If non-parallel sides of an isosceles trapezium are prolonged, an equilateral triangle with sides of 6cm would be formed. Knowing that the trapezium is half the height of the triangle,the area of the trapezium is { fill_number}.

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Detailed Solution

It is given that BDEC is a trapezium and ABC is an equilateral triangle formed by extending the sides BD and EC.
Each side of triangle = 6cm
Height of triangle = 2(Height of trapezium).
To find the area of the trapezium, we have to find its height.
So first of all, we will find the height of the triangle.
AG is the altitude of triangle ABC.
We know that the altitude of an equilateral triangle divides the triangle into two equal halves. So, AG is also the median of the triangle.
AG is the altitude of triangle ABC.
We know that the altitude of an equilateral triangle divides the triangle into two equal halves. So, AG is also the median of the triangle.
∴ BG = GC = 62 =3cm
Now, in △ABG, we will apply pythagoras theorem to find the length of AG.
On applying pythagoras theorem in △ABG, we have:
AB²=AG²+BG²
⇒AG = 5.2cm
Area of trapezium = ½ (sum of parallel sides)(height)
⇒Area = ½ * 9 * 2.6
= 11.70 cm².

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