ABC is a right-angled triangle in which `angleB=90^(@) and BC=a.` If n points `L_(1),L_(2),…,L_(n)` on AB is divided in n+1 equal parts and `L_(1)M_(1), L_(2)M_(2),…,L_(n)M_(n)` are line segments paralllel 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

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)`