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 =
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(i) 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]
∠
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