Question Number 219844 by cherokeesay last updated on 02/May/25
Answered by mr W last updated on 02/May/25
Commented by mr W last updated on 02/May/25
$${CD}=\sqrt{\left({b}+{a}\right)^{\mathrm{2}} +\left({b}−{a}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$${R}=\frac{{CD}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{45}°}=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{2}}}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${OF}=\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }−{a}={b}−{a} \\ $$$${h}^{\mathrm{2}} =\left({R}−{OF}\right)\left({R}+{OF}\right)={R}^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} =\mathrm{2}{ab} \\ $$$$\Rightarrow{h}=\sqrt{\mathrm{2}{ab}}=\sqrt{\mathrm{2}×\mathrm{9}×\mathrm{11}}=\mathrm{3}\sqrt{\mathrm{22}} \\ $$
Commented by cherokeesay last updated on 02/May/25
$${so}\:{nice}\:! \\ $$$$\:{thank}\:{you}\:{master}\:! \\ $$$$ \\ $$
Answered by mehdee7396 last updated on 02/May/25
$${h}^{\mathrm{2}} =\left(\mathrm{9}+{x}\right)\left(\mathrm{11}+{y}\right) \\ $$$$\mathrm{81}={x}\left(\mathrm{20}+{y}\right) \\ $$$$\mathrm{121}={y}\left(\mathrm{20}+{x}\right) \\ $$$$\mathrm{40}=\mathrm{20}\left({y}−{x}\right)\Rightarrow{y}={x}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{81}={x}\left(\mathrm{22}+{x}\right)\Rightarrow{x}=−\mathrm{11}+\sqrt{\mathrm{202}} \\ $$$$\Rightarrow{h}^{\mathrm{2}} =\left(−\mathrm{2}+\sqrt{\mathrm{202}}\right)\left(\mathrm{2}+\sqrt{\mathrm{202}}\right)=−\mathrm{4}+\mathrm{202}=\mathrm{198} \\ $$$${h}=\sqrt{\mathrm{198}}=\mathrm{3}\sqrt{\mathrm{22}} \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 02/May/25
Commented by cherokeesay last updated on 02/May/25
I followed the same approach. Thanks!