Question

prove that the area of the equilateral describe on the side of a square is 1/2 the area of the equilateral triangles described on its diangonal ​

Answer 1

Answer:

your answer is as follow:-

Step-by-step explanation:

▶ 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 ) . }

2

1

ar(△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}}. }

ar(△ACF)

ar(△BCE)

=

AC

2

BC

2

=

2(BC)

2

BC

2

.

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

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

ar(△ACF)

ar(△BCE)

=

2

1

.

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

ar(△BCE)=

2

1

×ar(△ACF).

Hence, it is proved.