question_answer

ABC is a right angled triangle in which \[\angle B=90{}^\circ \]and \[BC=a\]. If n points \[{{L}_{1}},\,{{L}_{2}},....{{L}_{n}}\] on AB are such that AB is divided in n + 1 equal parts and \[{{L}_{1}}{{M}_{1}},\,\,{{L}_{2}}{{M}_{2}},....,\,\,{{L}_{n}}{{M}_{n}}\] are line segments parallel to BC and \[{{M}_{1}},\,{{M}_{2}},...{{M}_{n}}\] are on AC, then the sum of the lengths of \[{{L}_{1}}{{M}_{1}},{{L}_{2}}{{M}_{2}},....{{L}_{n}}{{M}_{n}}\] is

A) \[\frac{a(n+1)}{2}\]

B) \[\frac{a(n-1)}{2}\]

C) \[\frac{an}{2}\]

D) None of these

Correct Answer: C

Solution :

[c] \[\frac{A{{L}_{1}}}{AB}=\frac{{{L}_{1}}{{M}_{1}}}{BC}\]

\[\therefore \frac{1}{n+1}=\frac{{{L}_{1}}{{M}_{1}}}{a}\]

\[\therefore {{L}_{1}}{{M}_{1}}=\frac{a}{n+1};\]

\[\frac{A{{L}_{2}}}{AB}=\frac{{{L}_{2}}{{M}_{2}}}{BC}\]

\[\therefore \frac{2}{n+1}=\frac{{{L}_{2}}{{M}_{2}}}{a}\]

\[\therefore {{L}_{2}}{{M}_{2}}=\frac{2a}{n+1},\,\]etc.

\[\therefore \] The required sum\[=\frac{a}{n+1}+\frac{2a}{n+1}+\frac{3a}{n+1}+\]\[...+\frac{na}{n+1}\]

\[=\frac{a}{n+1}(1+2+....+n)=\frac{a}{n+1}.\frac{n(n+1)}{2}=\frac{an}{2}\]