Mathematics Question and Solution
Let a be the area of the ascribed circle.
b = ½(a) units.
c = (a+2) units.
d = ½(c)
d = ½(a+2) units.
d is the radius of the big inscribed circle.
e = (a-2) units.
f = ½(e)
f = ½(a-2) units.
f is the radius of the small inscribed circle.
Therefore;
Calculating a, radius of the ascribed circle.
½(a)*½(a) = 2*(½(a+2)+½(a+2)-2)
¼(a²) = 2(a)
It implies;
a² = 8a
a = 8 units.
Again, a is the radius of the ascribed square.
d = ½(a+2)
And a = 8 units.
d = ½(8+2)
d = 5 units.
Again, d is the radius of the big inscribed circle.
f = ½(a-2)
And a = 8 units.
f = ½(8-2)
f = 3 units.
Again, f is the radius of the small inscribed circle.
Blue Area is;
(π*a²)-(π*d²)-(π*f²)
= (8²)π-(5²)π-(3²)π
= 64π-25π-9π
= 64π-34π
= 30π square units.
