Finding the difference between areas of triangles

Question

The sum of areas of triangle A and B is 2600. What is the difference between the areas of B and D?

Answer 1

Hint: Let $$h_1,h_2,h_3,h_4$$ be the heights in the triangles $A,B,C,D$ in that order. Then we get the following equations: $$h_1+h_4=180$$ $$h_2+h_3=80$$ $$8h_1+18h_2=520$$ $$(h_1+h_4)(h_2+h_3)=80\cdot 180$$ Can you proceed?

Answer 2

Mark $h$ and $w$ as below so we can find a formula for each triangle:

$$ \left. \begin{aligned} A &= \frac{1}{2}\times 80\times w \\ B &= \frac{1}{2}\times 180\times h \\ C &= \frac{1}{2}\times 80\times (180-w) \\ D &= \frac{1}{2}\times 180\times (80-h) \end{aligned} \right\} \begin{aligned} A &= 40w \\ B &= 90h \\ C &= 7200-40w \\ D&= 7200-90h \end{aligned} $$

Now using the area of the rectangle $A + B + C + D = 80\times 180$ and substituting leads to a dead end since $h$s and $w$s cancel each other out, leaving the obvious identity: $7200 + 7200 = 14400$.

To solve two unknowns we need two non-equivalent simultaneous equations—we have only one ($40w + 90h = 2600$)

So as far as I can see you can only reduce $ B - D = 180h - 7200 = -80w - 2000$.

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In fact, to prove that there is no definite answer, here's a to-scale diagram of a line segment determining all the possible places the intersection can be:

The area of the triangles $A$ and $B$ is always $2600$ above.