Question

Proove that area of equitorial triangle on side of the square is equals to half the area of equitorial triangle described on one of its triangle

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 )

$hence \: \: proved$