Question

$ABC$ is a right angled triangle in which $\angle B=90^{\circ }$ and $BC=a$. If $n$ points $L_1,L_2,\dots ,L_n$ on $AB$ are such that $AB$ is divided in $n+1$ equal parts and $L_1{M}_1,L_2{M}_2,\dots ,L_n{M}_n$ are line segments parallel to $BC$ and $M_1,M_2,\dots ,M_n$ are on $AC$, then the sum of the lengths of $L_1{M}_1,L_2{M}_2,\dots ,L_n{M}_n$ is

• A. $\frac{a(n+1)}{2}$
• B. $\frac{a(n-1)}{2}$
• C. $\frac{an}{2}$
• D. None of these

Answer 1

Answer: (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.
The The required sum $=\frac{a}{n+1}+\frac{2a}{n+1}+\frac{3a}{n+1}+\dots +\frac{na}{n+1}$
$=\frac{a}{n+1}(1+2+\dots +n)=\frac{a}{n+1}\cdot \frac{n(n+1)}{2}=\frac{an}{2}$