Krishna
0

Step 1: Assume an imaginary rhombus and write the properties of the rhombus.

                         

                             In rhombus ABCD, AB = BC = CD = DA...............................(1)

                             We know that diagonals of a rhombus bisect each other perpendicularly.

                             That is AC ⊥ BD, ∠AOB = ∠BOC = ∠COD = ∠AOD =90° and


                             OA = OC = \frac{AC}{2}, OB = OD = \frac{BD}{2}..............................(2)


Step 2: Apply the Pythagoras theorem to right angles triangle in rhombus.

              EXAMPLE: Consider right angled triangle AOB

                                 AB^2 = OA^2 + OB^2  [By Pythagoras theorem]

                                 AB^2 = (\frac{AC}{2})^2 + (\frac{BD}{2})^2        [Since equation (2)]

                                 AB^2 = (\frac{(AC)^2}{4}) + (\frac{(BD)^2}{4})

                                 4 AB^2 = AC^2 + BD^2

                                          

                                 AB^2 + AB^2 + AB^2 + AB^2 = AC^2 + BD^2

                                                

                                 AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2      [Since equation (1)]

                          Hence proved

                          Thus the sum of the squares of the sides of a rhombus is equa

                          l to the sum of the squares of its diagonals.