In a circle AB is a the diameter and CD is a chord AB and CD produced meet at a point P outsid...........

Quantitative Aptitude (CGL) Geometry

In a circle, AB is a the diameter and CD is a chord. AB and CD produced meet at a point P, outside the circle. If PD = 15.3 cm, CD = 11.9 cm and AP = 30.6 cm,then the radius of the circle is is:

This questions was previously asked in

SSC CGL 9th March 2020 Shift-3

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8 cm

8.5 cm

9cm

7.5 cm

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Explanation:

From the property,
PA $$\times PB = PC \times PD$$
30.6 $$\times PB = (PD + CD) \times 15.3$$
30.6 $$\times PB = (15.3 + 11.9) \times 15.3$$
30.6 $$\times PB = 27,2 \times 15.3$$
PB = 416.16/30.6 = 13.6 cm
Diameter (AB) = PA - PB = 30.6 - 13.6 = 17 cm
Radius = AB/2 = 17/2 = 8.5 cm

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