Question

In the figure (not drawn to scale) given below,P is a point on AB such that AP:PB = 4:3 .PQ is parallel to AC and QD is parallel to CP. In triangle ARC, angle ARC = 90 and in triangle PSQ = 90. The length of Qs is 6cm. If the ratio AP:PD = k then 3k =​

Answer 1

Explanation:

In Δ ABC, PQ|| AC

Thus, AP: PB= QC: QB= 4:3. ——–(i)

Also, In Δ PCB, QD|| PC

Thus, DB:PD = QB:QC.

Using (i), we can see that

DB:PD = 3:4

Also, we can write, PB: PD= (3+4):4= 7:4. ——–(ii)

Multiplying (i) and (ii) we get,

(AP/PB) * (PB/ PD) = (4/3) * (7/4)

=> AP:PD = 7:3

k= 7:3