Question Number 193581 by Rupesh123 last updated on 16/Jun/23
Answered by Subhi last updated on 16/Jun/23
$${a}^{\mathrm{2}} =\left({c}+{d}\right)^{\mathrm{2}} +\left({c}+{e}\right)^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{cd}+\mathrm{2}{ec}+{d}^{\mathrm{2}} +{e}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} ={e}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{cd}+\mathrm{2}{ec}\:\left({i}\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({c}+{d}\right)\left({c}+{e}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}{ed}=\frac{{c}^{\mathrm{2}} }{\mathrm{2}}+\frac{{ec}}{\mathrm{2}}+\frac{{dc}}{\mathrm{2}}+\frac{{de}}{\mathrm{2}}−\frac{{de}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({c}^{\mathrm{2}} +{ec}+{dc}\right)\:\left({ii}\right) \\ $$$$\therefore\:\mathrm{4}{S}\:=\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$
Commented by Subhi last updated on 16/Jun/23
Commented by Rupesh123 last updated on 16/Jun/23
Perfect ��
Commented by Mingma last updated on 17/Jun/23
Excellent!