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प्रश्न
In the following figure, a tangent segment PA touching a circle in A and a secant PBC is shown. If AP = 15, BP = 10, find BC.
उत्तर
PA is a tangent segment and PBC is the secant.
`therefore PB xx PC=PA^2` ........(Relationship to Tangent-Secant Theorem)
`therefore 10xxPC=15^2`
`PC=225/10=22.5`
`Now , PB+BC=PC`
`therefore 10+BC=22.5`
`therefore BC=22.5-10= 12.5`
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संबंधित प्रश्न
In the following figure, ray PA is a tangent to the circle at A and PBC is a secant. If AP = 15, BP = 10, then find BC.
In the given figure, P is the point of contact.
(1) If m(arc PR) = 140°, ∠ POR = 36°, find m(arc PQ)
(2) If OP = 7.2, OQ = 3.2, find OR and QR
(3) If OP = 7.2, OR = 16.2, find QR.
In the figure given above, ‘O’ is the centre of the circle, seg PS is a tangent segment and S is the point of contact. Line PR is a secant.
If PQ = 3.6, QR = 6.4, find PS.
Solution:
PS2 = PQ × `square` ......(tangent secant segments theorem)
= PQ × (PQ + `square`)
= 3.6 × (3.6 + 6.4)
= 3.6 × `square`
= 36
∴ PS = `square` .....(by taking square roots)
