Area to volume

Considering the sector of the circle:

let $L$ be the radius of the sector of the circle and $c$ be the arc length. Then

$\dfrac{1300}{9}\pi=\dfrac{130}{360}\pi(L^2)$ $\color{#D61F06}\large \implies$ $L=20$

$c=\dfrac{130}{360}(2)(\pi)(20)=\dfrac{130}{9}\pi$

Considering the right circular cone:

The arc length of the sector is equal to the circumference of the base of the cone , so we have $\dfrac{130}{9} \pi=2 \pi r$ $\color{#D61F06}\large \implies$ $r=\dfrac{65}{9}$

By the pythagorean theorem , we have $h=\sqrt{20^2-\left(\dfrac{65}{9}\right)^2}=\sqrt{\dfrac{28175}{81}}=\dfrac{35\sqrt{23}}{9}$

Hence, the volume of the cone is,

$V=\dfrac{1}{3} \pi r^2h=\dfrac{1}{3} \pi \left(\dfrac{65}{9}\right)^2\left(\dfrac{35\sqrt{23}}{9}\right)=\dfrac{147875}{2187} \pi \sqrt{23}$

Finally,

$a+b=147875+2187=$ $\color{#D61F06}\boxed{\large 150062}$