​ HALP!!!

In the figure, the area of trapezoid DBCE is 80 cm²
The ratio of the bases DE to BC is 3 : 5.
What is the area of triangle ADE, in square centimeters?

$\text{Let trapezoid $ DBCE = A_1 $}\\ \text{Let triangle $ ADE = {\color{red}A_2} $} \\ \text{Let triangle $ ABC = A = A_1+A_2 = \dfrac{BC\cdot h}{2} $}$

$\begin{array}{|rcll|} \hline \dfrac{DE}{BC} = \dfrac{3}{5} &=& \dfrac{h_2}{h} \\\\ \dfrac{3}{5} &=& \dfrac{h_2}{h} \\\\ \mathbf{h_2} & \mathbf{=} & \mathbf{\dfrac{3}{5}h} \\\\ \hline h &=& h_1+h_2 \\\\ h &=& h_1+\dfrac{3}{5}h \\\\ h_1 &=& h-\dfrac{3}{5}h \\\\ \mathbf{h_1} & \mathbf{=} & \mathbf{\dfrac{2}{5}h} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline A_2 &=& \dfrac{DE\cdot h_2}{2} \quad | \quad h_2 =\dfrac{3}{5}h \\\\ A_2 &=& \dfrac{DE}{2}\cdot \dfrac{3}{5}h \\\\ A_2 &=& DE\cdot h \cdot \dfrac{3}{10} \\\\ \mathbf{DE\cdot h } & \mathbf{=} & \mathbf{\dfrac{10} {3} A_2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline A_1 &=& \left(\dfrac{BC+DE}{2}\right) h_1 \quad | \quad h_1 =\dfrac{2}{5}h \\\\ A_1 &=& \left(\dfrac{BC+DE}{2}\right) \dfrac{2}{5}h \\\\ A_1 &=& \dfrac{BC}{2}\cdot \dfrac{2}{5}h + \dfrac{DE}{2}\cdot \dfrac{2}{5}h \\\\ A_1 &=& \dfrac{BC\cdot h}{2}\cdot \dfrac{2}{5} + DE\cdot h \cdot \dfrac{1}{5} \quad | \quad \dfrac{BC\cdot h}{2} = A_1 + A_2,\ \mathbf{DE\cdot h=\dfrac{10} {3} A_2} \\\\ A_1 &=& (A_1+A_2)\cdot \dfrac{2}{5} + \dfrac{10} {3} A_2\cdot \dfrac{1}{5} \\\\ A_1 &=& \dfrac{2}{5}A_1 +\dfrac{2}{5}A_2 + \dfrac{10} {15} A_2 \\\\ A_1-\dfrac{2}{5}A_1 &=& \dfrac{2}{5}A_2 + \dfrac{10} {15} A_2 \\\\ \dfrac{3}{5}A_1 &=& \dfrac{6}{15}A_2 + \dfrac{10} {15} A_2 \\\\ \dfrac{3}{5}A_1 &=& \dfrac{16} {15} A_2 \\\\ 3A_1 &=& \dfrac{16} {3} A_2 \\\\ A_2 &=& \dfrac{9} {16} A_1 \quad | \quad A_1 = 80\ cm^2 \\\\ A_2 &=& \dfrac{9} {16} \cdot 80\ cm^2 \\\\ A_2 &=& 9\cdot 5\ cm^2 \\\\ \mathbf{A_2} & \mathbf{=} & \mathbf{45 cm^2} \\ \hline \end{array}$

The area of triangle ADE is 45 cm²

Thanks!

You explained the problem really well and helped me understand!