Question Number 214220 by Ari last updated on 01/Dec/24
Answered by mr W last updated on 01/Dec/24
$${say}\:{a}>{b} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${a}−{b}=\mathrm{6}\:\:\:…\left({ii}\right) \\ $$$${h}_{{c}} =\frac{{ab}}{{c}}=\mathrm{8}\:\:\:…\left({iii}\right) \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{36} \\ $$$${c}^{\mathrm{2}} −\mathrm{16}{c}−\mathrm{36}=\mathrm{0} \\ $$$$\left({c}+\mathrm{2}\right)\left({c}−\mathrm{18}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}=\mathrm{18}\:\:\:\checkmark \\ $$$${ab}=\mathrm{8}×\mathrm{18}=\mathrm{144} \\ $$$${a},\:−{b}\:{are}\:{roots}\:{of}\:{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{144}=\mathrm{0} \\ $$$${x}=\mathrm{3}\pm\mathrm{3}\sqrt{\mathrm{17}} \\ $$$$\Rightarrow{a}=\mathrm{3}\left(\sqrt{\mathrm{17}}+\mathrm{1}\right),\:{b}=\mathrm{3}\left(\sqrt{\mathrm{17}}−\mathrm{1}\right) \\ $$
Commented by Ari last updated on 01/Dec/24
$${excuse}\:{me}\:{sir}\:{but}\:{why}\: \\ $$$${h}_{{c}} =\frac{{a}.{b}}{{c}}? \\ $$
Commented by mr W last updated on 01/Dec/24
Commented by mr W last updated on 01/Dec/24
$${area}\:{of}\:{triangle}\:=\frac{{ab}}{\mathrm{2}} \\ $$$${area}\:{of}\:{triangle}\:=\frac{{ch}_{{c}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{{ch}_{{c}} }{\mathrm{2}}=\frac{{ab}}{\mathrm{2}} \\ $$$$\Rightarrow{h}_{{c}} =\frac{{ab}}{{c}} \\ $$
Commented by Ari last updated on 02/Dec/24
$${thank}\:{you}\:{Sir}! \\ $$