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:
A. 8
B. 6
C. 6.25
D. 7.5
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & CD:DB = 3:2 \cr & CD = 5\sqrt {13} \times \frac{3}{5} \cr & CD = 3\sqrt {13} \cr & \sin \theta = \frac{{3\sqrt {13} }}{{DE}} = \frac{{5\sqrt {13} }}{{10}} \cr & DE = 6{\text{ cm}} \cr} $$
This Question Belongs to Arithmetic Ability >> Geometry
Related Questions on Geometry
A. $$\frac{{23\sqrt {21} }}{4}$$
B. $$\frac{{15\sqrt {21} }}{4}$$
C. $$\frac{{17\sqrt {21} }}{5}$$
D. $$\frac{{23\sqrt {21} }}{5}$$
In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is
A. 30°
B. 40°
C. 20°
D. 70°
Join The Discussion