Question

prove that the area of the equilateral triangle described on one side of a square is equal to the half the area of an equilateral triangle discribed on one of its diagonal​

Answer 1

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

Answer 2

Answer:

yes it will be half

Step-by-step explanation:

let the square be of the side s.

height of the equilateral triangle is a√3/2. where a is the side of the equilateral triangle .

then area of the triangle on it s side = 1/2 * s * s√3/2:

lenght of diagnol = s√2

height = s√2√3/2

area of the triangle on its diagonal = 1/2 * s√2 * s√6/2

by dividing the area of the triangle by the area of the triangle on its diagonal we get 1/2

hence proved

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