In ΔABC, ∠A = 90°, AD is the bisector of ∠A meeting BC at D, and DE ⊥ AC at E. If AB = 10 cm and AC = 15 cm, then the length of DE, in cm, is:

Solution (By Examveda Team)

$$\eqalign{ & CD:DB = 3:2 \cr & CD = 5\sqrt {13} \times \frac{3}{5} \cr & CD = 3\sqrt {13} \cr & \sin \theta = \frac{{3\sqrt {13} }}{{DE}} = \frac{{5\sqrt {13} }}{{10}} \cr & DE = 6{\text{ cm}} \cr} $$