Mathematics Question and Solution

By Ogheneovo Daniel Ephivbotor
5th June, 2025

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

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