Trapezoid finding length of dividing line

Question

Given a trapezoid with bases $437\mathrm{m}$ and $57\mathrm{m}$ with sides $300\mathrm{m}$ and $350\mathrm{m}$ , it is to be divided into two parts in the ratio $2:3$ by a line parallel to the parallel sides, the $\color{brown}{\text{larger part be adjacent to the smaller parallel side}}$. Find the length of the dividing line, the altitude of the trapezoid and the area of the trapezoid.

I managed to solve for the trapezoid's $\text{height}(296.012\mathrm{m})$ by equating the $h^2$ of the two triangles formed when a line is drawn parallel to the bases. Hence, found the necessary variables for $\text{area}(149782.072\mathrm{m})$.

I don't know what to use to determine the dividing line. I do know the area can be proportioned $2:3$, but given a third base(dividing line), and the height of the two resulting trapezoids to be different, I would have to find for the two altitudes first. How will I be able to do that? Or is there any easier way around?

Answer 1

(1) After constructing the dotted parallel line as shown, we have two //grams with sides of length = 437.

(2) The length of the base of the larger triangle = 138

(3) The length of the base of the smaller triangle = 138* (3/5) = x, say.

(4) The required = x + 437.

Answer 2

Let's call $h$ the height of the trapezoid.

Let's call $A,B,C,D$ the vortice of the trapezoid with $AB$ the smaller base and $CD$ the longer one.

Let's call $I,J$ the intersection of the dividing line and the two sides $AD$ and $BC$.

Let's call $h'$ the height of the upper trapezoid (adjacent to the smaller base) and $h''$ the height of the lower one.

$ABJI=\frac 23 ABCD$ and $JCDI=\frac 13 ABCD$

$ABJI=(AB+JI)*h'/2=(AB+CD)*h/3$

You can express $h'$ in term of the rest.

$JCDI=(CD+IJ)*h''/2=(AB+CD)*h/6$

Similar with $h''$.

But $h'+h''=h$ Thus you got a quadratic expression in $IJ$ that you can solve and get the answer.