What is the approximate area of a regular pentagon with a side length of 6 feet and a distance from the center to a vertex of 6 feet?

In order to find the area of a regular Pentagon, which is composed of 5 congruent isosceles triangles with bases (b) being the sides of the Pentagon (P).
So b = 6ft.
Distance from the center, where all 5 triangle apexes touch, to each vertex also makes the sides of the triangles. This = 6ft. We'll call it side (s). Now to find the area of each triangle, we need the apothem, which is the height of the Isosceles triangle. The height (h) or apothem bisects the polygon side/triangle base (b) into 2 equal lengths. So b/2 = 6/2 = 3ft.
Now we have 2 right triangles making up each of the 5 triangles, split by the apothem h.
We can find h by Pythagorean Theorem: h^2 = (b/2)^2 + s^2
So h^2 = 3^2 + 6^2 = 9 + 36 = 45
So square root (sr) of h^2 = h, which = square root of 45
--> h = sr (9) × sr (5) = 3sr5
Now the area of a triangle (At) = 1/2 b×h --> At = 3×3sr5 = 9sr5
Now that's one triangle, multiply it by 5 to get the area of all 5 triangles, which as we've said make up the Area of the regular Pentagon (P).
So P = 5 × 9sr5 = 45sr5, or approximately 100.62 ft. squared
Hope this helps!

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