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
Answers
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.