Question
Download Solution PDFChord AB and diameter CD of a circle meet at the point P, outside the circle when it is produced, If PB = 8 cm, AB = 12 cm and distance of P from the centre of the circle is 18 cm, the radius (in cm) of the circle is closest to: [√41 = 6.4]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Secant Property
If chord AB and chord CD of a circle intersect at a point P, then
PA × PB = PC × PD
Calculation:
Let the radius of the circle = r
PA = 12 + 8 = 20 cm, PB = 8 cm, PC = (18 + r) cm, PD = (18 - r) cm
PA × PB = PC × PD
⇒ 8 × 20 = (18 - r) × (18 + r)
⇒ 160 = 324 - r2
⇒ r2 = 164
⇒ r = 12.8062
∴ The radius of the circle is closest to 12.8 cm
Last updated on May 28, 2025
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