Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Radius of the inscribed circle, r is;
2r² = 4
r² = 2
r = √(2) units.
Therefore;
Area R in square unit to 3 d. p. is;
Area sector with radius √(2) units and angle atan(0.5√(2))° - Area triangle with height 0.8165 units and base √(2)sin(atan(0.5√(2))) units - Area sector with radius 2 units and angle 25.5286945392° + Area triangle with height 2 units and base (2sin25.5286945392) units.
= (atan(0.5√(2))π*√(2)²÷360) - (0.5*0.8165*√(2)sin(atan(0.5√(2)))) - (25.5286945392π*2²÷360) + (0.5*2*2sin25.5286945392)
= 0.61547970867 - 0.33333472916 - 0.89111954689 + 0.86192614063
= 0.25295157325
≈ 0.253 square units.
