In a triangle A B C , if the mid points of sides A B , B C , C A are ( 3 , 0 , 0 ) , ( 0 , 4 , 0 ) , ( 0 , 0 , 5 ) , respectively, then A B 2 + B C 2 + C A 2 =

Step-by-step solution

The midpoints of the sides are given as, A B , B C and C A are   3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 The coordinates of A , B and C A = 3 , - 4 , 5 B = 3 , 4 , - 5 C = - 3 , 4 , 5 And A B 2 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2 = ( 3 - 3 ) 2 + ( 4 + 4 ) 2 + ( - 5 - 5 ) 2 = 8 2 + 10 2 = 164 Similarly, B C 2 = ( - 3 - 3 ) 2 + ( 4 - 4 ) 2 + ( 5 + 5 ) 2 = 36 + 100 = 136 Similarly, C A 2 = ( - 3 - 3 ) 2 + ( 4 - 4 ) 2 + ( 5 + 5 ) 2 = 36 + 64 = 100 A B 2 + B C 2 + C D 2 = 164 + 136 + 100 = 400