Draw a line segment of length 7cm. Find a point P on it which divides it in the ratio 3:5.

Explanation

Steps of construction
1. Draw a line segment AB=7 cm.
2. Draw a ray AX , making an acute `angleBAX`.
3. Along AX, mark 3+5=8 points
`A_(1),A_(2),A_(3),A_(4),A_(5),A_(6),A_(6),A_(7),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)B`.
From `A_(3)`, draw `A_(3)C||A_(8)` B meeting AB at C. [ by making an angle equal to `angleBA_(8)A " at " A_(3)`]
Then, C is the point on AB which divides it in the ratio 3:5.
Thus, AC:CB=3:5

Justification
Let `A A_(1)=A_(1)A_(2)=A_(2)A_(3)=A_(3)A_(4)= ....=A_(7)A_(8)=x`
In `DeltaABA_(8)` , we have
`A_(3)C||A_(8)B`
`:. (AC)/(CB)=(A A_(3))/(A_(3)A_(8))=(3x)/(5x)=(3)/(5)`
Hence, `AC:CB=3:5`