Question:
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.
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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