Answer: [tex]2x^2+\frac{\pi x^2}{4}[/tex]
Step-by-step explanation:
The area of a rectangle can be calculated with this formula:
[tex]A_r=lw[/tex]
Where "l" is the lenght and "w" is the width.
The area of a semi-circle can be found with this formula:
[tex]A_{sc}=\frac{\pi r^2}{2}[/tex]
Where "r" is the radius.
Let be "x" the shorter side of the rectangle (which is also the diameter of the semi-circle).
Based on the information given in the exercise, you know that:
[tex]w=x\\\\l=2x[/tex]
Since the radius is half the diameter:
[tex]r=\frac{x}{2}[/tex]
Observe the figure attached, which shows the stadium.
The area of the stadium is the sum of the areas of the semi-cirlcles and the rectangle.
Therefore, you can construct the following expression for the area of the stadium is terms of "x":
[tex](2x)(x)+\frac{\pi (\frac{x}{2})^2}{2}+\frac{\pi (\frac{x}{2})^2}{2}[/tex]
Simplifying it, you get:
[tex](2x)(x)+\frac{\frac{\pi x^2}{4}}{2}+\frac{\frac{\pi x^2}{4}}{2}=2x^2+\frac{\pi x^2}{4}[/tex]
