question_answer

In a right angled \[\Delta ABC\] right angled at B, if P and Q are points on the sides AB and BC respectively, then:

A) \[A{{Q}^{2}}+C{{P}^{2}}=2{{(A{{C}^{2}}+PQ)}^{2}}\]

B) \[2(A{{Q}^{2}}+C{{P}^{2}})=A{{C}^{2}}+P{{Q}^{2}}\]

C) \[A{{Q}^{2}}+C{{P}^{2}}=A{{C}^{2}}+P{{Q}^{2}}\]

D) \[AQ+CP=\frac{1}{2}(AC+PQ)\]

Correct Answer: C

Solution :

[c] In right angled \[\Delta ABQ\] and \[\Delta CPB,\]

\[C{{P}^{2}}=C{{B}^{2}}+B{{P}^{2}}\]

and \[A{{Q}^{2}}=A{{B}^{2}}+B{{Q}^{2}}\]

\[C{{P}^{2}}+A{{Q}^{2}}=C{{B}^{2}}+B{{P}^{2}}+A{{B}^{2}}+B{{Q}^{2}}\]

\[=C{{B}^{2}}+A{{B}^{2}}+B{{P}^{2}}+B{{Q}^{2}}\]

\[=A{{C}^{2}}+P{{Q}^{2}}\]