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
We appreciate you contacting us. Our support will get back in touch with you soon!
Have a great day!
Please note that your query will be processed only if we find it relevant. Rest all requests will be ignored. If you need help with the website, please login to your dashboard and connect to support