question_answer

In trapezium\[ABCD\], if\[AB||CD\], then\[A{{C}^{2}}+B{{D}^{2}}\] is equal to:

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

B)  \[A{{B}^{2}}+C{{D}^{2}}+2AD.BC\]

C)  \[A{{B}^{2}}+C{{D}^{2}}+2AB.CD\]

D)  \[B{{C}^{2}}+A{{D}^{2}}+2BC.AD\]

Correct Answer: A

Solution :

 In \[\Delta ABC,\,\,\angle A\] is acute. So\[,\]\[B{{D}^{2}}=A{{D}^{2}}+A{{B}^{2}}-2AB.AQ\]     ? (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) \[\therefore \]      \[A{{C}^{2}}+B{{D}^{2}}=(B{{C}^{2}}+A{{D}^{2}})+2AB\]             \[(AB-BP-AQ)\]             \[=(B{{C}^{2}}+A{{D}^{2}})+2AB.PQ\]             \[=B{{C}^{2}}+A{{D}^{2}}+2AB.CD\]             \[\left[ \because \,\,PQ=DC \right]\]