The radii of two concentric circles are 12 cm and 13 cm. AB is a diameter of the bigger circle. BD is a tangent to a smaller circle touching it at D. Find the length (in cm) of AD? (correct to one decimal place)

24.5

By pythagoras theorem, in triangle ODB

\( {OB }^{2 } \) = \( {OD }^{2 } \) + \( {BD }^{2 } \)

= 169 = 144 + \( {BD }^{2 } \)

= BD = 5 cm

BD = DE = \(\frac{1}{2}\) x BE    [Perpendicular drawn from the center on a chord bisects it in two equal parts]

= DE = BD = 5 cm

= BE = 10 cm

In triangle ABE,

\(\angle\)AEB = 90   [Angle made in semicircle]

= \( {AB }^{2 } \) = \( {EB }^{2 } \) + \( {AE }^{2 } \)

= 676 = 100 + \( {AE }^{2 } \)

= AE = \(\sqrt {576 }\)

= AE = 24 cm

In triangle ADE

= \( {AD }^{2 } \) = \( {ED }^{2 } \) + \( {AE }^{2 } \)

= \( {AD }^{2 } \) = 25 + 576

= \( {AB }^{2 } \) = 601

= AD = \(\sqrt {601 }\)

= AD = 24.51 cm

Therefore, length of AD is 24.5 cm.