Question

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.​

Answer 1

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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