question_answer

If ABCD is a rhombus then \[A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+A{{D}^{2}}\] is equal to

A)  \[A{{D}^{2}}+B{{C}^{2}}\]           

B)  \[A{{O}^{2}}+O{{C}^{2}}\]

C)  \[A{{C}^{2}}+B{{D}^{2}}\]  

D)         \[2(A{{O}^{2}}+O{{B}^{2}})\]

Correct Answer: C

Solution :

 As diagonals of a rhombus bisect each other at right angles. \[\Rightarrow \]   \[AO=OC\]and\[BO=OD\] Applying Pythagoras theorem \[\Delta AOB\,\,\Delta AOD,\,\,\Delta DOC,\,\,\Delta BOC\]and on adding \[A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+A{{D}^{2}}\]             \[=2[A{{O}^{2}}+O{{C}^{2}}+B{{O}^{2}}+D{{O}^{2}}]\]             \[=2[2A{{O}^{2}}+2B{{O}^{2}}]\]             \[=4[A{{O}^{2}}+B{{O}^{2}}]\]             \[\left[ \because \,\,AO=\frac{AC}{2},\,\,BO=\frac{BD}{2} \right]\]             \[=A{{C}^{2}}+B{{D}^{2}}\] f