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Question -

ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one sixteenth of the area of ∆ABC. (C.B.S.E. 2005)



Answer -

In ∆ABC, P and Q are two points on AB and AC respectively such that
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

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