Solution:  

Areas of Similar Triangles NCERT Solutions Tenth Grade

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

To Prove:     ar(△DBF)ar(△AEB)=21

Proof:  Any two equilateral triangles are similar because all angles are of 60 degrees.

Therefore, by AAA similarity criterion, △DBF ~ △AEB

ar(△DBF)ar(△AEB)=DB2AB2        (1)

{The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides}

But, we have DB=2–√AB     {Diagonal of square is 2–√ times of its side}         (2)

Putting equation (2) in equation (1), we get

ar(△DBF)ar(△AEB)=(2√AB)2AB2=2AB2AB2 = 2

Therefore, 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.

Hence Proved