Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the square side be x.
Calculating x.
25 = (x²-9)+2x²-2√(2)x*√(x²-9)cos(45+y)
Where y is an angle.
Notice;
cosy = √(x²-9)/x and siny = 3/x.
It implies;
25 = (x²-9)+2x²-2√(2)x*√(x²-9)((√(2x²-18)-3√(2))/2x)
34 = x²+18+6√(x²-9)
16 = x²+6√(x²-9)
x = √(10) units.
Therefore;
Area A exactly in square units is;
Area sector with radius 5 units and angle asin(4/5) - Area triangle with height 2√(5) units and base sin(45+atan3) units - Area triangle with height and base √(10) units respectively - Area triangle with height √(10) units and base 4sin(atan⅓) units.
= (asin(4/5)*π*5²÷360) - (½*2√(5)*sin(45+atan3) - (½*(√(10))²) - (½*4*√(10)sin(atan⅓))
= (5πasin(4/5)/72) - 2 - 5 - 2
= (5πasin(4/5)/72) - 9
= (5πasin(4/5) - 648)/72 square units.
