Math, asked by NUKALAKARTHIK, 1 day ago

AB² = AD²+ 1/2AB²
A. 5AB² = 4AD²
B. 3AB² = 4AD²
C. 4AB² = 3AD²
D. 2AB² = 3AD²
how to solve this​

Answers

Answered by tiwaripoonam9032
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:

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