Відношення площ і трапеція

Assume that a and b are the bases and c and d are the other two sides:

Next, how do you find out the lengths of the diagonals?

Then, what's next?

Look at this trapezoid:

Make the intersection of the four lines (two diagonals), O.

Therefore, we know that $\angle AOD = \angle BOC$ and $\angle DOC = \angle AOB$.

From this, the ratio of the areas of each pair of adjacent triangles is equal to that between the lengths of the parallel sides.

Trapezoid $ABCD$ has parallel edges $AB$ and $CD$, and diagonals $AC$ and $BD$ intersect at $O$. Suppose that $[\triangle AOB]=1$, $[\triangle COD]=25$. Find the area of $[ABCD]$.

We are trying to find the area of the whole trapezoids. It looks like that the diagonals divide it into $4$ parts. What's nice is that we already have $2$ of them! Now what do we do to find the remaining $2$?

Maybe ratios will help, since trapezoids have $2$ parallel sides, and parallel lines induce similar triangles which give ratios. But how?

We can observe that $\triangle AOB$ is similar to $\triangle COD$. Thus $\frac{AO}{OC}=\frac{BO}{OD}$. Now this obviously helps us relate the areas of $\triangle AOB$ and $\triangle COD$.

Since they're similar, their areas are in the ratio $AO^2:OC^2=1:25$, so the ratio of areas of $\triangle AOB$ and $\triangle COD$ is $\sqrt{\frac{1}{25}}=\frac{1}{5}$. We're now really close!

But the ratio of areas of $AOB$ and $AOD$ is just $BO:OC$, which is also $1:5$. Hence $AOD=5$. Similarly $BOC=5$, so we're almost there!

$ABCD=AOB+BOC+COD+DOA=1+5+25+5=\boxed{36}$. We're done!