Question Number 206060 by cortano21 last updated on 06/Apr/24
Answered by mr W last updated on 06/Apr/24
Commented by mr W last updated on 06/Apr/24
$${S}={area}\:{of}\:{parallelogram} \\ $$$${S}=\mathrm{20}+\mathrm{22}+\mathrm{16}+{X} \\ $$$$\mathrm{22}+{X}−{A}=\frac{{S}}{\mathrm{2}}\:=\frac{\mathrm{20}+\mathrm{22}+\mathrm{16}+{X}}{\mathrm{2}} \\ $$$$\Rightarrow{X}−\mathrm{2}{A}\:=\mathrm{14}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\frac{{B}}{{A}}=\frac{\mid{BD}\mid}{\mid{AC}\mid}=\frac{\mathrm{16}}{\mathrm{20}}=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow{B}=\frac{\mathrm{4}{A}}{\mathrm{5}} \\ $$$$ \\ $$$$\mathrm{20}+{A}+\mathrm{22}−{B}=\frac{{S}}{\mathrm{2}} \\ $$$$\mathrm{20}+{A}+\mathrm{22}−\frac{\mathrm{4}{A}}{\mathrm{5}}=\frac{\mathrm{20}+\mathrm{22}+\mathrm{16}+{X}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}{X}−\mathrm{2}{A}=\mathrm{130}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\mathrm{4}{X}=\mathrm{130}−\mathrm{14} \\ $$$$\Rightarrow{X}=\mathrm{29}\:\checkmark \\ $$