PSA_LEVEL-UP_QNR17. 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 ΔARC, ∠ARC=90 and in ΔPQS, ∠PSQ=90. The length of QS is 6 cm. What is the ratio of AP:PD?
Ans: 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. Option 3.
Note: In this question, the information that are not used and thus are ‘junk’ are:
a)In ΔARC, ∠ARC=90 and in ΔPQS, ∠PSQ=90.
b)The length of QS is 6 cm
The only concept of “ratio of sides” that has been used to solve this question pertains to” Similar Triangles “!