Mathematics Question and Solution
#Sir Mike Ambrose is the author of the question.
Let the side length of the regular Pentagon be 1 unit.
Therefore Area Pentagon is;
(5*1)÷(4tan(180/5))
= 5/(4tan36) space units.
= 1.72047740059 space units.
Area star is;
5(area triangle with height 0.61803398875 units and base 0.61803398875sin36 units) + Area regular Pentagon with side 0.38196601125 units.
= 5(0.5*0.61803398875*0.61803398875sin36) + ((5*0.38196601125²)/(4tan(180/5)))
= 0.56128497072 + 0.25101426986
= 0.81229924058 space units.
It implies;
Area Star ÷ Area Pentagon to 3 d. p. is;
= 0.81229924058 ÷ 1.72047740059
= 0.472135955
≈ 0.472
