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:

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& CD:DB=3:2\\ & CD=5\sqrt{13}\times \frac{3}{5}\\ & CD=3\sqrt{13}\\ & \sin \theta =\frac{3\sqrt{13}}{DE}=\frac{5\sqrt{13}}{10}\\ & DE=6\text{ cm}\end{array}$$