Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the radius of the ascribed semicircle be 2 units.
Therefore;
Radius of the inscribed circle is 1 unit.
It implies;
Area R is;
Area sector with radius 2 units and angle 45° - Area triangle with height and base √(2) unit each.
= (45*2*2π÷360)-½(√(2))²
= ½(π)-1
= ½(π-2) square unit.
Area S is;
Area triangle with height and base √(2) unit each - Area sector with radius 1 unit and angle 90° + Area triangle with height and base 1 unit respectively.
= (½(√(2))²) - (90*1*1*π÷360) + (½*1*1)
= 1 - ¼(π) + ½
= ¼(4-π+2)
= ¼(6-π) square unit.
Therefore;
R Area ÷ S Area is;
½(π-2) ÷ ¼(6-π)
= 2(π-2)/(6-π)
