266. In the given figure, ABC is a right-angled triangle with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle.

It is given that ABC is a right angle triangle with AB = 6 cm and AC = 8 cm and a circle with centre O has been inscribed.
Using Pythagoras theorem, we get
BC² = AC² + AB²
= (8)² + (6)²
= 64 + 36 = 100
⇒ BC = 10 cm



 Tangents at any point of a circle is perpendicular to the radius through the point of contact 





=  3r + 4r + 5r
= 12r                                              ...(ii)

Comparing (i) and (ii), we get
24 = 12r ⇒ r = 2 cm
Method – II:
In quadrilateral APOR,
∵ ∠OPA = ∠ORA = 90°
∠PAR = 90°
⇒ ∠OPA = ∠ORA = ∠PAR = ∠POR = 90° ....(i)
and AP = AR (ii) (length of tangents drawn from an external point are equal)
Using result (i) and (ii), we get
APOR is a square
Therefore,
OR = AR = r [Sides of square] and OR = AP = r    [Sides of square]
Now, BP = AB – AP = 6 – r and, CR = AC – AR = 8 – r
Since tangents from an external point are equal
CR = CQ = 8 – r and    BP = BQ = 6 – r
Now, In ΔABC,
BC² = AC² + AB²
⇒ (CQ + BQ)² = (8)² + (6)²
⇒ (8 – r + 6 – r)² = 64 + 36
⇒ (14 – 2r)² = 100
⇒ (14 – 2r)² = (10)²
⇒ 14 – 2r = 10
⇒ –2r = –4
⇒    r = 2
Hence radius of circle (r) = 2 cm.