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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

$$diagonallength=(1+\sqrt{2})(7+9)=16(1+\sqrt{2})$$$$sidelength=\frac{16(1+\sqrt{2})}{\sqrt{2}}=8(2+\sqrt{2})$$$$TM=8(2+\sqrt{2})-2\times 7=8\sqrt{2}+2$$$$SM=8(2+\sqrt{2})-2\times 9=8\sqrt{2}-2$$$$areayellow=(8\sqrt{2}+2)(8\sqrt{2}-2)=124$$

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

$$\mathrm{Sir},\mathrm{please}\mathrm{do}\mathrm{you}\mathrm{also}\mathrm{have}\mathrm{process}\mathrm{for}{\mathrm{ax}}^4+{\mathrm{bx}}^3+{\mathrm{cx}}^2+\mathrm{dx}+\mathrm{e}=0$$$$\mathrm{just}\mathrm{like}\mathrm{you}\mathrm{gave}\mathrm{in}\mathrm{degree}3\mathrm{sometimes}.$$

Commented by mr W last updated on 01/Aug/21

$$no$$

Commented by Tawa11 last updated on 01/Aug/21

$$\mathrm{Alright}.\mathrm{Thank}\mathrm{you}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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