Question

In the figure QR=5cm,QT=3cm.What is the ratio between the area of triangle PQT and area of triangle PRT ​

Answer 1

Solution :-

given that,

→ QR = 5 cm

→ QT = 3 cm

so,

→ TR = QR - QT = 5 - 3 = 2 cm .

now, let us assume that, ∠QTP is equal to θ . so ∠PTR will be (180 - θ) since QR is a straight line .

then,

→ Area of ∆PQT = (1/2) * QT * PT * sin (∠QTP)

→ Area of ∆PQT = (1/2) * QT * PT * sin θ

and,

→ Area of ∆PRT = (1/2) * QT * PT * sin (∠PTR)

→ Area of ∆PRT = (1/2) * TR * PT * sin (180 - θ)

→ Area of ∆PRT = (1/2) * TR * PT * sin θ

therefore,

→ Area of ∆PQT : Area of ∆PRT = (1/2) * QT * PT * sin θ : (1/2) * TR * PT * sin θ

→ Area of ∆PQT : Area of ∆PRT = QT : TR

→ Area of ∆PQT : Area of ∆PRT = 3 : 2 (Ans.)

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