Question

Prove that the area of an equalateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals ​

Answer 1

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Answer 2

Answer:

Correct Question:-

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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

Given :-

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→ A square ABCD an equilateral triangle ABC and ACF have been described on side BC and diagonal AC respectively.

To Prove :-

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$\qquad\sf \longrightarrow arc\big( ∆BCE \big) = \sf{ \frac{1}{2} \: \: arc\big( \triangle ACF \big) . }$

Proof :-

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→ Since each of the ∆ABC and ∆ACF is an equilateral triangle, so each angle of his strength is one of them is 60° . So, the angles are equiangular, and hence similar.

$\sf \longrightarrow \: \: \: \: \: \underline{ \boxed{ \sf \purple{∆ BCE \sim ∆ACF}}}$

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$\qquad \sf \underline{ Note:-}$

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$\qquad \longrightarrow\sf{ \frac{ arc\big( \triangle BCE \big) }{ arc\big( \triangle ACF \big) } = \frac{{BC}^{2} }{ {AC}^{2}} = \frac{{BC}^{2}}{2{(BC)}^{2}}. }$

[ Because, AC is hypotenuse

=> AC = √2BC.]

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$\\ \\ \longrightarrow\sf{ \frac{ arc \big( \triangle BCE \big) }{ arc \big( \triangle ACF \big) } = \frac{1}{2} . }$

$\sf \: \qquad \longrightarrow \underline{ \boxed{ \purple{ \sf \: ar( \triangle BCE ) = \frac{1}{2} \times ar\big( \triangle ACF \big)}}}$

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• Hence proved.