I've received the following geometry problem:
A trapezium (not necessarily isosceles) composed of two parallel lines of lengths x and y is joined at each end by two lines.
A line (the orange one in the attached image), parallel to the two joining horizontal lines, bisects the area of the trapezium. What is the length of this line?
I've attempted to work the problem using similar triangles and breaking the trapezium into composite shapes, with little success.
Any help is much appreciated.
Image of the trapezium:
Let $h$ be the height of the original trapezium, so its area is $\frac12h(x+y)$. Let $z$ be the length of your new line, and $h'$ the height of the upper sub-trapezium. This upper sub-trapezium has area $\frac 12h'(x+z)$ so we get the equation $$2h(x+y)=h'(x+z).\tag{1}$$ We need another equation involving $h'$.
I'll assume that $y>x$, as in the diagram you've drawn. Draw a line through the top left vertex of the trapezium parallel to the right edge. This forms a triangle $R$, with base $y-x$ and height $h$ and also your orange line makes a chord in $T$ of length $z-x$. Comparing similar triangles gives $$\frac{h'}h=\frac{z-x}{y-x}.\tag{2}$$ Eliminating $h'$ from $(1)$ and $(2)$ gives $$2(x+y)(y-x)=(x+z)(z-x).\tag{3}$$ This is an easy quadratic equation for $z$.