First, we find the $x$-coordinates of the points where the lines intersect the $y$-axis.
For $y = x$, the $x$-coordinate is $0$.
For $y = 10$, the $x$-coordinate is $10$.
For $y = 5$, the $x$-coordinate is $5$.
Next, we find the $x$-coordinates of the points where the lines intersect each other.
For $y = x$ and $y = 10$, the $x$-coordinate is $10$.
For $y = x$ and $y = 5$, the $x$-coordinate is $5$.
For $y = 5$ and $y = 10$, the $x$-coordinate is $5$.
We can now draw the trapezoid and calculate its area. The formula for the area of a trapezoid is $\frac{1}{2}(b_1 + b_2)h$, where $b_1$ and $b_2$ are the lengths of the parallel sides and $h$ is the height.
In this case, $b_1 = 10$, $b_2 = 5$, and $h = 5$. Plugging these values into the formula, we get $\frac{1}{2}(10 + 5)(5) = \frac{1}{2}(15)(5) = 37.5$.
The area of the trapezoid is therefore $\boxed{37.5}$ square units.The answer is: 37.5
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
What is the area, in square units, of a trapezoid bounded by the lines $y = x$, $y = 10$, $y = 5$ and the $y$-axis? Express your answer as a decimal to the nearest tenth.

### Answer:
