Question

Prove that area of a equilateral trianglr describe one side of square equal to half of the area of equilateral triangle described an arc of its diagonal

Answer 1

[ Figure in the attachment ]

To prove :-

area of ∆BCF = 1/2 ( area of ∆ACE )

[ both triangle are Equilateral triangle ]

Solution :-

Let the side of square be a unit !

then ,the side of ∆BCF will be a unit !! 


In ∆ADC ( right angle at D ) Using Pythagoras theorem :-

AC² = AD² + DC²

AC² = a² + a²

AC² = 2a²

AC = √2a²

AC = a√2

So, the sides of ∆ACE will be Equal to a√2 unit !


• All equilateral triangle are similar
So, 
∆ACE ≈ ∆BCF 

• The ratio of area of similar triangle are equal to the ratio of square of their corresponding sides.

So, 

area of ∆ACE / area of ∆BCF = ( AC/BC)²


area of ∆ACE / area of ∆BCF = (a√2/a)²


area of ∆ACE / area of ∆BCF = ( √2/1)²

area of ∆ACE / area of ∆BCF = 2/1

area of ∆ACE = 2 ( area of ∆ BCF )

area of ∆ BCF = 1/2 ( area of ∆ACE )


$hence \: \: \: \: proved$