ABCD is a trapezium in which AB || DC and its diagonals intersect at P. If AP = (3x - 1) cm. PC = (5x - 3) cm. BP = (2x + 1) cm and PD = (6x - 5) cm, then the length of DB is:

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{\text{AP}}}}{{{\text{PC}}}} = \frac{{{\text{BP}}}}{{{\text{PD}}}}\,\,\,\,\,\,\left( {{\text{Similarity}}} \right) \cr & \frac{{3{\text{x}} - 1}}{{5{\text{x}} - 3}} = \frac{{2{\text{x}} + 1}}{{6{\text{x}} - 5}} \cr} $$
18x² - 6x - 15x + 5 = 10x² + 5x - 6x - 3
8x² - 20x + 8 = 0
x = 2, $$\frac{1}{2}$$
at x = 2,
DB = 6x - 5 + 2x + 1
= 8x - 4
= 8 × 2 - 4
= 12