ΔPQR is inscribed in a circle. The bisector of ∠P cuts QR at S and the circle at T. If PR = 5 cm, PS = 6 cm and ST = 4 cm, then the length (in cm) of PQ is

12

We have,

PR = 5 cm

PS = 6 cm

ST = 4 cm

According to the question

PT = (PS + ST)

PT = 6 + 4= 10 cm

Now, In ΔPQS and ΔPTR

∠P = ∠P

∠PQR = ∠PTR (Angles made by same chord on circumference)

So, ΔPQS ∼ ΔPTR 

= \(\frac{PS}{PR}\) = \(\frac{PQ}{PT}\) 

= \(\frac{6}{5}\) = \(\frac{PQ}{10}\) 

= 5PQ = 60

= PQ = 12 cm