ABC is a right angled triangle in which `/_B=90^(@)` 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_(3),"......,"M_(n)` are on AC, the sum of the lenghts of `L_(1)M_(1),L_(2)M_(2),"......,"L_(n)M_(n)` is

Options

A. `(n(n+1))/((2)`
B. `(a(n-1))/(2)`
C. `(an)/(2)`
D. Impossible to find from the given data

Answer

C

Explanation

`:. (AL_(1))/(AB)=(L_(1)M_(1))/(BC) implies L_(1)M_(1)=(a)/(n+1)`
Similarly, `L_(2)M_(2)=(2a)/(n+1)`
`L_(3)M_(3)=(3a)/(n+1)`
`vdots " " vdots " " vdots " "`
`L_(n)M_(n)=(na)/(n+1)`
`L_(1)M_(1)+L_(2)M_(2)+"…."+L_(n)M_(n)`
`=(a)/((n+1))(1+2+3+"....."+n)`
`=(a)/((n+1))*(n(n+1))/(2)=(na)/(2)`