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Question Number 208465 by Tawa11 last updated on 16/Jun/24

Answered by A5T last updated on 17/Jun/24

((3(√2))/s)=((3(√5))/(10))⇒s=((10(√(10)))/5)=2(√(10))⇒s^2 =40

$$\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{s}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\Rightarrow{s}=\frac{\mathrm{10}\sqrt{\mathrm{10}}}{\mathrm{5}}=\mathrm{2}\sqrt{\mathrm{10}}\Rightarrow{s}^{\mathrm{2}} =\mathrm{40} \\ $$

Commented by Tawa11 last updated on 17/Jun/24

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 17/Jun/24

How do you get ((3(√2))/s) sir?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\:\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{s}}\:\:\:\mathrm{sir}? \\ $$

Commented by A5T last updated on 17/Jun/24

Commented by A5T last updated on 17/Jun/24

Construct say BD=BE=3⇒DE=3(√2), and construct DF=FG=EG=3(√2) ⇒BF=3(√5);BH=r=10 Homothety that maps FG to HI gives the desired square and maps the constructed square above to another square with the desired coincident properties. This homothety has ratio ((BF)/(BH))=((3(√5))/r)=((3(√5))/(10)) which is also equal to ((FG)/(HI))=((3(√2))/s) ⇒((3(√5))/(10))=((3(√2))/s)⇒s=((10(√(10)))/5)=2(√(10))

$${Construct}\:{say}\:{BD}={BE}=\mathrm{3}\Rightarrow{DE}=\mathrm{3}\sqrt{\mathrm{2}},\: \\ $$$${and}\:{construct}\:{DF}={FG}={EG}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{BF}=\mathrm{3}\sqrt{\mathrm{5}};{BH}={r}=\mathrm{10} \\ $$$${Homothety}\:{that}\:{maps}\:{FG}\:{to}\:{HI}\:{gives}\:{the} \\ $$$${desired}\:{square}\:{and}\:{maps}\:{the}\:{constructed}\:{square} \\ $$$${above}\:{to}\:{another}\:{square}\:{with}\:{the}\:{desired}\: \\ $$$${coincident}\:{properties}.\:{This}\:{homothety}\:{has}\:{ratio} \\ $$$$\frac{{BF}}{{BH}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{{r}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\:{which}\:{is}\:{also}\:{equal}\:{to}\:\frac{{FG}}{{HI}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{s}} \\ $$$$\Rightarrow\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{{s}}\Rightarrow{s}=\frac{\mathrm{10}\sqrt{\mathrm{10}}}{\mathrm{5}}=\mathrm{2}\sqrt{\mathrm{10}} \\ $$

Commented by A5T last updated on 17/Jun/24

Commented by A5T last updated on 17/Jun/24

One can also decide to use any variable x. Construct say BD=BE=x⇒DE=x(√2) and construct say DF=FG=EG=x(√2) ⇒BF=x(√5);BH=r=10 ((BF)/(BH))=((x(√5))/(10))=((FG)/(HI))=((x(√2))/s)⇒((x(√5))/(10))=((x(√2))/s)⇒s=2(√(10))

$${One}\:{can}\:{also}\:{decide}\:{to}\:{use}\:{any}\:{variable}\:{x}. \\ $$$${Construct}\:{say}\:{BD}={BE}={x}\Rightarrow{DE}={x}\sqrt{\mathrm{2}} \\ $$$${and}\:{construct}\:{say}\:{DF}={FG}={EG}={x}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{BF}={x}\sqrt{\mathrm{5}};{BH}={r}=\mathrm{10} \\ $$$$\frac{{BF}}{{BH}}=\frac{{x}\sqrt{\mathrm{5}}}{\mathrm{10}}=\frac{{FG}}{{HI}}=\frac{{x}\sqrt{\mathrm{2}}}{{s}}\Rightarrow\frac{{x}\sqrt{\mathrm{5}}}{\mathrm{10}}=\frac{{x}\sqrt{\mathrm{2}}}{{s}}\Rightarrow{s}=\mathrm{2}\sqrt{\mathrm{10}} \\ $$

Commented by Tawa11 last updated on 17/Jun/24

Wow, I appreciate sir.

$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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