Geometry – Practice – Similar triangles and what else!!!

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?

  1. 10:3
  2. 2:1
  3. 7:3
  4. 8:3

geometry 3_1

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 “!


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