Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side length of the square be 10 unit.
Area square is;
10²
= 100 square unit.
Therefore;
Area R is;
Area semi circle with radius 5 unit - Area sector with radius 5 unit and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) unit and √(200-200cos(2atan(½))) unit + Area sector with radius 10 unit and angle 2atan(½)° - 2(area quarter circle with radius 5 unit - area triangle with height and base 5 unit respectively) - area sector with radius 5 unit and angle (2atan2-90)° + area triangle with two side 5 unit and angle (2atan2-90)° - area triangle with height 3 unit and base 1 unit - area triangle with height 5 unit and base 1 unit + area sector with radius 5 unit and angle atan(1/5)°
½(25π) - (180-2atan(2))÷360*25π - ½(√(10²+5²)*√(200-200cos(2atan(½)))) + 2atan(½)÷360*100π - 2(¼(25π)-½(25)) - (2atan2-90)÷360*25π + ½(25)sin(2atan2-90) - ½(3) - ½(5) + atan(1/5)÷360*25π
= 12.5π - 11.591190225 - 50 + 46.3647608999 - 14.2699081699 - 8.04376385992 + 7.5 - 1.5 - 2.5 + 2.46744449812
= 9.77357067487 - 2.0763193618
= 7.69725131307 square unit
It implies;
Area R ÷ Area square is;
7.69725131307 ÷ 100
= 0.07697251313
≈ 0.077 to 3 decimal places.
