In a circle of radius 10 cm, with center O.  PQ and PR are two chords each of length 12 cm.  PO intersects chord QR at point S.  The length of OS is?

2.8 cm

RS = SQ   PQ = PR = 12 cm

PM = 20 cm

QM =  \(\sqrt {20^2 - 12^2 }\) = \(\sqrt {400 - 144 }\)

     = \(\sqrt {256 }\)

QM = 16 cm

MS = 20 - MS

SQ² = PQ² - PS² = 12² - PS²

solving ⇒ 144 - y² + (20 - PS)² = 16²

⇒ PS = \(\frac{288}{40}\) = 7.2

∴ OS = OP - PS = 10 - 7.2 = 2.8 cm