Q19 of 168 Page 5

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 m, prove that area of is one-sixteenth of the area of .

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AP=1 cm, PB=3 cm,AQ=1.5cm,and QC=4.5 m


In APQ and ABC


∠A=∠A [Common]


AP/AB=AQ/AC [Each equal to 1/4]


APQABC [By SAS]


By area of similar triangle theorem


Area () =12


Area () 42


Area () =1


Area () 16 x ar()


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