Question

To divide a line segment AB in the ratio 5:6, draw a ray AX such that `angleBAX` is an acute angle, the draw a ray BY parallel to AX and the points `A_(1),A_(2),A_(3),….." and " B_(1),B_(2),B_(3),…..` are located to equal distances on ray AX and BY, respectively. Then, the points joined are
A. `A_(5) " and " B_(6)`
B. `A_(6) " and " B_(5)`
C. `A_(4) " and " B_(5)`
D. `A_(5) " and " B_(4)`

Answer 1

Correct Answer - A
(a) Given , a line segment AB and we have to divide it in the ratio 5:6.

Steps of construction
1. Draw a AX making an acute `angleBAX`.
2. Draw a ray BY parallel to AX by making `angleABY` equal to `angleBAX`.
3. Now, locate the points `A_(1),A_(2),A_(3),A_(4) " and " A_(5)`(m=5) on AX and `B_(1),B_(2),B_(3),B_(4),B_(5) " and " B_(6)`s(n=6) such that all the points are at equal distance from each other.
4. Join `B_(6)A_(5)`. Let it intesect AB at a point C.
Then, AC:BC=5:6