question_answer

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

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

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

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

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

Correct Answer: A

Solution :

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