Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the length of the inscribed rectangle be 2 units.
Let the width of the inscribed rectangle be √(3) units.
Therefore the single side length of the inscribed square x will be;
x+((x)/√(3))+√(3)x=2
x=2√(3)/(4+√(3)) units.
Area R is;
x² = (2√(3)/(4+√(3)))²
= 12/(19+8√(3)) square units.
= 0.36522557677 square units.
Area S will be;
Area triangle of height √(3) units and base 1 unit + Area triangle of two side 2 units and (2-(6+2√(3))/(4+√(3))) units, and angle 60°, the angle they form.
= ½√(3)+½(2x(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
