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Trapezium

ABC is a righ...

Question

ABC is a right angle triangle in which $∠ B = 90^{0}$ and BC=a. If n points $L_{1}$ , $L_{2}$ , ..., $L_{n}$ on AB are such that AB 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}

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B

\frac{a (n - 1)}{2}

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C

\frac{a n}{2}

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D

impossible to find from given data

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Solution

The correct option is C $\frac{a n}{2}$
$\frac{{A L}_{1}}{A B} = \frac{L_{1} M_{1}}{B C}; ∴ \frac{1}{n + 1} = \frac{L_{1} M_{1}}{a}; ∴ L_{1} M_{1} \frac{a}{n + 1}$
$\frac{{A L}_{2}}{A B} = \frac{L_{2} M_{2}}{B C}; ∴ \frac{2}{n + 1} = \frac{L_{2} M_{2}}{a}; ∴ L_{1} M_{2} \frac{2 a}{n + 1},$
$∴$ the required sum
$= \frac{a}{n + 1} + \frac{2 a}{n + 1} + \frac{3 a}{n + 1} + . . . + \frac{n a}{n + 1}$
$= \frac{a}{n + 1} (1 + 2 + 3 + . . . + n) = \frac{a}{n + 1} . \frac{n (n + 1)}{2} = \frac{a n}{2}$

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