Since chord $C D$ and tangent at point $T$ intersect each other at $P$ , $\therefore\; PC \times PD = PT^2$ .........(1)Since chord $A B$ and tangent at point $T$ intersect each other at $P$ , $\therefore\; PA \times PB = PT^2$ ..........(2) From (1) and (2), $PC \times PD = PA \times PB$ .........(3) Given : $CD = 7.8\;\mathrm{cm}; PD = 5\;\mathrm{cm}, PB =$ $4\;\mathrm{cm}$. $\; \therefore PA = PB + AB = 4 + AB,$ $\; PC = PD + CD = 5+7.8 = 12.8\;\mathrm{cm}.$ Putting these values in eq. (3) $12.8 \times 5\; = (4+AB) \times 4$ $\Rightarrow \; 4+AB\; =\; \frac{12.8 \times 5}{4}$ $\Rightarrow \; 4+AB\; =16$ $\Rightarrow \; AB\; =12\;\mathrm{cm}.$ $\; \text{ Hence, }\; \; AB\; =12\;\mathrm{cm}.$