Question

ABCD is a trapezium in which AB II CD and P, Q are points on AD and BC
respectively such that PQ II DC. If PD=18 cm, BQ=35 cm and QC=15 cm, find AD
(See figure)
B
35 cm
Q
18 cm
15 cm
с
(A) 55 cm
(B) 57 cm
(C) 60 cm
(D) 62 cm
​

Answer 1

$\huge \mathscr{\orange {\underline{\pink{\underline {★Answer★:-}}}}}$

In △ABD,  PO∥AB        [∵ PQ∥AB ]

⇒  DP/AP = DO/OB      [ By basic proportionality theorem ]          ------ ( 1 )

⇒  In △BDC, OQ∥DC    [ ∵ PQ∥DC ]

BQ/QC=OB/OD     [ By basic proportionality theorem ]

⇒  QC/BQ =OD/OB         ----- ( 2 )

⇒  DP/AP=QC/BQ        [ From ( 1 ) and ( 2 ) ]

⇒  18/AP = 15/35

∴   AP = 18/15 × 35 = 42

∴   AD = AP+DP = 42+18 = 60cm

Hence,

Option C) 60 Cm