Question Number 13307 by tawa tawa last updated on 18/May/17
Answered by ajfour last updated on 18/May/17
$${A}=\pi{RL}−\pi{rl}+\mathrm{2}\pi{rh} \\ $$$$\:\:\:\:=\pi\left(\mathrm{9}{cm}\right)\left(\frac{\mathrm{12}}{\mathrm{5}}×\mathrm{9}{cm}\right)−\pi\left(\mathrm{4}{cm}\right)\left(\frac{\mathrm{12}}{\mathrm{5}}×\mathrm{4}{cm}\right)+\mathrm{2}\pi\left(\mathrm{4}{cm}\right)\left(\mathrm{10}{cm}\right) \\ $$$$\:\:\:\:=\left[\pi\left(\frac{\mathrm{12}}{\mathrm{5}}\right)\left(\mathrm{81}−\mathrm{16}\right)+\mathrm{80}\pi\right]\:{cm}^{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{236}\pi\:{cm}^{\mathrm{2}} . \\ $$
Commented by tawa tawa last updated on 18/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$