (Common)
OBD 
OCD (SSS congruence criterion)
BOD = COD (CPCT)
BOC = 2 BAC
= 2 × 60°
= 120° ( The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle)
In BOD,
Sin BOD 
Thus, the radius of the circle is 
.
ABC is an equilateral triangle.
AB = BC = CA = 9 cm
O is the circumcentre of ABC.
OD is the perpendicular bisector of the side BC.
(O is the point of intersection of the perpendicular bisectors of the sides of the triangle)
In OBD and OCD,
OB = OC (Radius of the circle)
BD = DC (D is the mid point of BC)
OD = OD (Common)
OBD OCD (SSS congruence criterion)
BOD = COD (CPCT)
BOC = 2 BAC
= 2 × 60°
= 120° ( The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle)
In BOD,
Sin BOD
.Answered by | 3rd Jan, 2013, 11:32: AM
Related Videos
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change