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

Answer:

In △BPQ and △BAC

∠BPQ=∠BAC[∵PQ∥AC]

∠B=∠B[common]

∴ △BPQ∼△BAC  (By AA similarity)

ACPQ=BABP[BySSST]⟶(2)

Also,  BPAP=34⇒BPAP+1=34+1

⇒PBAP+PB=37⇒PBAB=37⇒ABPB=73⟶(2)

from (1) and (2), ACPQ=73

(ii) In △RAC and △PSQ

    ∠ARC=∠PAQ[900]

    ∠