Answer:
64
in²
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 =
![\pi](https://img.qammunity.org/2023/formulas/mathematics/high-school/s4a9atp2j2mg6r59j0huycjcajdeqo7hpu.png)
![r^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/7llb7qkbn5xzomhq0yri10x1phwgy9lxdn.png)
And if we have 4 circles that means the area of one circle is 1/4th the total
So,
A=
![(\pi r^(2) )/(4)](https://img.qammunity.org/2023/formulas/mathematics/middle-school/ftnp3qp8o0j6uyvy6hm64a0s2fy699nbo3.png)
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=
=
=4
![\pi](https://img.qammunity.org/2023/formulas/mathematics/high-school/s4a9atp2j2mg6r59j0huycjcajdeqo7hpu.png)
Now lets get it back to
by square rooting the 4
A=
=2
![\pi](https://img.qammunity.org/2023/formulas/mathematics/high-school/s4a9atp2j2mg6r59j0huycjcajdeqo7hpu.png)
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
which simplifies to 64
![\pi](https://img.qammunity.org/2023/formulas/mathematics/high-school/s4a9atp2j2mg6r59j0huycjcajdeqo7hpu.png)
And a final answer of
64
in²
Hope this helps :)