In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.

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Solution

Given: In parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠SPQ meets SR at M.

Let ∠SPQ = 2x.

⇒ ∠SRQ = 2x and ∠TPQ = x.

Also, PQ ∥ SR

⇒ ∠TMR = ∠TPQ = x.

In △TMR, ∠SRQ is an exterior angle.

⇒ ∠SRQ = ∠TMR + ∠MTR

⇒ 2x = x + ∠MTR

⇒ ∠MTR = x

⇒ △TPQ is an isosceles triangle.

⇒ TQ = PQ = 12 cm

Now,

RT = TQ − QR

= TQ − PS

= 12 − 9

= 3 cm

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