Online Course Discussion Forum

Geometry IIA 3.9 and 5.27

 
 
Picture of John Lensmire
Re: Geometry IIA 3.9 and 5.27
by John Lensmire - Thursday, April 14, 2022, 1:52 PM
 

As in your picture, we can start by finding the distance of $AC$, which is$$\sqrt{(5-1)^2 + (5-1)^2} = \sqrt{32} = 4\sqrt{2}.$$Be careful however, because each of the 4 circles has a radius of $2$. So when we're finding the diameter of the big circle, it's equal to the distance from the bottom left edge of circle $A$ to the center (which is length $2$), plus the length of $AC$ (which is $4\sqrt{2}$), plus the distance from the center of circle $C$ to it's top right edge (also length $2$). This gives a diameter of $4+4\sqrt{2}$.

It looks like you may have mixed up the outside radiuses of the circles a bit. Hope this helps!