circlesgeometry
There is a trapezium inside a circle with parallel side 6 and 10 cm with other sides are 4 and 4 cm. Find the radius of the circle. 
My attempt :-
Imagine the center of the circle. The distance between the center and the vertices of the trapeze that touch the circle is r. Now imagine 2 triangles, where the hypotenuse is the radius and the legs is half the horizontal lines of the trapeze. After this I calculate the height of trapezium which come out to be $ \sqrt{12}$
There are four possible variatoins for the given situation:
$\hspace{2cm}$
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
Distance between parallel sides of trapezium $MN=\sqrt{4^2-2^2}=2\sqrt{3}$
Let $R$ be the radius of circle & $OM=x$
Using Pythagoras theorem in right $\Delta OMA$ $$R^2=x^2+5^2\iff R^2=x^2+25\tag 1$$
Similarly, using Pythagoras theorem in right $\Delta OND$ $$R^2=(x+2\sqrt3)^2+3^2\iff R^2=x^2+4x\sqrt3+21\tag2$$ subtracting (1) from (2) we get $$0=4x\sqrt3-4\implies x=\frac{1}{\sqrt3}$$ setting value of $x$ in (1), we get $$R=\sqrt{\frac13+25}=2\sqrt{\frac{19}{3}}\approx 5.033222957\ cm$$