Question 1204506
<font color=black size=3>
Diagram
{{{
drawing(400,400,-5,5,-5,5,

line(-2.11,-2.14,-2.14,2.11),line(-2.14,2.11,2.11,2.14),line(2.11,2.14,2.14,-2.11),line(2.14,-2.11,-2.11,-2.14),

circle(0,0,3),circle(0,0,sqrt(4.5)),

locate(-4,-4,matrix(1,4,"Diagram","is","to","scale"))

)
}}}


Radius of larger circle = 72 cm
Diagonal of square = 2*72 = 144 cm
Side length of square = 144/sqrt(2) = 72*sqrt(2) cm
Radius of smaller circle = 72*sqrt(2)/2 = 36*sqrt(2) cm


Area of large circle = pi*r^2 = pi*(72)^2 = 5184pi
Area of smaller circle = pi*r^2 = pi*(36*sqrt(2))^2 = 2592pi


Area of wasted material = (Area of large circle) - (Area of small circle)
Area of wasted material = 5184pi - 2592pi
Area of wasted material = <font color=red>2592pi</font>


It's not a coincidence that the smaller circle area and the amount of wasted material is the same.
In other words, the area of the smaller circle is half that of the larger circle.
Here's scratch work to prove this claim.
r = large circle radius
2r = square diagonal
2r/sqrt(2) = r*sqrt(2) = square side length
r*sqrt(2)/2 = smaller circle radius
smaller circle area = pi*(smaller radius)^2
smaller circle area = pi*(r*sqrt(2)/2)^2 
smaller circle area = (1/2)*pi*r^2
smaller circle area = (1/2)*(larger circle area)



Answer: <font color=red size=4>Choice C) 2592pi</font>
</font>