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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 m, prove that area of ΔAPQ is one- sixteenth of the area of ABC.
Solution
We have,
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 m
In ΔAPQ and ΔABC
∠A = ∠A [Common]
`"AP"/"AB"="AQ"/"AC"` [Each equal to 1/4]
Then, ΔAPQ ~ ΔABC [By SAS similarity]
By area of similar triangle theorem
`("area"(triangleAPQ))/("area"(triangleABC))=1^2/4^2`
`rArr("area"(triangleAPQ))/("area"(triangleABC))=1^2/16xx"area"(triangleABC)`
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