1. Inscribed Area = n/2 Sin[2 Pi / n].

Code and Figures.

Input := 

circle = PolarPlot[1, {th, 0, 2 Pi},
 DisplayFunction -> Identity];

Input := 

norm1 = Sqrt[1 + Tan[Pi/5]^2];
norm3 = Sqrt[1 + Tan[3 Pi/5]^2];
norm5 = Sqrt[1 + Tan[5 Pi/5]^2];
norm7 = Sqrt[1 + Tan[7 Pi/5]^2];
norm9 = Sqrt[1 + Tan[9 Pi/5]^2];

Input := 

lines = ParametricPlot[{
   {t/norm1, Tan[Pi/5] t/norm1},
   {-t/norm3, Tan[3 Pi/5] t/norm3},
   {-t/norm5, Tan[5 Pi/5] t/norm5},
   {-t/norm7, Tan[7 Pi/5] t/norm7},
   {t/norm9, Tan[9 Pi/5] t/norm9}}, {t, 0, 1},
   DisplayFunction -> Identity];

Input := 

secants = ParametricPlot[{
 {Cos[Pi/5] + (Cos[3 Pi/5] - Cos[Pi/5]) t,
    Sin[Pi/5] + (Sin[3 Pi/5] - Sin[Pi/5]) t},
 {Cos[3 Pi/5] + (Cos[5 Pi/5] - Cos[3 Pi/5]) t,
    Sin[3 Pi/5] + (Sin[5 Pi/5] - Sin[3 Pi/5]) t},
 {Cos[5 Pi/5] + (Cos[7 Pi/5] - Cos[5 Pi/5]) t,
    Sin[5 Pi/5] + (Sin[7 Pi/5] - Sin[5 Pi/5]) t},
 {Cos[7 Pi/5] + (Cos[9 Pi/5] - Cos[7 Pi/5]) t,
    Sin[7 Pi/5] + (Sin[9 Pi/5] - Sin[7 Pi/5]) t},
 {Cos[9 Pi/5] + (Cos[Pi/5] - Cos[9 Pi/5]) t,
    Sin[9 Pi/5] + (Sin[Pi/5] - Sin[9 Pi/5]) t}},  {t, 0, 1},
 DisplayFunction -> Identity];

Input := 

p1 = Show[circle, lines, secants,
 DisplayFunction -> $DisplayFunction,
 Prolog -> AbsoluteThickness[3]]

Output =

-Graphics-

Input := 

morenorm = Sqrt[1 + Tan[Pi/5]^2];

Input := 

morelines = ParametricPlot[{{Cos[Pi/5] t, 0},
   {Cos[Pi/5] t, Sin[Pi/5] t},
   {Cos[Pi/5] t, -Sin[Pi/5] t},
   {Cos[Pi/5], Sin[Pi/5] - Sin[Pi/5] t},
   {Cos[Pi/5], -Sin[Pi/5] + Sin[Pi/5] t}}, {t, 0, 1},
   PlotStyle -> {AbsoluteThickness[5]},
   DisplayFunction -> Identity];

Input := 

Show[p1, morelines,
 DisplayFunction -> $DisplayFunction]

Output =

-Graphics-

The area of one of the bold triangles is (1/2)*Cos[theta]*Sin[theta] (where in this case theta=Pi/5. Thus, the area of one of the n (or 5 in this case) is Cos[theta] Sin[theta] leaving an

Area = n Cos[theta] Sin[theta].

Remark: One can compactly represent the area equivalently by

Area = n/2 Sin[2 Pi/n]