What is the area of trapezoid DBCE?

Question

Question

I'm preparing for the upcoming AMC 10 competition, and I've stumbled across this problem while completing practice tests. This is the only question that has given me such great difficulty. I cannot find a way to get to the solution; I must be missing something. Any help would be appreciated.

(Question linked above)

Answer 1

Draw a few more small triangles:

Now you can see $16$ small congruent triangles of area $1$. The remaining area (of the trapezoid $DBCE$) is $40-16\cdot 1=24$. The answer $E$ is right.

Answer 2

If proportion between areas of two similar triangle is $p$, then proportion of corresponding sides is $\sqrt{p}$. By using this fact, we can say if $|DF| = |FG| = |GH| = |HE| = k$, then we have $|BC| = 2k\sqrt{10}$ (comes from $\sqrt{40}$). Then by using the similarity of the triangles $\Delta ADE$ and $\Delta ABC$, we have

$$\bigg(\frac{|DE|}{|BC|}\bigg)^2 = \frac{A(ADE)}{A(ABC)} \implies \frac{4}{10} = \frac{A(ADE)}{A(ABC)}$$ where $A(ABC) = 40$ is given. Therefore, $A(ADE) = 16$ and hence $A(DBCE) = A(ABC)-A(ADE) = 40-16 = 24$.

Answer 3

Consider the lengths \begin{eqnarray*} \frac{x}{y} =\frac{3}{4}. \end{eqnarray*} Consider the are areas \begin{eqnarray*} \frac{x^2}{y^2} =\frac{9}{16}=\frac{A}{A+7}. \end{eqnarray*} Linear algebra rapidly gives $A=9$, hence the trapezoid is $40-9-7$.

Answer 4

In the figure below, the vertices of one of the "smallest" triangles are labeled $D,$ $F,$ and $G.$

Triangle $\triangle ADE$ is similar to triangle $\triangle DFG,$ and side $DE$ of $\triangle ADE$ is exactly $4$ times as long as the corresponding side $DF$ of $\triangle DFG.$ Denoting the areas of the triangles by $\mathit{Area}(\triangle ADE)$ and $\mathit{Area}(\triangle DFG),$ and using the fact that the areas of similar polygons are proportional to the squares of corresponding sides, $$\mathit{Area}(\triangle ADE) = \left(\frac{DE}{DF}\right)^2 \mathit{Area}(\triangle DFG) = 4^2 \cdot 1 = 16.$$

Since triangle $ABC$ is partitioned by $DE$ into triangle $\triangle ADE$ and trapezoid $DBCE,$ $$\mathit{Area}(\triangle ABC) = \mathit{Area}(\triangle ADE) + \mathit{Area}(DBCE)$$ and therefore $$\mathit{Area}(DBCE) = \mathit{Area}(\triangle ABC) - \mathit{Area}(\triangle ADE) = 40 - 16 = 24.$$