Question

Line PQ is such that P and Q are points on sides AB and AC respectively of triangleABC.If AP=1cm, PB=3cm, AQ=1.5cm and QC=4.5cm, prove that ar(triangleAPQ)=1/16 ar(triangleABC).​

Answer 1

Given :   P and Q are points on sides AB and AC respectively of triangle ABC.  AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm,

To find :  prove that ar(triangleAPQ)=1/16 ar(triangleABC).​

Solution:

AP = 1 cm

PB = 3 cm

AQ  = 1.5 cm

QC = 4.5 cm

AP/PB = 1/3

AQ/QC = 1.5/4.5 = 1/3

=> AP/PB = AQ/QC = 1/3

=> PQ  ║ AB

=>Δ APQ ≈ ΔABC

AB = AP + PB = 1 + 3 = 4  cm

AC = AQ +  QC = 1.5 + 4.5 =  6 cm

AP/AB = 1/4   or  AQ/AC = 1.5/6  = 1/14

Ratio  of Area of Similar triangle = (Ratio  of sides)²

=> Area Δ APQ / Area of Δ ABC  =      (1/4)²

=>Area Δ APQ / Area of Δ ABC  =      1/16

=> Area Δ APQ = (1/16)Area of Δ ABC

QED

Hence proved

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