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
Answer & explanation
Correct answer: option 2
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