Question Number 198477 by a.lgnaoui last updated on 20/Oct/23
$$\mathrm{find}\:\mathrm{the}\:\mathrm{Area}\:\left(\mathrm{ABCD}\right) \\ $$
Commented by a.lgnaoui last updated on 20/Oct/23
Commented by mr W last updated on 21/Oct/23
$${Area}\left({ABCD}\right)=\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{\mathrm{4}} \\ $$
Answered by mr W last updated on 21/Oct/23
Commented by mr W last updated on 21/Oct/23
![b^2 =d^2 +f^2 ...(i) a^2 =c^2 +(f+c+d)^2 =c^2 +f^2 +c^2 +d^2 +2fc+2fd+2cd ...(ii) (ii)−(i): a^2 −b^2 =2c^2 +2fc+2fd+2cd ⇒c^2 +fc+fd+cd=((a^2 −b^2 )/2) [ABCD]=(df/2)+((c(f+c+d))/2) =((c^2 +df+cf+cd)/2)=((a^2 −b^2 )/4) ✓](https://www.tinkutara.com/question/Q198520.png)
$${b}^{\mathrm{2}} ={d}^{\mathrm{2}} +{f}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${a}^{\mathrm{2}} ={c}^{\mathrm{2}} +\left({f}+{c}+{d}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} +{f}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}{fc}+\mathrm{2}{fd}+\mathrm{2}{cd}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{fc}+\mathrm{2}{fd}+\mathrm{2}{cd} \\ $$$$\Rightarrow{c}^{\mathrm{2}} +{fc}+{fd}+{cd}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left[{ABCD}\right]=\frac{{df}}{\mathrm{2}}+\frac{{c}\left({f}+{c}+{d}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{c}^{\mathrm{2}} +{df}+{cf}+{cd}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}}\:\checkmark \\ $$
Commented by a.lgnaoui last updated on 21/Oct/23
$$\mathrm{exactly}\:\:\mathrm{thanks} \\ $$