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

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

BC^{2} = AC^{2} + AB^{2}⇒ (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.

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The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

Given : Two concentric circles C_{1} and C_{2 }of radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C_{1}) and BD is a tangent to the smaller circle (C_{2}).

∴ 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

OB^{2} = OD^{2} = BD^{2} [ Using Pythagoras theorem]

(13)^{2 }= (8)^{2 }+ BD^{2 }

169 = 64 + BD^{2}

BD^{2} = 169 - 64

BD^{2} = 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

AD^{2} = AE^{2} + DE^{2 }

= 256 + 105

= 361

AD = 19 cm.

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From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle?

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

OQ^{2} = OP^{2} + PQ^{2}[Using Pythagoras theorem]

⇒ (17)^{2} = OP^{2} + (15)^{2}OP^{2} = (17)^{2 }– (15)^{2}= 289 – 225 = 64 ⇒ OP = 8 cm

Hence, radius of the circle = 8 cm.

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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.

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