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ABE and BCD are equilateral triangles, therefore AE and BD are parallel to each other. This makes ABDE a trapezoid. Area of BDE is : area of ABDE minus area of ABE, hence 0.5 *(2+1)*2*sin60 - 0.5*2*2*sin60 =3*sin60 - 2*sin60 = sin60= 0.866 which is also (sqrt(3))/2 .

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The pink triangle is a right triangle, so using the Pythagorean theorem, the height of the pink triangle is: sqrt (2^2 - 1^2) = sqrt (3).

Area of the triangle = base * height / 2 = 1 * sqrt (3) / 2 = sqrt (3) / 2.

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The result is correct! - but how we are sure that the pink triangle is a right triangle?

The pink triangle is a right triangle, because both yellow triangles are equilateral triangles (1,1,1) and (2,2,2), which constrains them to be equi-angular, too. The sum of all angles in a triangle is always 180, and an equilateral triangle has angles of (60,60,60). The straight line, ABC, displays an angle of 180 degrees. Thus, angle EBD is 60 degrees. Cos (EBD) = 0.5, but only if the pink triangle is a right triangle. As we can see, (BD)/(EB) = 0.5, so the Cos works and the pink triangle is, indeed, a right triangle.

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Very elegant conclusion!

ABC being a straight line then all angles at point b add up to 180. Angle ABE=Angle DBC=60 degrees since both are angles of an equilateral trianlge.

Therefore, angle EBD= 180 - (60+60)= 60 degrees.

Using the formula: Area= 1/2 ab sin C where a=BE=2 and b=BD=1 and C=angleEBD=60 degrees, the area thus becomes:

1/2 X 2 X 1 X sin60 = 1.7321 sq. units.

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the simple answer is(total area-the area of rest two triangles )

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Its not that simple..

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Working on the maths I was taught at school in the latter half of the last century, I was able to correctly calculate the area! But then I was using Pythagoras' theory-going back to the ancient Greeks! Tony

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