prove that the equilateral triangle described on the two sides of a right angle triangle are together equal to the equilateral triangle on the hypotenuse in terms of their areas
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heya!!!.☺
it's too easy ..☺
given -A square ABCD and Equilateral triangle BCE and ACF have been described on side BC and diagonal on side BC and Diagonal AC respectively
to proove ---Area(∆BCE)=1/2ar∆ACE)
=) proof -since each of ∆BCE and ∆ACE is an equilateral triangle so ,the each one of them is 60° so, the triangle are equiangular,and hence similar.
so,∆BCE~∆ACE
we know that the ratio of the areas of two similar triangle is equal to the square of their correspomding sides.
.s
=)ar∆BCE/ar∆ACF=BC^2/AC^2=BC^2/2(BC)^2
so,ar(∆BCE)=1/2ar( ∆ACF)
it's too easy ..☺
given -A square ABCD and Equilateral triangle BCE and ACF have been described on side BC and diagonal on side BC and Diagonal AC respectively
to proove ---Area(∆BCE)=1/2ar∆ACE)
=) proof -since each of ∆BCE and ∆ACE is an equilateral triangle so ,the each one of them is 60° so, the triangle are equiangular,and hence similar.
so,∆BCE~∆ACE
we know that the ratio of the areas of two similar triangle is equal to the square of their correspomding sides.
.s
=)ar∆BCE/ar∆ACF=BC^2/AC^2=BC^2/2(BC)^2
so,ar(∆BCE)=1/2ar( ∆ACF)
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