Q. 424.7( 6 Votes )

In a right

Answer :

Given:


BC = 12 cm


AB = 5 cm



Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.


Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


Property 3: Sum of all angles of a quadrilateral = 360°.


By property 1,


AP = AQ (Tangent from A)


BP = BR (Tangent from B)


CR = CQ (Tangent from C)


ABC is a right-angled triangle, by Pythagoras Theorem


AC2 = AB2 + BC2


AC2 = 52 + 122


AC2 = 25 + 144


AC2 = 169


AC = 169


AC = 13 cm


Clearly,


AQ + QC = AC = 13 cm


AP + RC = 13 cm [ AQ = AP and QC = RC]


Also,


AB + BC = 5 cm + 12 cm = 17 cm


AP + PB + BR + RC = 17 cm [ AB = AP + PB and BC = BR + RC]


AP + RC + PB + BR = 17 cm


13 cm + BR + BR = 17 cm [ AP + RC = 10 cm and PB = BR]


13 cm + 2BR = 17 cm


2BR = 17 cm 13 cm = 4 cm



BR = 2 cm


Now,


BPO = 90° [By property 2]


BRO = 90° [By property 2]


PBM = 90° [Given]


Now by property 3,


BPO + BRO + PBM + ROP = 360°


ROP = 360° - (BPO + BRO + PBM)


ROP = 360° - (90° + 90° + 90°)


ROP = 360° - 270°


ROP = 90°


Now, ROP = 90° and BP = BR which are adjacent sides


Quadrilateral PBRO is a square


PO = BR = 2 cm


Hence, Radius = 2 cm

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