Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the bigger inscribed square be 2 units.
BC = 1+2+x = (3+x) units.
Calculating x.
(3+x)²=1²+8-4√(2)cos135
3+x=√(13)
x = (√(13)-3) units.
Therefore;
BC (side of the ascribed square ABCD) is;
BC = 3+√(13)-3
BC = √(13) units.
Therefore;
Area ABCD is;
√(13)²
= 13 square units.
Area Yellow is;
Area trapezium with parallel sides √(13) units and (√(13)-3) unit respectively, and height (√(13)-2) unit - Area triangle with height and base (√(13)-3) unit respectively.
= ½*(√(13)-2)*(√(13)+(√(13)-3)) - ½(√(13)-3)²
= ½(32-7√(13)) - ½(22-6√(13))
= ½(10-√(13)) units.
Area Yellow ÷ Area ABCD exactly as a single fraction is;
½(10-√(13)) ÷ 13
= (10-√(13))/26
