Mathematics Question and Solution
a = 4+4+x
a = (8+x) cm.
a is the diameter of the bigger inscribed circle and also the side length of the square.
b = ½(a)
b = ½(8+x) cm.
6+y = b
6+y = ½(8+x)
12+2y = 8+x
x = 2y+4 --- (1).
c = y+b
c = y+½(8+x)
c = ½(2y+8+x) cm.
c is the diameter of the smaller inscribed circle.
d = ½(c)
d = ¼(2y+8+x) cm.
d is the radius of the smaller inscribed circle.
e = d-y
e = ¼(2y+8+x)-y
e = ¼(8+x-2y) cm.
Therefore;
(¼(2y+8+x))² = (x/2)²+(¼(-2y+8+x))²
(4y²+16y+2xy+16y+64+8x+2xy+8x+x²)/16 = (x²/4)+(4y²-16y-2xy-16y+64+8x-2xy+8x+x²)/16
(32y+32y+8xy)/16 = (x²/4)
64y+8xy = 4x²
16y+2xy = x² --- (2).
Substituting (1) in (2) to get y.
16y+2y(2y+4) = (2y+4)²
16y+4y²+8y = 4y²+16y+16
8y = 16
y = 2 cm.
Calculating x.
At (1).
x = 2y+4
And y = 2 cm.
x = 2(2)+4
x = 4+4
x = 8 cm.
Notice!
a = (x+8) cm.
Where a is the square side length and also the diameter of the bigger inscribed circle.
And x = 8 cm.
Therefore;
a = 8+8
a = 16 cm.
Again, a is the side length of the square and also the diameter of the bigger inscribed circle.
Area of the square is;
a²
= 16²
= 256 cm²
