Triangle ABC is right angled at B. BD is an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm, then find the measure of AB (in cm).

$3\sqrt{6}$

Taking two similar triangle \(\Delta \)ABC & \(\Delta \)BDC,

\( {BC }^{2 } \) = DC x AC

= \( {BC }^{2 } \) = CD x AC

= \( {BC }^{2 } \) = 3 x 9

= \( {BC }^{2 } \) = 27

From \(\Delta \)ABC,

\( {AB }^{2 } \) + \( {BC }^{2 } \) = \( {AC }^{2 } \)

= \( {AB }^{2 } \) = \( {AC }^{2 } \) - \( {BC }^{2 } \)

= \( {AB }^{2 } \) = \( {9 }^{2 } \) - 27

= \( {AB }^{2 } \) = 54

= AB = 3\(\sqrt {6 }\) cm.