geometry
Q.Is equilateral Trapezium possible?
The defination of Trapezium :A quadrilateral with one pair of parallel sides.
If we have a trapezium that have equal sides,it becomes a rhombus (& contradicts it's own defination).
This implies that equilateral trapezium is not possible.
There are four possible variatoins for the given situation:
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Thus, depending on which of these variations holds, we get the perimeter to be one of the four following values:
$$ 2a+b+c\pm x_b\pm x_c=2a+b+c\pm\sqrt{b^2-h^2}\pm\sqrt{c^2-h^2} $$
The midline divides the trapezoid into two equally high trapezoids. Its length is also the mean of the length of the two parallel sides. (You can see that by extending the nonparallel sides until they intersect in a point. The two parallel sides and the midline then forms the base line in three similar triangles).
Call the long parallel side $AB$, and the short one $CD$ with $AD$ and $BC$ being the two other sides of the trapezoid. Further let $r$ be the radius of the circle. The trapezoid then has height $2r$. Further, let $E, F, G, H$ be the places the circle touches the trapezoid at $AB$, $BC$, $CD$ and $AD$ respectively.
The hypothesis about the midline dividing the area (note that the two parts are also trapezoids) then dictates that $$ 3\frac{AB + \frac{AB + CD}{2}}{2}r = 5\frac{CD + \frac{AB + CD}{2}}{2}r \\ {3}AB + 3\frac{AB + CD}{2} -5CD -5\frac{AB+CD}{2} = 0\\ 2AB - 6CD = 0\\ AB = 3CD $$ so the midline has length $2CD$.
Since the sides touch the circle we must have the following: $$ AE = AH\\ BE = BF\\ CF = CG\\ DG = DH $$ Since $DH + AH + BF + CF = 16$, then we must have $AE+BE + CG + DG = 16$. But this also equals $4CD$, so $CD = 4$. The long parallel side $AB$ must therefore be $12$.
Best Answer
A quadrilateral that is both a trapezium (in those countries that call football soccer: a trapezoid) and a rhombus is a square. Note that "one pair" i the definition of trapezium is understood as "at least one pair"
Apart from that, the term "equilateral (or: isosceles) tapezium" is used for a trapezium where (only) the non-parallel (or not necessarily parallel) pair of sides are of the same langth.