Mathematics Question and Solution
Area of the ascribed regular Pentagon is;
5/(4tan(180/5))
= 1.72047740059 square units.
Single side length of the formed small inscribed regular Pentagon is;
Let it be x.
x² = (1/(1-cos108)) - (1/(1-cos108)cos36)
x = √(0.14589803375)
= 0.38196601125 units.
Therefore area of the small inscribed regular Pentagon is;
5(0.38196601125)²/(4tan(180/5))
= 0.25101426986 square units.
It implies;
Ratios small inscribed regular pentagon to the ascribed big regular pentagon is;
Area small inscribed regular Pentagon ÷ Area ascribed regular Pentagon is;
0.25101426986 ÷ 1.72047740059
= 0.14589803375
