Question

Prove that the area ofban equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of itsdiagonal

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 !!

Answer 2

plz refer to this attachment