Question

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. If BC = 10, then the sum of the lengths of the seven line segments is 

(a) 33 

(b) 34 

(c) 35 

(d) 45

Answer 1

(c) 35

All the triangles formed with A as vertex of AB₁C₁, AB₂C₂, AB₃C₃, AB₄C₄, AB₅C₅, AB₆C₆, AB₇C₇, and ABC are similar. Let hk be the length of the side parallel to BC at a distance \(\frac{k}{8}\). AC from A, 

where k = 1, 2, ...., 7, i.e.,

B₁C₁ = h₁ = \(\frac{AC}{8}\), B₂C₂ = h₂ = \(\frac{2}{8}\) AC, B₃C₃ = h₃ = \(\frac{3}{8}\) AC, .................B₇C₇ = h₇ = \(\frac{7}{8}\) = AC

∴ ∆AB1C1 ~ ∆ABC, their sides are proportional

⇒ \(\frac{B_1C_1}{AC_1}\) = \(\frac{BC}{AC}\) ⇒ \(\frac{h_1}{\frac{1}{8}AC}\) = \(\frac{10}{AC}\)

Similarly, ΔAB₂C₂ ~ ∆ABC, ΔAB₃C₃ ~ ∆ABC, ..........., ΔAB₇C₇ ~ ∆ABC

∴ In general, \(\frac{h_k}{\frac{k}{8}AC}\) = \(\frac{10}{AC}\) = h_k = \(\frac{10k}{8}\)

∴ Required sum = h₁ + h₂ + h₃ + ......... + h₇

= \(\frac{10}{8}\) (1+2+3+.......+7) = \(\frac{10}{8}\) x \(\frac{7}{2}\) x (1+7) = 35

\(\big(\)∵ Sum of n terms = \(\frac{n}{2}(a+l)\)\(\big)\)