question_answer

P and Q are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then

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

B) \[\text{A}{{\text{Q}}^{\text{2}}}\text{ + C}{{\text{P}}^{\text{2}}}\text{ = }\frac{4}{5}\,\,\text{A}{{\text{C}}^{\text{2}}}\]

C) \[\text{A}{{\text{Q}}^{\text{2}}}-\text{C}{{\text{P}}^{\text{2}}}\text{ = }\frac{4}{5}\,\,\text{A}{{\text{C}}^{\text{2}}}\]

D) \[\text{A}{{\text{Q}}^{\text{2}}}\text{+ C}{{\text{P}}^{\text{2}}}\text{ = }\frac{5}{4}\,\,\text{A}{{\text{C}}^{\text{2}}}\]

Correct Answer: D

Solution :

\[A{{Q}^{2}}=A{{B}^{2}}+B{{Q}^{2}}=A{{B}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}\] \[=A{{B}^{2}}+\frac{B{{C}^{2}}}{4}\]                        ?.(i) Similarly   \[C{{P}^{2}}=B{{C}^{2}}+B{{P}^{2}}\]                 \[=B{{C}^{2}}+{{\left( \frac{AB}{2} \right)}^{2}}\]                 \[=\frac{A{{B}^{2}}}{4}+B{{C}^{2}}\]                        ?..(ii) \[\therefore \]   \[A{{Q}^{2}}+C{{P}^{2}}=A{{B}^{2}}\left( 1+\frac{1}{4} \right)+B{{C}^{2}}\left( 1+\frac{1}{4} \right)\]                 \[=\frac{5}{4}\,(A{{B}^{2}}+B{{C}^{2}})=\frac{5}{4}A{{C}^{2}}\]