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  4. ABC is a right angled triangle in which angle B=90° and BC=a. If n points L1, L2, ldots, Ln on AB are such that AB is divided in n+1 equal parts and L1 M1, L2 M2, ldots, Ln Mn are line segments parallel to BC and M1, M2, ldots, Mn are on AC, then the sum of the lengths of L1 M1, L2 M2, ldots, Ln Mn is

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

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Solution:

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$\frac{A L_{1}}{A B}=\frac{L_{1} M_{1}}{B C} \therefore \frac{1}{n+1}=\frac{L_{1} M_{1}}{a}$
$\therefore L_{1} M_{1}=\frac{a}{n+1} ; \frac{A L_{2}}{A B}=\frac{L_{2} M_{2}}{B C}$
$\therefore \frac{2}{n+1}=\frac{L_{2} M_{2}}{a}$
$\therefore L_{2} M_{2}=\frac{2 a}{n+1}$, etc.
The The required sum $=\frac{a}{n+1}+\frac{2 a}{n+1}+\frac{3 a}{n+1}+\ldots+\frac{n a}{n+1}$
$=\frac{a}{n+1}(1+2+\ldots+n)=\frac{a}{n+1} \cdot \frac{n(n+1)}{2}=\frac{a n}{2}$