Question

prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilateral triangle described on one of its diagonals.

Answer 1

$<b> <I>$
Hey there !!

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.


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


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



✔✔ Hence, it is proved ✅✅.

____________________________________




$\huge \boxed{ \mathbb{THANKS}}$



$\huge \bf{ \# \mathbb{B}e \mathcal{B}rainly.}$

Answer 2

∆BCE ~ ∆ACF.
the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.


ar(triangle BCE ) / ar(triangle ACF ) = BC^2/AC^2=BC^2/2(BC)^2


ar(triangle BCE ) /ar( triangle ACF ) = 1/2/.

=ar(triangle BCE ) =1/2× ar(triangle ACF )