$\begin{array}{l}\text{In\hspace{0.17em}a triangle }ABC\text{\hspace{0.17em}, the lengths of sides }AC\text{\hspace{0.17em} and }AB\text{\hspace{0.17em} are 12cm and 5 cm respectively. }\\ \text{If the area of triagnle }ABC{\text{\hspace{0.17em} is 30\hspace{0.17em}cm}}^2\text{\hspace{0.17em}and }R\text{\hspace{0.17em} and \hspace{0.17em}}r\text{\hspace{0.17em} are respectively the radii of circumcircle }\\ \text{and incircle of triangle\hspace{0.17em}}ABC\text{\hspace{0.17em} then the value of }2R+r\text{\hspace{0.17em}is equal to\hspace{0.17em}}\end{array}$

$\begin{array}{l}\text{Given the lengths of sides }AC\text{\hspace{0.17em} and }AB\text{\hspace{0.17em} are 12cm and 5 cm respectively. }\\ \text{If the area of triagnle }ABC{\text{\hspace{0.17em} is 30\hspace{0.17em}cm}}^2\\ \text{Area of triangle is equal to half of product of two sides}\\ \text{it means the triangle is right angled triangle}\\ \angle A=90°\\ \text{The hypotenuse of the right angled triangle is }BC=13\text{\hspace{0.17em}cm. }\left(\text{pythagoras Theorem}\right)\\ \text{Circum radius of the triangle is half of hypotenuse}\\ R=\frac{13}{2}\\ r=\frac{\Delta }{s}=\frac{30}{\left(\frac{12+13+5}{2}\right)}=\frac{60}{30}=2\\ 2R+r=13+2=15\end{array}$