Prove that the area of an equilateral triangle described on one sidw of the square is equal to half the area of the equilateral triangle described on one of its diagonal.
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Given : A square ABCD , and ∆BCE and ∆ACF are triangles on it's side BC and diagonal AC respectively.
To Prove : ar(∆BCE) : ar(∆ACF) = 1:2.
Proof :- Let the side of square be 'a' unit.
Diagonal of square = √2a
Since, ∆BCE and ∆ACF are equilateral triangles.
Therefore, all angles are equal to 60°.
So, ∆BCE ~ ∆ACF (AAA-Similarity).
Since, ratio of areas of similar triangles are equal to square of their corresponding sides.
Therefore, ar(∆BCF) / ar(∆ACF) = BC^2 / AC^2
(a)^2 / (√2a)^2
= a^2 / 2a^2
= 1/2
Hence, ar(∆BCF) : ar(∆ACF) = 1 :2.
To Prove : ar(∆BCE) : ar(∆ACF) = 1:2.
Proof :- Let the side of square be 'a' unit.
Diagonal of square = √2a
Since, ∆BCE and ∆ACF are equilateral triangles.
Therefore, all angles are equal to 60°.
So, ∆BCE ~ ∆ACF (AAA-Similarity).
Since, ratio of areas of similar triangles are equal to square of their corresponding sides.
Therefore, ar(∆BCF) / ar(∆ACF) = BC^2 / AC^2
(a)^2 / (√2a)^2
= a^2 / 2a^2
= 1/2
Hence, ar(∆BCF) : ar(∆ACF) = 1 :2.
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