Question Number 220499 by Jubr last updated on 13/May/25
Commented by Jubr last updated on 13/May/25
$${Find}\:{x}\:{and}\:{y} \\ $$
Answered by Ghisom last updated on 14/May/25
$$\left(\mathrm{1}\right)\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:{x}^{\mathrm{2}} +\left(\mathrm{15}−{y}\right)^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:\:\:\:\:\mathrm{30}{y}−\mathrm{225}=\mathrm{63}\:\Rightarrow\:{y}=\frac{\mathrm{48}}{\mathrm{5}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{48}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }=\mathrm{12}^{\mathrm{2}} \:\Rightarrow\:{x}=\frac{\mathrm{36}}{\mathrm{5}} \\ $$
Commented by Jubr last updated on 14/May/25
$${Thanks}\:{sir}. \\ $$
Answered by mr W last updated on 14/May/25
$${area}\:{of}\:\Delta{ADB}=\frac{\mathrm{15}{x}}{\mathrm{2}}=\frac{\mathrm{9}×\mathrm{12}}{\mathrm{2}}\: \\ $$$$\Rightarrow{x}=\frac{\mathrm{9}×\mathrm{12}}{\mathrm{15}}=\frac{\mathrm{36}}{\mathrm{5}} \\ $$$$\frac{{y}}{\mathrm{12}}=\frac{\mathrm{12}}{\mathrm{15}}\:\Rightarrow{y}=\frac{\mathrm{12}^{\mathrm{2}} }{\mathrm{15}}=\frac{\mathrm{48}}{\mathrm{5}} \\ $$
Commented by Jubr last updated on 14/May/25
$${Thanks}\:{sir}. \\ $$$${So},\:{it}\:{has}\:{the}\:{same}\:{area}. \\ $$
Commented by Jubr last updated on 14/May/25
$${please}\:{sir}. \\ $$$${Q}\mathrm{220495} \\ $$
Answered by efronzo1 last updated on 14/May/25
$$\:\mathrm{12}^{\mathrm{2}} \:=\:\mathrm{y}.\mathrm{15}\:\Rightarrow\mathrm{y}=\frac{\mathrm{12}×\mathrm{12}}{\mathrm{15}}\:=\:\frac{\mathrm{48}}{\mathrm{5}} \\ $$$$\:\mathrm{x}^{\mathrm{2}\:} =\:\mathrm{y}.\left(\mathrm{15}−\mathrm{y}\right) \\ $$$$\:\mathrm{x}\:=\sqrt{\frac{\mathrm{48}}{\mathrm{5}}.\left(\mathrm{15}−\frac{\mathrm{48}}{\mathrm{5}}\right)}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}\:.\:\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$$$\:\mathrm{x}\:=\:\frac{\mathrm{36}}{\mathrm{5}} \\ $$
Commented by Jubr last updated on 14/May/25
$${Thanks}\:{sir}. \\ $$