Answer:
64[tex]\pi[/tex] in²
Step-by-step explanation:
I did not ever do a problem like this, however this is the beauty of math, you can easily reverse engineer it.
Remember,
A = [tex]\pi[/tex][tex]r^{2}[/tex]
And if we have 4 circles that means the area of one circle is 1/4th the total
So,
A=[tex]\frac{\pi r^{2} }{4}[/tex]
Assuming that 16 is the radius squared times 4 lets ignore that squared for now because when going backwards we would get rid of the squared last as that was the first step.
A= [tex]\frac{16\pi }{4}[/tex]=[tex]\frac{4\pi }{1}[/tex]=4[tex]\pi[/tex]
Now lets get it back to [tex]r^{2}[/tex] by square rooting the 4
A=[tex]\sqrt{4\pi }[/tex]=2[tex]\pi[/tex]
The radius of one small circle is 2. Therefore; the diameter would be 4 for each. This in mind we know that two small circles diameters make up the radius of the larger circle we will multiply it by two again.
This gives us a final radius of the bigger circle of 8
Therefore, the area of the bigger circle is 8²[tex]\pi[/tex] which simplifies to 64[tex]\pi[/tex]
And a final answer of
64[tex]\pi[/tex] in²
Hope this helps :)
