PL explain Me by a easy method!!
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It's given that ABC is an equilateral triangle.
Let it's side be 6x.
Draw AE⊥BC.
AE=h
Now, BE=EC [∵It is also an isoceles triangle]
Using Pythagoras theorem,
AB²=AE²+BE²
=9x²+h²___(1)
AD²=DE²+BE²
=x²+h²___(2)
From∆AEC,
h²=(6x)²-EC²
=36x²-9x²=27x²
From (1),
AB²=36x²
From (2),
AD²=28x²
(AB²/AD²)=(36x²/28x²)
(AB²/AD²)=(9/7)
7AB²=9AD²
Thus proved.
(^_^)b
Let it's side be 6x.
Draw AE⊥BC.
AE=h
Now, BE=EC [∵It is also an isoceles triangle]
Using Pythagoras theorem,
AB²=AE²+BE²
=9x²+h²___(1)
AD²=DE²+BE²
=x²+h²___(2)
From∆AEC,
h²=(6x)²-EC²
=36x²-9x²=27x²
From (1),
AB²=36x²
From (2),
AD²=28x²
(AB²/AD²)=(36x²/28x²)
(AB²/AD²)=(9/7)
7AB²=9AD²
Thus proved.
(^_^)b
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