Answer:
maximum difference of radii =[tex](r-R)=\frac{1}{2\pi ^{2}}[/tex]
Step-by-step explanation:
We know that area of circle is given by
[tex]A=\pi \times (radius)^{2}[/tex]
For circle with radius 'r' we have
[tex]A_{1}=\pi \times (r)^{2}[/tex]
For circle with radius 'R' we have
[tex]A_{2}=\pi \times (R)^{2}[/tex]
Now according to given condition we have
[tex]A_{1}-A_{2}\leq \frac{5}{\pi }[/tex]
[tex]\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}[/tex]
Thus maximum difference of radii =[tex](r-R)=\frac{1}{2\pi ^{2}}[/tex]
