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

AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = 5^{2} + 12^{2}

⇒ AC^{2} = 25 + 144

⇒ AC^{2} = 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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