To divide a line segment AB in the ratio 5:6, draw a ray AX such that \angle BAX is an acute angle, then draw a ray BY parallel to AX and the points A_{1},A_{2},A_{3}\cdotsand B_{1},B_{2},B_{3}\cdotsare located at equal distances on ray AX and BY, respectively. Then the points joined are
(A) A5 and B6            (B) A­ and B5               (C) A4 and B5             (D) A5 and B4

Answers (1)

Answer(A) A5 and B6            
Solution

  Given: \angle BAX and \angle ABY both are acute angles and AX parallel to BY
 The required ratio is 5:6 
 Let m = 5 , n = 6

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY\parallel AX.
3. Locate the points A_{1},A_{2},A_{3},A_{4},A_{5} on AX at equal distances 
4. Locate the points B_{1},B_{2},B_{3},B_{4},B_{5} on BY at distance equal to the distance between points on AX line.
5. Join A_{5}B_{6}.
Let it intersect AB at a point C in figure.
Then AC:CB= 5:6
Here \bigtriangleup AA_{5}C  is similar to \bigtriangleup BB_{6}C
Then \frac{AA_{5}}{BB_{6}}= \frac{5}{6}= \frac{AC}{BC}             
\thereforeby construction \frac{AA_{5}}{BB_{6}}= \frac{5}{6}
  \therefore \frac{AC}{BC}= \frac{5}{6}
 \therefore Points joined one A5 and B6.

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