Question

Prove the area of equilateral triangle described on the side of a square is half the area of the equilateral triangle describe on its diagonal

Answer 1

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Answer 2

▶ Answer :-

▶ Step-by-step explanation :-

Given :-

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

To Prove :-

→ ar( ∆BCE ) = $\bf{ \frac{1}{2} ar( \triangle ACF ) . }$

Proof :-

→ 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.

==> ∆BCE ~ ∆ACF.


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


$\bf{ \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{{BC}^{2} }{ {AC}^{2}} = \frac{{BC}^{2}}{2{(BC)}^{2}}. }$


[ Because, AC is hypotenuse => AC = √2BC. ]


$\bf{ \implies \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{1}{2} . }$


Hence, $\boxed{ \sf ar( \triangle BCE ) = \frac{1}{2} \times ar( \triangle ACF ) . }$


Hence, it is proved.