In △ABC, ∠A = 90º, AD is the bisector of ∠A meeting BC at D, and DE Ʇ AC at E. If AB = 10 cm and AC = 15 cm, then the length of DE, in cm , is:
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 6
Concept
Usage of Similarity
Application
In ΔEAD
∠AED + ∠EAD + ∠EDA = 180°
90° + 45° + ∠EDA = 180°
∠EDA = 180° - 135° = 45°
Now,
EA = ED
Let ED be K, EA = K
EC = 15 - K
In ΔCED and ΔCAB (Applying Similiarity
∠C = ∠C (common)
∠CED = ∠CAB = 90°
So, ΔCED ~ ΔCAB (AA)
(ED)/(AB) = (CE)/(CA)
x/10 = (15 - x)/15
15x = 10(15 - x)
upon solving
x = DE = 6cm