Question Number 181439 by mnjuly1970 last updated on 25/Nov/22
Answered by mr W last updated on 25/Nov/22
Commented by mr W last updated on 25/Nov/22
![[ABCD]=[AB′D′] =[AB′C]+[ACD′] =((d_1 ×B′C×sin α)/2)+((d_1 ×CD′×sin (π−α))/2) =((d_1 ×(B′C+CD′)×sin α)/2) =((d_1 d_2 sin α)/2) ✓](https://www.tinkutara.com/question/Q181483.png)
$$\left[{ABCD}\right]=\left[{AB}'{D}'\right] \\ $$$$=\left[{AB}'{C}\right]+\left[{ACD}'\right] \\ $$$$=\frac{{d}_{\mathrm{1}} ×{B}'{C}×\mathrm{sin}\:\alpha}{\mathrm{2}}+\frac{{d}_{\mathrm{1}} ×{CD}'×\mathrm{sin}\:\left(\pi−\alpha\right)}{\mathrm{2}} \\ $$$$=\frac{{d}_{\mathrm{1}} ×\left({B}'{C}+{CD}'\right)×\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$=\frac{{d}_{\mathrm{1}} {d}_{\mathrm{2}} \:\mathrm{sin}\:\alpha}{\mathrm{2}}\:\checkmark \\ $$
Commented by SEKRET last updated on 26/Nov/22
$$\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{always}}\:\boldsymbol{\mathrm{give}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{very}}\:\:\boldsymbol{\mathrm{nice}}\:\boldsymbol{\mathrm{solution}} \\ $$
Commented by mr W last updated on 26/Nov/22
$${thanks}! \\ $$