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 `AX"||"BY` such that `angleAand angleB` are acute angles.
3. Divide AX and BY in 3 and 5 parts equally by compass and mark `A_(1),A_(2),A_(3),B_(1),B_(2),B_(3),B_(4)and B_(5)` respectively.
4. Join `A_(3)B_(5)` which intersect AB at P and divides `AP: PB = 3:5.`
Hence P is the required point on AB which divide it in `3:5.`
Verification (Justicicatio) : `In DeltaAA_(3)P and DeltaBB_(5)P`
`" "AX"||"BY" "("by construction")`
`" "angleA=angleB" "("alt.angles")`
`" "angleA_(3)PA=angleB_(5)PB" "("vertically opp. angles")`
`therefore" "DeltaA A_(3)P~DeltaBB_(5)P" "("by AA criterion of similarity")`
`implies" "(A A_(3))/(BB_(5))=(AP)/(BP)" "("let ecah equal part = c cm" thereforeA A_(1)=A_(1)A_(2)=B_(1)B_(2)...=x)`
`implies" "(3x)/(5x)=(AP)/(BP)
implies" "AP:BP=3:5`