Question

ABC is a right-angled triangle in which $\angle B={90}^{\circ }$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 parallel to BC and $M_1,M_2......,M_n$ are on AC then the sum of lengths of $L_1{M}_1,L_2{M}_2,......L_n{M}_n$ is

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

Answer 1

The correct option is C $\frac{an}{2}$

$\tan \theta =\frac{AB}{BC}=\frac{AB}{a}B{L}_1=L_1{L}_2=L_2{L}_3=k(n+1)k=AB\tan \theta =\frac{(n+1)k}{a}\frac{a\tan \theta }{n+1}=k{k}_i=\frac{\left(i\right)\left(a\tan \theta \right)}{n+1}Also,\tan \theta =\frac{k_i}{x_i}=\frac{\left(i\right)\left(a\tan \theta \right)}{(n+1)x_i}{x}_i=\frac{a_i}{(n+1)}$

Length Segment $L_i{M}_i=a-\frac{a_i}{(n+1)}$

$\sum L_i{M}_i=a\stackrel{n}{\sum _{i=1}}\left(1\right)-\frac{a}{n+1}\stackrel{n}{\sum _{i=1}}i=an-\frac{a}{2}\frac{n(n+1)}{(n+1)}=\frac{an}{2}$