Trapezium drawn in the circle A circle is drawn inside a trapezium such that it touches all the sides of trapezium. The line joining the midpoints of the non parallel sides divides the trapezium in two parts with the area in the ratio of 3:5. If the length of the non parallel sides are 6 cm and 10 cm, then what is the length of the longer parallel side?
 A: 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 $12$.
A: Draw a picture. Let the vertices of the trapezium be, in counterclockwise order, $A,B,C,D$, where $AB$ is the longer of the two parallel sides.
Note that the two tangents from $A$ to the circle have the same length. Call them both $a$, and write $a$ next to each of them. Do something similar with $B$, $C$, and $D$.
Note that $AB+CD=a+b+c+d=BC+AD=16$.
Now work with the bisectors. Let $x$ be the big base, and $y$ the side parallel to it. we have seen that $x+y=16$.  
The area of the bottom half is one quarter the height times $x+(x+y)/2$. The area of the top half is one quarter the height times $y+(x+y)/2$. Thus $\frac{y+8}{x+8}=\frac{3}{5}$. Using this and $x+y=16$, we get two linear equations in $x$ and $y$, and we can find $x$. 
