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Question Number 73919 by necxxx last updated on 16/Nov/19

Commented by necxxx last updated on 16/Nov/19

$$Gooddaysirs.Thisquestionwasformed$$$$andsolvedbysomeofushere.Ireally$$$$donotrememberthequestionor$$$$approachesapplied.Pleasehelp.$$$$Thanksinadvance.$$

Commented by mr W last updated on 16/Nov/19

$$creatureofajfoursir$$

Commented by necxxx last updated on 16/Nov/19

$$yessir$$

Answered by mind is power last updated on 16/Nov/19

$$\frac{\mathrm{ab}}{2}=\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{2}+\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{2}$$$$\Rightarrow \mathrm{ab}=\mathrm{a}+\mathrm{b}+\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}...\mathrm{E}$$$$\Rightarrow {(\mathrm{ab}-\mathrm{a}-\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2$$$$\Rightarrow {\mathrm{a}}^2{\mathrm{b}}^2+2\mathrm{ab}-{2\mathrm{a}}^2\mathrm{b}-{2\mathrm{ab}}^2=0$$$$\Rightarrow \mathrm{ab}(\mathrm{ab}+2-2\mathrm{a}-2\mathrm{b})=0$$$$\Rightarrow \mathrm{ab}+2-2\mathrm{a}-2\mathrm{b}=0$$$$\Rightarrow \mathrm{b}=\frac{2\mathrm{a}-2}{\mathrm{a}-2}...\mathrm{G}$$$$$$$$\mathrm{DT}=\mathrm{DP}=\mathrm{b}-1$$$$\mathrm{by}\mathrm{thales}\Rightarrow \frac{\mathrm{b}-1}{\mathrm{b}}=\frac{\mathrm{b}-1}{\sqrt{{\mathrm{b}}^2+{\mathrm{a}}^2}-\mathrm{b}+1}$$$$\Rightarrow \mathrm{b}=\sqrt{{\mathrm{b}}^2+{\mathrm{a}}^2}-\mathrm{b}+1\Rightarrow 2\mathrm{b}-1=\sqrt{{\mathrm{b}}^2+{\mathrm{a}}^2}$$$$\mathrm{E}\iff \mathrm{ab}=\mathrm{a}+\mathrm{b}+2\mathrm{b}-1\Rightarrow \mathrm{b}(\mathrm{a}-3)=\mathrm{a}-1\Rightarrow \mathrm{b}=\frac{\mathrm{a}-1}{\mathrm{a}-3}$$$$\mathrm{witheG}\Rightarrow \frac{\mathrm{a}-1}{\mathrm{a}-3}=\frac{2\mathrm{a}-2}{\mathrm{a}-2}\Rightarrow \mathrm{a}-2=2\mathrm{a}-6\Rightarrow \mathrm{a}=4$$$$\mathrm{b}=\frac{4-1}{4-3}=3$$$$$$$$$$$$$$

Commented by necxxx last updated on 17/Nov/19

$$ThankyousomuchbutwhatdoGandE$$$$represent?$$

Commented by mr W last updated on 17/Nov/19

$$hemeansequationGandequationE.$$$$otherpeoplemayuseeqn.\left(i\right)and\left(ii\right)$$$$insuchcases.$$

Commented by necxxx last updated on 17/Nov/19

$$okThanks$$$$$$

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