Solved Question of Exercise 10.1 of class 11th

Q. No 2  :
                        The base of an equilateral triangle with side 2a lies along the y-axis such that the mid point of the base is at origin.Find the vertices of triangle.



Sol  :

      
              BC is the base of ∆ABC such that
                                 
                                    BO=OC=a

                                    AB = AC = 2a

             Now            AO² =  AB²  -   BO²       


                                                                                                    
                                             

                  ∴               AO = (4a² – a² )^1/2

                                                               

                                                             = 3^1/2  a


                 ∴         The point A lying on x-axis is (3^1/2 a, 0).
The point B is (0, a) and the point C is (0, -a). When A lies on the left of y-axis, the vertices of the triangle are (-3^1/2 a, 0), (0,a) and (0, -a).