Question Number 92807 by Crabby89p13 last updated on 09/May/20
Commented by i jagooll last updated on 09/May/20
$$\mathrm{28}\:\mathrm{cm}^{\mathrm{2}} \\ $$
Answered by M±th+et+s last updated on 09/May/20
Answered by M±th+et+s last updated on 09/May/20
$${x}+{Z}+{Z}+{R}+{R}+{y}=\mathrm{32}+\mathrm{20}+\mathrm{16} \\ $$$${x}+{y}+\mathrm{2}{Z}+\mathrm{2}{R}=\mathrm{68} \\ $$$${x}+{y}=\mathrm{68}−\mathrm{2}\left({Z}−{R}\right) \\ $$$$=\mathrm{68}−\left(\mathrm{2}×\mathrm{20}\right)=\mathrm{28}{cm}^{\mathrm{2}} \\ $$$$ \\ $$