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.

Since, the tangent of any point of a line is perpendicular to the radius through the point of contact.

Therefore, ∠OPO =90°
It is given that PQ =15 cm
and    OQ = 17 cm
In right ΔOPQ
OQ² = OP² + PQ²
[Using Pythagoras theorem]
⇒ (17)² = OP² + (15)²
OP² = (17)² – (15)²
= 289 – 225 = 64 ⇒ OP = 8 cm
Hence, radius of the circle = 8 cm.

Given : Two concentric circles C₁ and C₂ of radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C₁) and BD is a tangent to the smaller circle (C₂).
∴ The tangent at any point of a circle is perpendicular to the radius through the point of contact.



Now, in right triangle BOD,  we have

OB² = OD² = BD²                [ Using Pythagoras theorem]

   (13)² = (8)² + BD² 

   169 = 64 + BD² 
   BD² = 169 - 64 
  BD² = 105


Since, perpendicular drawn from the centre to the chord bisects the chord.

(∵ ∠BEA = 90°, angle in semicircle is right angle] and, ∠OBD = ∠ABE     (common)

Therefore, using AAS similar condition ΔBOD ~ ΔBAE



[Proportional sides of two similar triangles]



Now, in right triangle ADE, we have

AD² = AE² + DE² 


                = 256 + 105
                = 361

     AD  = 19   cm.

C.

So,    BP = BQ
(Tangents from external point B)
But    BP = 27 cm
⇒    BQ = 27 cm
It is given that BC = 38 cm
⇒    BQ + CQ = 38
⇒    27 + CQ = 38
⇒    CQ = 11 cm
⇒ CQ = CR (Tagents from an external point C)
But    CQ = 11 cm
⇒    CR = 11 cm

It is given that : CD = 25 cm
⇒    CR + DR = 25
⇒    11 + DR = 25
⇒    DR = 14 cm
Since, tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠ORD = ∠OSD = 90°
It is given that
∠D = 90°
Now, in quadrilateral ORDS,
∠ORD = ∠OSD = ∠RDS = ∠ROS = 90°
and    OR = OS [radii of circle]
Therefore, ORDS is a square
So,    OR = DR = 14 cm
Hence    r = 14 cm.