Question

$ABC$ 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. impossible to find from the given data

Answer 1

Answer: (C)

$\frac{an}{2}$

Let the length of each $(n+1)$ equal parts on $AB=x$
And total length of $AB=b$
Consider length $L_i{M}_i=p_i$ where $i=1,2,...,n$
Length $L_{1}A=nx=\frac{bn}{(n+1)}$
By congruity of $\Delta ABC$ and $\Delta A{L}_1{M}_1,\frac{a}{b}=\frac{p_1(n+1)}{bn}$
$\Rightarrow p_1=\frac{na}{n+1}$
Similarly, $p_2=\frac{(n-1)a}{n+1}$
$p_3=\frac{(n-2)a}{n+1}$
$⋮$
$⋮$
$p_n=\frac{a}{n+1}$
$\Rightarrow p_1+p_2+\cdots +p_n=\frac{na}{n+1}+\frac{(n-1)a}{n+1}+\cdots +\frac{2a}{n+1}+\frac{a}{n+1}$
$=\frac{a}{n+1}\{n+(n-1)+\cdots +2+1\}=\frac{a}{n+1}\left\{\frac{n(n+1)}{2}\right\}$
$=\frac{an}{2}$