Question Number 148739 by Tawa11 last updated on 30/Jul/21
Commented by Tawa11 last updated on 30/Jul/21
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:\:\:\:\:\:\:\mathrm{JSMT} \\ $$
Commented by Tawa11 last updated on 30/Jul/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Answered by mr W last updated on 30/Jul/21
$${diagonal}\:{length}\:=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{7}+\mathrm{9}\right)=\mathrm{16}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${side}\:{length}\:=\frac{\mathrm{16}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}}=\mathrm{8}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$${TM}=\mathrm{8}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)−\mathrm{2}×\mathrm{7}=\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$${SM}=\mathrm{8}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)−\mathrm{2}×\mathrm{9}=\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{2} \\ $$$${area}\:{yellow}\:=\left(\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{2}\right)\left(\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{2}\right)=\mathrm{124} \\ $$
Commented by Tawa11 last updated on 31/Jul/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$
Commented by Tawa11 last updated on 01/Aug/21
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{do}\:\mathrm{you}\:\mathrm{also}\:\mathrm{have}\:\mathrm{process}\:\mathrm{for}\:\:\:\:\:\:\mathrm{ax}^{\mathrm{4}} \:\:+\:\:\mathrm{bx}^{\mathrm{3}} \:\:+\:\:\mathrm{cx}^{\mathrm{2}} \:\:+\:\:\mathrm{dx}\:\:+\:\:\mathrm{e}\:\:=\:\:\mathrm{0} \\ $$$$\mathrm{just}\:\mathrm{like}\:\mathrm{you}\:\mathrm{gave}\:\mathrm{in}\:\mathrm{degree}\:\mathrm{3}\:\:\mathrm{sometimes}. \\ $$
Commented by mr W last updated on 01/Aug/21
$${no} \\ $$
Commented by Tawa11 last updated on 01/Aug/21
$$\mathrm{Alright}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$