Like you would a regular prism.
I assume it is the space (in 3 dimensions) that something occupies
Mostly true - you cannot tessellate only regular pentagons in two dimensions, since you cannot sum up the intersection of the angles to 360 degrees. If you tessellate a regular pentagon in three dimensions, you end up with a dodecahedron.
This question cannot be answered sensibly. A square foot is a measure of area, with dimensions [L2]. A [regular] foot is a measure of distance, with dimensions [L]. Basic dimensional analysis teaches that you cannot convert between measures with different dimensions such as these without additional information.
A pentagon is a plane figure and has only two dimensions. It cannot have three dimensions.
What you are looking for are driven dimensions. Derived dimensions must be a typo. Driven dimensions are enclosed in parentheses to distinguish them from regular dimensions in inventor. These dimensions do not contrain a sketch they simply reflect dimensioned geometry which is most likely under some geometric constraint.
Impossible to figure out. You did not mention any dimensions of the figure. You mush have the dimensions.
In 2 dimensions, a regular 5 sided figure is a pentagon. In 3 dimensions, there is no regular sided 5 sided figure. The tetrahedron is a regular 4 sided figure and a cube is a regular 6 sided figure.
it will melt
Like you would a regular prism.
A 2-dimensions regular polygon is one which has all its sides equal AND all its angles equal.
I assume it is the space (in 3 dimensions) that something occupies
Mostly true - you cannot tessellate only regular pentagons in two dimensions, since you cannot sum up the intersection of the angles to 360 degrees. If you tessellate a regular pentagon in three dimensions, you end up with a dodecahedron.
This question cannot be answered sensibly. A square foot is a measure of area, with dimensions [L2]. A [regular] foot is a measure of distance, with dimensions [L]. Basic dimensional analysis teaches that you cannot convert between measures with different dimensions such as these without additional information.
190 in is a length, not an area. Even if you were given an area, unless it was in the form of a regular shape you would not be able to determine its dimensions.
A pentagon is a plane figure and has only two dimensions. It cannot have three dimensions.
Figure it like this: Because each side of the rectangular plot will have a one-meter wide sidewalk adjoining it, you know that the sidewalk, on each side of the plot will be two meters longer than the plot's dimensions. However, if you figured the length by adding the areas of each length of sidewalk, you would be wrong, because each length of sidewalk shares two corners with the adjacent lengths. Each corner is one square meter, and each length of sidewalk without two corners is the same as the dimension of the plot, times one meter. So essentially, you just add the plot dimensions and then add four. In other words, 2(13m * 1m) + 2(20m * 1m) + 4m2 = 26m2 + 40m2 + 4m2 = 70m2. Another way to figure it would be to figure the area which will be enclosed by the outer edge of the sidewalk (again, two meters longer than the plot on each side), and subtract the area of the plot itself, or: (15m * 22m) - (13m * 20m) = 330m2 - 260m2 = 70m2.