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:

Answer: Option B

Solution

Answer: Option B

Solution:

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