In a circle of radius 10 cm, PQ and PR are two chords such that PQ = PR = 12 cm and centre O lies outside of ∆PQR. Find the length of the chord QR.
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Given : In a circle of radius 10 cm, PQ and PR are two chords such that PQ = PR = 12 cm
centre O lies outside of ∆PQR.
To Find : length of the chord QR.
Solution:
PQ = PR = 12 cm
OM ⊥ PQ
=> PM = QM = 12/2 = 6 cm
Cos ∠QPO = PM/OP
=> Cos ∠QPO = 6/10
=> Cos ∠QPO = 3/5
=> Sin ∠QPO = 4/5 ( sin²x = 1 - cos²x)
Draw OTP , T mid point of QR
∠QPO = ∠QPT
Sin ∠QPT = QT/PQ
=> 4/5 = QT/12
=> QT = 48/5
QR = 2 * QT
=> QR = 2 * 48/5
= 19.2 cm
length of the chord QR. = 19.2 cm
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