# Area of a trapezoid given the areas of triangles A and B whose bases are the parallel sides of the trapezoid

So for example if we have trapezoid ABCD then we could draw diagonal BD and AC. They would intersect at a point and create four triangles. So say that we knew the area of the two triangles whose bases are the parallel sides of the trapezoid. Let's call the areas A and B. The area of the trapezoid would then be A+B+2sqrt(AB). My question is how would you derive this formula?

Right now, I'm thinking of using the fact that the two triangles A and B are similar to each other and so their heights and bases would be similar. So we know their heights are similar and the triangles' heights also add up to the height of the trapezoid. That's my lead, but I'm not sure if that will go anywhere.

Anyone know how to derive the formula?

So that we will be looking at the same picture, let the vertices of the trapezoid be $A,B,C,D$, in counterclockwise order, with $AB$ and $CD$ the parallel sides, and $AB$ horizontal at the bottom. Let diagonals $AC$ and $BD$ meet at $X$.
Let the area of $\triangle ABX$ be $s^2$ and let the area of $\triangle CXD$ be $t^2$. Then corresponding sides in these two similar triangles are in the ratio $s$ to $t$. This is because if we scale linear dimensions by the factor $\rho$, areas get scaled by the factor $\rho^2$.
By similarity, we have $\frac{AX}{XC}=\frac{AB}{CD}=\frac{s}{t}$. So the area of $\triangle AXD$ is $\frac{s}{t}$ times the area of $\triangle XCD$. That's because these two triangles have the same height with respect to their two bases $AX$ and $XC$.
Thus $\triangle AXD$ has area $st$. Similarly, $\triangle BXC$ has area $st$, so the trapezoid has area $s^2+t^2+2st$.