In the figure below, the right trapezoids $A B C D$ and $A E F G$, with $B C \| E F$ and $C D \| F G$, have the same area. Knowing that $B C=4, A D=7, C T=1$ and $T D=2$, determine the measure of the segment $D G$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-23.jpg?height=354&width=942&top_left_y=594&top_left_x=657)

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Solution

Draw a perpendicular from $B$ to side $D A$, intersecting $E F$ at $Y$ and $A D$ at $X$. Let $D G$ be $x$ and $E Y$ be $y$. See the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-23.jpg?height=412&width=1028&top_left_y=1296&top_left_x=614)

The triangles $\triangle E Y B$ and $\triangle A X B$ have the same angles and are therefore similar. Thus, $\frac{y}{3}=\frac{1}{1+2}$, and consequently $y=1$.

Let's calculate the areas of the given trapezoids:

$$
\begin{aligned}
{[A B C D] } & =\frac{C D \cdot(B C+A D)}{2} \\
& =\frac{3 \cdot(4+7)}{2} \\
& =\frac{33}{2}
\end{aligned}
$$

and

$$
\begin{aligned}
{[A E F G] } & =\frac{F G \cdot(E F+A G)}{2} \\
& =\frac{2 \cdot((x+5)+(x+7))}{2} \\
& =2 x+12
\end{aligned}
$$

Since $[A B C D]=[A E F G]$, we have $33=4 x+24$ and $D G=x=9 / 4$.