Find a point on AB which divides AB in the ratio 4:5.AB = 7cm

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Solution

1. Draw a line segment AB = 7 cm.

2. Draw a ray AX making an acute angle $∠ B A X$ with AB.

3. Along AX, mark off 8(= 3 + 5) points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6} A_{7}$ and $A_{8}$ such that $A A_{1}$

$= A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5} = A_{5} A_{6} = A_{6} A_{7} = A_{7} A_{8}$

4. Join $A_{8}$ to B.

5. Through $A_{3}$ , draw a line $A_{3} | | A_{8}$ B intersecting AB at P.
The point P is the desired point which divides AB internally in the ratio of 4 : 5.

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