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Question-193581




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

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