Question Number 140874 by cherokeesay last updated on 13/May/21
Commented by ok_19 last updated on 13/May/21
$$\mathrm{12}\:\mathrm{cm}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 13/May/21
$$\mathrm{6}×\mathrm{9}=\mathrm{54} \\ $$$$\mathrm{54}−\mathrm{12}−\mathrm{15}=\mathrm{27} \\ $$$$\frac{\mathrm{12}}{{x}}=\frac{\mathrm{27}−{x}}{\mathrm{15}} \\ $$$${x}^{\mathrm{2}} −\mathrm{27}{x}+\mathrm{180}=\mathrm{0} \\ $$$$\left({x}−\mathrm{12}\right)\left({x}−\mathrm{15}\right)=\mathrm{0} \\ $$$${x}=\mathrm{12}\:{or}\:\mathrm{15} \\ $$
Commented by mr W last updated on 13/May/21
Commented by cherokeesay last updated on 13/May/21
$${thank}\:{you}\:{sir}\:! \\ $$$${very}\:{nice}\:! \\ $$$$ \\ $$
Answered by MJS_new last updated on 14/May/21
$$\left(\mathrm{9}−{x}\right){y}=\mathrm{15} \\ $$$${x}\left(\mathrm{6}−{y}\right)=\mathrm{12} \\ $$$${y}=\frac{\mathrm{15}}{\mathrm{9}−{x}} \\ $$$${y}=\frac{\mathrm{6}\left({x}−\mathrm{2}\right)}{{x}} \\ $$$$\mathrm{15}{x}=\mathrm{6}\left({x}−\mathrm{2}\right)\left(\mathrm{9}−{x}\right) \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{2}}{x}+\mathrm{18}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{4}\:\Rightarrow\:{y}_{\mathrm{1}} =\mathrm{3}\:\Rightarrow\:{A}_{\mathrm{1}} =\mathrm{12} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow\:{y}_{\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{3}}\:\Rightarrow\:{A}_{\mathrm{2}} =\mathrm{15} \\ $$
Commented by peter frank last updated on 14/May/21
$${thank}\:{you} \\ $$
Commented by cherokeesay last updated on 14/May/21
$${thank}\:{you}\:{sir}\:! \\ $$