Question

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

(A) 19 cm

(B) 20 cm

(C) 16 cm

(D) √105 cm

Answer 1

Correct option is (A) 19 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.

∴ OD \(\perp\) BD

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

⇒ BD = √105 cm

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

⇒ BD = DE = √105 cm

⇒ BE = 2BD

⇒ BE = 2√105 cm

Now, In ΔBOD and ΔBAE,

∠BDO = ∠BEA = 90°

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

Therefore, using AAS similar condition ΔBOD ~ ΔBAE

⇒ \(\frac{BD}{BE} = \frac{OD}{AE}\)

[Proportional sides of two similar triangles]

⇒ \(\frac{\sqrt{105}}{2\sqrt{105}} = \frac 8{AE}\)

⇒ \(\frac 12 = \frac 8{AE}\)

⇒ AE = 16 cm

Now, in right triangle ADE, we have

AD² = AE² + DE² 

⇒ AD² = 16² + (√105)²

= 256 + 105

= 361

⇒ AD = 19 cm

Answer 2

The correct option is: (A) 19 cm

Explanation:

Produce BD to meet the bigger circles at E. Join AE. Then

∠AEB = 90°   [Angle in a semicircle]

OD ⊥ BE

['.' BE is tangent to the smaller circle at D and OD is its radius]

BD = DE ['.^. BE is a chord of the circle and OD ⊥ BE]

. ^.. OD || AE                    [. ^.. ∠AEB = ∠ODB = 90°]

In ΔAEB O and D are mid-points of AB and BE. Therefore, by mid-Point theorem, we have

=> AD = 19 cm