question_answer

If ABC is a triangle, right angled at B and M, N are mid-points of AB and BC respectively, then what is the value of \[4\,\,(A{{N}^{2}}+C{{M}^{2}})\]?

A)  \[3\,\,A{{C}^{2}}\]

B)  \[4\,\,A{{C}^{2}}\]

C)  \[5\,\,A{{C}^{2}}\]

D)  \[6\,\,A{{C}^{2}}\]

Correct Answer: A

Solution :

In right angled \[\Delta ABC\]

In \[\Delta ABN,\]

\[A{{N}^{2}}=A{{B}^{2}}+B{{N}^{2}}=A{{B}^{2}}+\frac{B{{C}^{2}}}{4}\]              ?(i)

In \[\Delta CBM\,\,C{{M}^{2}}=B{{C}^{2}}+B{{M}^{2}}=B{{C}^{2}}+\frac{A{{B}^{2}}}{4}\] ?(ii)

From Eqs. (i) and (ii), we get

\[A{{N}^{2}}+C{{M}^{2}}=A{{B}^{2}}+\frac{A{{B}^{2}}}{4}+B{{C}^{2}}+\frac{B{{C}^{2}}}{4}\]\[=\frac{5\,(A{{B}^{2}}+B{{C}^{2}})}{4}\]\[\Rightarrow \]\[4\,\,(A{{N}^{2}}+C{{M}^{2}})=5A{{C}^{2}}\]