question_answer

In a right angled \[\Delta ABC,\]\[\angle C=90{}^\circ \]and CD is perpendicular to AB. If \[AB\times CD=CA\times CB,\]then \[\frac{1}{C{{D}^{2}}}\]is equal to

A)  \[\frac{1}{A{{B}^{2}}}-\frac{1}{C{{A}^{2}}}\]                   

B)  \[\frac{1}{A{{B}^{2}}}-\frac{1}{C{{B}^{2}}}\]

C) \[\frac{1}{B{{C}^{2}}}+\frac{1}{C{{A}^{2}}}\]                  

D)  \[\frac{1}{B{{C}^{2}}}-\frac{1}{C{{A}^{2}}},\]if \[CA>CB\]

Correct Answer: D

Solution :

According to question,

In \[\Delta ABC,\]

\[\angle C=90{}^\circ ;CD\bot AB\]

In\[\Delta ABC,\frac{1}{2}\times CD\times AB=\frac{1}{2}\times AC\times BC\]     ?(i)

\[A{{B}^{2}}=B{{C}^{2}}+C{{A}^{2}}\]               ?(ii) (by Pythagoras theorem)

\[AB\times CD=AC\times BC\](given)

\[\left\{ \begin{align}   & A{{C}^{2}}=A{{D}^{2}}+C{{D}^{2}} \\  & B{{C}^{2}}=C{{D}^{2}}+B{{D}^{2}} \\ \end{align} \right\}\]

\[\sqrt{(B{{C}^{2}}+C{{A}^{2}})}\times CD=AC\times BC\]

\[(B{{C}^{2}}+C{{A}^{2}})\times C{{D}^{2}}=A{{C}^{2}}\times B{{C}^{2}}\]

\[\frac{1}{C{{D}^{2}}}=\frac{B{{C}^{2}}+C{{A}^{2}}}{A{{C}^{2}}\times B{{C}^{2}}}\]

\[\frac{1}{C{{D}^{2}}}=\frac{1}{A{{C}^{2}}}+\frac{1}{B{{C}^{2}}}\]