Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side of the inscribed square be 2 units.
Therefore;
Area S is;
Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45°
= 0.5*1.03528*0.53589838486sin45
= 0.1961531464 space units.
Area R is;
Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609°
= 0.5*(2√(2)-2)*0.65961372961sin72.367664609
= 0.26038499708 space units.
Area T is;
Area sector with radius 2√(2) units and angle 30° - Area triangle with two side 2√(2) units and 2 units, and angle 30° - Area R
= (30(8π)÷360) - (0.5*2√(2)*2sin30) - 0.26038499708
= 2.09439510239 - 1.41421356237 - 0.26038499708
= 0.41979654294 space units.
Therefore;
Area T ÷ Area (R+S) to 2 d. p. is;
0.41979654294 ÷ (0.26038499708+0.1961531464)
= 0.41979654294 ÷ 0.45653814348
= 0.91952129069
≈ 0.92
