Please solve the NCERT question No.4 Exercise-13.5 (p-258)

Calculations for lateral surface area: --------------------- +P -------------------- | /|\ | | / | \ | | / | \ S-s |H-h | / | \ | | / | \ | | /*****|**r**\ | | ** C+-----*+D-------------- H| S / *********** \ | | / | \ | | / | \ | | / | \ s | | / | \ |h | / | \ | | / | \ | | / **********|********** \ | | /***** | R *****\ | -----* A+----------------+B---- ****** ****** *********************The formula for a complete cone is: A = pi R Swhere R is the radius and S is the slant height of the whole cone. For the frustum, we will subtract the area of the cut-off cone (whose slant height is S-s) from the whole: A = pi R S - pi r (S-s) = pi (RS - rS + rs) = pi ((R-r)S + rs)By the same similar triangles as before, we can write AB CD R r -- = -- or - = --- PB PD S S-sAgain solving for S, R(S-s) = rS RS - Rs = rS RS - rS = Rs (R-r)S = Rs Rs S = --- R-rNow the area is Rs A = pi ((R-r)--- + rs) = pi (Rs + rs) = pi(R+r)s R-r