Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Area R ÷ Area S to 2 decimal places.
Let the base of the inscribed rectangle be 2 units.
tan60 = a/(½(2))
√(3) = a
a = √(3) units.
a is the height of the inscribed rectangle.
Calculating the single side length of the inscribed square, let it be x.
It implies;
x+((x)/√(3))+√(3)x=2
x=2√(3)/(4+√(3)) units.
Therefore;
Area R is;
x² = (2√(3)/(4+√(3)))²
= 12/(19+8√(3)) square units.
= 0.36522557677 square units.
Area S is;
Area triangle with height √(3) units and base 1 unit + Area triangle with length 2 units and base (2-(6+2√(3))/(4+√(3)))sin60 units.
½√(3)+½*(2*(2-(6+2√(3))/(4+√(3)))*sin60
= 1.16819488303 square units.
Therefore;
Area R ÷ Area S to 2 d.p.
is;
(0.36522557676) ÷ (1.16819488303)
= 0.31264096605
≈ 0.31
