Prove that the area of an equilateral triangle described on one side of a square is equal
to half the area of the equilateral triangle described on one of its diagonals.
Answers
Answer:
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.
∴ , 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
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
∴ 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