Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the single side length of the square be 2 unit.
Therefore;
Area blue is;
Area triangle of height (12+2√(3))/9 unit and base 2 unit - Area triangle of height 1 unit and base 2 unit.
= ½(2x(12+2√(3))/9) - ½(2x1)
= (12+2√(3))/9) - 1
= ⅑(3+2√(3)) square unit.
Area red is;
Area triangle with height (√(3)/3) unit and base (5+√(3))/22 unit.
= ½(√(3)/3)((√(3)+5)/22)
= (3+5√(3))/132 square unit.
Therefore;
Area Red/Area Blue is;
(3+5√(3))/132 ÷ ⅑(3+2√(3))
= (15√(3)+9)/(132+88√(3))
