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 the area of ∆APQ is one-sixteenth of the area of ∆ABC.

Answer 1

SOLUTION :  

Given : In ΔABC, PQ is a line segment intersecting AB at P and AC at Q. AP = 1 cm , PB = 3cm, AQ= 1.5 cm and QC= 4.5cm.

In ∆APQ and ∆ABC,

∠A = ∠A               [Common]

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

∆APQ ~ ∆ABC      [By SAS similarity]

We know that the ratio of the two similar triangles is equal to the ratio of the squares of their corresponding sides

ar∆APQ /ar∆ABC = (AP/AB)²

ar∆APQ /ar∆ABC = ( ¼)²

ar∆APQ /ar∆ABC = 1/16

ar∆APQ =  1/16 × ar∆ABC  

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Answer 2

$<b>$Solution

given : AP = 1 cm , PB = 3cm , AQ= 1.5 cm , QC = 4.5cm

to prove: ar(APQ) = 1/16 ar (ABC)

proof: AP = 1 cm  and PB = 3 cm then AB = 4 cm

since the sides are in proportion then the line PQ is parallel to BC

in triangle APQ and triangle ABC

angle A = angle A                                  (common)

angle APQ = angle B                            (alt. int. angles)

angle AQP = angle C                            (alt. int. angles)

therefore,                 triangle APQ ~ triangle ABC 

ar(APQ)/ ar(ABC) = AP²/ AB²         (area of two similar triangles is equal to the square of their                                                                                                                                                                                     coressponding side)

ar(APQ) / ar(ABC) = 12/ 42

ar (APQ) / ar (ABC) = 1/16

therefore                ar(APQ) = 1/ 16 ar(ABC)

hence proved...........