Mathematics Question and Solution
Let the single side length of the square be x.
Therefore;
Area rectangle of length x unit and width 1 cm + area triangle of height x cm and width (x-2) cm + 2(area triangle of height (x-1) cm and width 1 cm) + 2(area triangle of height ½√(x²+(x-2)²) cm and width 5 cm) = Area square of single side length x cm.
It will be;
x+½(x(x-2))+½(2(x-1))+¼(10√(x²+(x-2)²)) = x²
2x+x²-2x+2x-2+5√(2x²-4x+4) = 2x²
x²+2x-2+5√(2x²-4x+4) = 2x²
5√(2x²-4x+4) = x²-2x+2
5√(2(x²-2x+2)) = x²-2x+2
Let x²-2x+2 = y
Therefore;
5√(2y) = y
25(2y) = y²
50y = y²
Therefore;
y = 50
It implies;
x²-2x+2 = 50
x²-2x-48 = 0
(x-1)² = 48+1²
x = 1±√(49)
x = 1±7
x ≠ -6
x = 8
Therefore area blue, a trapezoid will be;
½(1+(x-1))x
= ½(1+(8-1))*8
= ½(8*8), half the area of the square.
= ½(64)
= 32 cm²
