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ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in ΔABC. The radius of the circle is 

A. 1 cm 

B. 2 cm 

C. 3 cm 

D. 4 cm

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Answer is B.  2 cm

Given: 

BC = 6 cm 

AB = 8 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 = 82 + 6

⇒ AC2 = 64 + 36 

⇒ AC2 = 100 

⇒ AC = √100 

⇒ AC = 10 cm 

Clearly, 

AQ + QC = AC = 10 cm 

⇒ AP + RC = 10 cm [∵ AQ = AP and QC = RC] 

Also,

AB + BC = 8 cm + 6 cm = 14 cm 

⇒ AP + PB + BR + RC = 14 cm [∵ AB = AP + PB and BC = BR + RC] 

⇒ AP + RC + PB + BR = 14 cm 

⇒ 10 cm + BR + BR = 14 cm [∵ AP + RC = 10 cm and PB = BR] 

⇒ 10 cm + 2BR = 14 cm 

⇒ 2BR = 14 cm – 10 cm = 4 cm

⇒ BR = \(\frac{4}{2}\) cm

⇒ BR = 2 cm 

Now,

∠BPO = 90° [By property 3] 

∠BRO = 90° [By property 3] 

∠PBM = 90° [Given] 

Now by property 2, 

∠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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