The dimension that would give the maximum area is 20.8569
<h3>How to solve for the maximum area</h3>
Let the shorter side be = x
Perimeter of the semi-circle is πx
Twice the Length of the longer side
![[70-(\pi )x -x]](https://tex.z-dn.net/?f=%5B70-%28%5Cpi%20%29x%20-x%5D)
Length = ![[70-(1+\pi )x]/2](https://tex.z-dn.net/?f=%5B70-%281%2B%5Cpi%20%29x%5D%2F2)
Total area =
area of rectangle + area of the semi-circle.
Total area =
![x[[70-(1+\pi )x]/2] + [(\pi )(x/2)^2]/2](https://tex.z-dn.net/?f=x%5B%5B70-%281%2B%5Cpi%20%29x%5D%2F2%5D%20%2B%20%5B%28%5Cpi%20%29%28x%2F2%29%5E2%5D%2F2)
When we square it we would have
![70x +[(\pi /4)-(1+\pi)]x^2](https://tex.z-dn.net/?f=70x%20%2B%5B%28%5Cpi%20%2F4%29-%281%2B%5Cpi%29%5Dx%5E2)
This gives
![70x - [3.3562]x^2](https://tex.z-dn.net/?f=70x%20-%20%5B3.3562%5Dx%5E2)
From here we divide by 2

The maximum side would be at

This gives us 20.8569
Read more on areas and dimensions here:
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2 1/4+1 1/4=3 1/2
or 3.5 (3 and a half yards. )
I do not completely understand your question, but if you mean area by "murals", then when the width increases, area does, too. For example:
I have a rectangular board that is 5 feet by 3 feet. The area is 15 square feet. I have another board with a longer width, but the same length. It is 5 feet by 4 feet. Its area is 20 square feet.
Answer:
40 degrees
Step-by-step explanation:
Look at the attachment