AB² = AD²+ 1/2AB²
A. 5AB² = 4AD²
B. 3AB² = 4AD²
C. 4AB² = 3AD²
D. 2AB² = 3AD²
how to solve this
Answers
Answered by
0
Answer:
A ∆ ABC, in which sides are AB=BC= AC= a units
and AD is perpendicular to BC ,
In ∆ADB ,
AB²= AD²+ BD² (by Pythagoras theorem)
a² = AD² + (a/2)² [BD= 1/2BC, since in an equilateral triangle altitude AD is perpendicular bisector of BC ]
a²- a²/4 =AD²
=( 4a²-a²)/4 = AD²
= 3a² /4 = AD²
3AB²/4= AD²
[ AB= a]
3AB²= 4AD²
Step-by-step explanation:
Similar questions