Mathematics Question and Solution
Calculating area ADBC.
Let let x be the side length of the small square.
a = (2-x) cm.
b² = x²+x²
b = √(2x²)
b = √(2)x cm.
b is BD, the diagonal of the small square.
c² = x²+a²
c² = x²+(2-x)²
c² = x²+4-4x+x²
c = √(2x²-4x+4) cm.
c is AD, the diagonal of the big square.
Notice.
d = 2b
d = 2√(2)x cm.
d is the side length of the big square.
√(2x²-4x+4)² = (2√(2)x)²+(2√(2)x)²
2x²-4x+4 = 16x²
7x²+2x-2 = 0
(x+(1/7))² = (2/7)+(1/7)²
(x+(1/7))² J
= (15/49)
x = -(1/7)±√(15/49)
x = -(1/7)±(√(15)/7)
x = -(1/7)+(√(15)/7)
x = ⅐(√(15)-1)
x = 0.41042619232 cm.
Again, x is the side length of the small square.
Recall.
c = √(2x²-4x+4) cm.
And x = 0.41042619232 cm.
c = √(2(0.41042619232)-4(0.41042619232)+4)
c = 1.64170476926 cm.
Again, c is AD, the diagonal of the big square.
cose = 0.41042619232/1.64170476926
e = acos(0.41042619232/1.64170476926)
e = 75.5224878139°
e is angle BDC = angle CBD.
f = 180-2e
f = 180-2(75.5224878139)
f = 28.9550243722°
f is angle BCD.
g²+g² = c²
2g² = 1.64170476926²
g² = 1.34759727471
g = √(1.34759727471)
g = 1.16086057505 cm.
g is the side length of the big square.
Or
Recall.
d = 2√(2)x cm.
And x = 0.41042619232 cm.
d = 2√(2)*0.41042619232
d = 1.16086057506 cm.
It implies;
d = g.
Therefore, area quadrilateral ADBC is;
½(d²)+½(d²)sinf
= 0.5(1.16086057506*1.16086057506)+0.5(1.16086057506*1.16086057506)sin28.9550243722
= 0.67379863736+0.32620136265
= 1 square units.
