We want to find the area of the given trapezoid. We can think of the trapezoid as a rectangle and a triangle to find the area. Remember to find the length of any unknown sides when breaking up the shape!
We do not have enough information to find x_1 and x_2 as the figure is right now. Let's rearrange the figure again to find the area of the triangles.
Now we can find the area of the triangles as the area of one bigger triangle with a height of $\colIII{5}$ $\colIII{\text{centimeters}}$ and a base of $\colIV{x_1+x_2}$ $\colIV{\text{centimeters}}.$ Let's find $\colIV{x_1+x_2}!$
\begin{gathered}
\colIV{x_1}+\colV{3}+\colIV{x_2}=\colIII{7} \Rightarrow \StackEqIIb{\colIV{x_1+x_2}=\colIII{7}-\colV{3}}{\colIV{x_1+x_2}=4}
\end{gathered}
We now have all the information we need to get the area of the total figure. First, let's think about the rectangle.
The length $\ell$ of the rectangle is $\colIII{5}$ $\colIII{\text{centimeters}}$ and the width $w$ is $\colV{3}$ $\colV{\text{centimeters}}.$ We can substitute these values into the formula for the area of a rectangle.
The area of the rectangle is $15$ square centimeters. Next, let's look at the triangles.
The base for the triangle is $\colIV{4}$ $\colIV{\text{centimeters}}.$ The height of the triangle is $\colIII{5}$ $\colIII{\text{centimeters}}.$ We can substitute these values into the formula for the area of a triangle.
The area of the triangle is $10$ square centimeters. Finally, let's add both areas to find the total area of the trapezoid.
\begin{gathered}
\textbf{Area of the Trapezoid}\\
15+10=25\text{ square centimeters}
\end{gathered}
