Given that ,AB= 5 cm and AC= 7 cm.
Also,`AP=(3)/(4)AB` and `AQ=(1)/(4)AC` …..(i)
From Eq. (i), `AP=(3)/(4).AB=(3)/(4)xx5=(15)/(4)cm`
Then, `PB=AB-AP=5-(15)/(4)=(20-15)/(4)=(5)/(4) cm` [ `:.` P is any point on the AB]
`:. AP:PB=(15)/(4):(5)/(4)implies AP:PB=3:1`
i.e, scale factor of line segment AB is `(3)/(1)`.
Again from Eq. (i), `AQ=(1)/(4)AC=(1)/(4)xx7=(7)/(4)cm`
Then, `QC=AC-AQ=7-(1)/(4)`
`=(28-7)/(4)=(21)/(4) cm` [`:.` Q is any point on the AC]
`:. AQ:QC=(7)/(4):(21)/(4)=1:3`
`implies AQ:QC =1:3`
i.e., scale factor of line segment AQ is `(1)/(3)`.
Steps of construction
1. Draw a line segment AB=5 cm.
2. Now draw a ray AZ making an acute `angleBAZ=60^(@)`.
3. Whit A as centre and radius equal to 7 cm draw an arc cutting the line AZ at C.
4. Draw a ray AX, making an acute ` angle BAX`.
5. Along AX , mark 1+3=4 points `A_(1),A_(2),A_(3), " and " A_(4)`.
Such that `A A_(1)=A_(1)A_(2)=A_(2)A_(3)=A_(3)A_(4)`
6. Join `A_(4)B` ltbr. 7. From `A_(3)` draw `A_(3)P||A_(4)B` meeting AB at P. [ by making an angle equal to `angleAA_(4)B`]
Then, P is the point on AB which divided it in the ratio 3:1.
So, AP:PB=3:1 ,brgt 8. Draw a ray AY, making an acute `angleCAY`.
9. Along AY, mark 3+1=4 points `B_(1),B_(2),B_(3) " and " B_(4)`
Such that `AB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)`
10. Join `B_(4)C`.
11. From `B_(1)` draw `B_(1)Q||B_(4)C` meeting AC at Q. [by making an angle equal to `angleAB_(4)C`]
Then, Q is the point on AC which divides in the ratio 1:3.
So, AQ:QC=1:3 ltbtgt 12. Finally, join PQ and its measurement is 3.25 cm.
