Question

prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle discribed on its diagonal

Answer 1

Here ABCD is a square, AEB is an equilateral triangle described on the side of the square and DBF is an equilateral triangle described on diagonal BD of  the square.


 To Prove:    Ar(ΔDBF) / Ar(ΔAEB) = 2 / 1 


Proof:  If two equilateral triangles are similar then all angles are = 60 degrees.

Therefore, by AAA similarity criterion , 

△DBF ~ △AEB

 Ar(ΔDBF) / Ar(ΔAEB) = DB2 / AB2 

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(i)We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides i .e.But, we have DB = √2AB     {But diagonal of square is √2 times of its side} 


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(ii).Substitute equation (ii) in equation (i),


we getAr(ΔDBF) / Ar(ΔAEB)

= (√2AB )2 / AB2  

= 2 AB2 / AB2 = 2

 ∴ Area of equilateral triangle described on one side os square is equal to half the area of the equilateral triangle described on one of its diagonals.