question_answer

In a trapezium ABCD If \[\mathbf{AB}\parallel \mathbf{CD}\], thee \[\mathbf{A}{{\mathbf{C}}^{2}}+\mathbf{B}{{\mathbf{D}}^{2}}\]is equal to:

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

B)  \[A{{B}^{2}}+C{{D}^{2}}+2AD.BC\]

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

D)  \[B{{C}^{\mathbf{2}}}+A{{D}^{2}}+2BC.AD\]

Correct Answer: A

Solution :

(a): In \[\Delta ABD,\angle A\] is acute. So \[B{{D}^{2}}=A{{D}^{2}}+\text{ }A{{B}^{2}}-2AB.AQ\]              ---(i) In \[\Delta ABC,\angle B\] is acute. So, \[A{{C}^{2}}=B{{C}^{2}}+A{{B}^{2}}-2AB.AD\]            ?.(ii) Adding (i) and (ii) \[A{{C}^{2}}+B{{D}^{2}}=(B{{C}^{2}}+A{{D}^{2}})+2AB\{AB-BP-AQ\}\]\[=(B{{C}^{2}}+A{{D}^{2}})+2AB.PQ\] \[=B{{C}^{2}}+A{{D}^{2}}+2AB.CD~~~~~\left[ \therefore PQ=DC \right]\]