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

Given:

• Radius of outer circle $(R) = 13$ cm
• Radius of inner circle $(r) = 8$ cm
• AB is diameter of bigger circle
• BD is tangent to smaller circle at D

We need to take the triangle OBD into consideration for this, as shown in the diagram above.

Important properties used:

1. Since $OD \perp BD$ (radius $\perp$ tangent):

• D is midpoint of BE
  • O is midpoint of AB
  • Therefore, $OD = \frac{1}{2}AE$

2. Using midpoint theorem:

Since OD = $8$ cm (inner radius)

$AE = 2 × 8 = 16$ cm

3. For right triangle OBD:

$BD = \sqrt{OB^2 - OD^2}$

$BD = \sqrt{13^2 - 8^2}$

$BD = \sqrt{169 - 64}$

$BD = \sqrt{105}$

Also, $DE = BD = \sqrt{105}$

4. For right triangle ADE:

$AD^2 = AE^2 + DE^2$

$AD^2 = 16^2 + (\sqrt{105})^2$

$AD^2 = 256 + 105$

$AD^2 = 361$

Therefore, $AD = 19$ cm

The answer is 19 cm.