ΔABC, AB = 20 cm, BC = 7 cm and CA = 15 cm. Side BC is produced to D such that ΔDAB ∽ ΔDCA. DC is qual to:

Solution (By Examveda Team)

$$\eqalign{ & \Delta DAB \sim \Delta DCA \cr & \frac{{DA}}{{DC}} = \frac{{AB}}{{AC}} = \frac{{BD}}{{AD}} \cr & \frac{{DA}}{{DC}} = \frac{{20}}{{15}} = \frac{4}{3} \cr & AD = \frac{4}{3} \times DC\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{{20}}{{15}} = \frac{{BD}}{{AD}} = \frac{4}{3} \cr & AD = \frac{3}{4} \times BD{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{4}{3} \times DC = \frac{3}{4} \times BD \cr & \frac{{DC}}{{BD}} = \frac{9}{{16}} \cr & \frac{{DC}}{{BC + DC}} = \frac{9}{{16}} \cr & \frac{{DC}}{{7 + DC}} = \frac{9}{{16}} \cr & 16DC = 63 + 9DC \cr & 7DC = 63 \cr & DC = 9{\text{ cm}} \cr} $$