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  • To divide a line segment AB in the ratio 5 ratio 7 first a ray AX is drawn so that angle BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is (a) 8 (b) 10 (c) 11 (d) 12

To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that\angle BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is
(A) 8                            (B) 10                          (C) 11                          (D) 12

Answers (1)

Answer(D) 12
Solution
 Given: \angle BAX is an acute angle.
  The required ratio is 5:7           
    Let m = 5, n = 7

Steps of construction
   1. Draw any ray AX making an acute angle with AB.
  2. Locate 12 points on AX at equal distances.   (Because here m+n= 12)
  3. Join A_{12}B 
4. Through the point A_{5} draw a line parallel to  A_{12}B intersecting AB at the point P.
  Then AP:PB= 5:7
  \because A_{5}P\parallel A_{12}B
  \therefore \frac{AA_{5}}{A_{5}A_{12}}= \frac{AP}{PB}  (By Basic Proportionality theorem)

       By construction \frac{AA_{5}}{A_{5}A_{12}}= \frac{5}{7}
    \therefore \frac{AP}{PB}= \frac{5}{7}
Hence the number of points is 12.

 

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