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Question Number 187968 by normans last updated on 24/Feb/23

Commented by a.lgnaoui last updated on 24/Feb/23

21?

$$\mathrm{21}? \\ $$

Commented by a.lgnaoui last updated on 24/Feb/23

I think in top=21=mesure of line and down =(Area)

$${I}\:{think}\:\:{in}\:{top}=\mathrm{21}={mesure}\:{of}\:{line} \\ $$$${and}\:{down}\:=\left({Area}\right) \\ $$

Commented by a.lgnaoui last updated on 24/Feb/23

or the inverse

$${or}\:\:{the}\:{inverse} \\ $$

Commented by normans last updated on 25/Feb/23

no sir

$${no}\:{sir} \\ $$

Commented by mr W last updated on 25/Feb/23

wrong! even blind people can clearly see that blue area must be much bigger than the gray area 21. so 21 is clearly wrong. 100 is the right answer.

$${wrong}! \\ $$$${even}\:{blind}\:{people}\:{can}\:{clearly}\:{see}\:{that} \\ $$$${blue}\:{area}\:{must}\:{be}\:{much}\:{bigger}\:{than} \\ $$$${the}\:{gray}\:{area}\:\mathrm{21}.\:{so}\:\mathrm{21}\:{is}\:{clearly}\:{wrong}. \\ $$$$\mathrm{100}\:{is}\:{the}\:{right}\:{answer}. \\ $$

Answered by mr W last updated on 25/Feb/23

Commented by mr W last updated on 25/Feb/23

((d+c)/d)=((28)/a) ⇒(c/d)=((28)/a)−1 ...(i) ((d+c)/c)=((2b)/a) ⇒(d/c)=((2b)/a)−1 ...(ii) (i)×(ii): 1=(((28)/a)−1)(((2b)/a)−1) ⇒14(2b−a)=ab ...(I) ((d+e)/d)=((21)/b) ⇒(e/d)=((21)/b)−1 ...(iii) ((d+e)/e)=(a/b) ⇒(d/e)=(a/b)−1 ...(iv) (iii)×(iv): 1=(((21)/b)−1)((a/b)−1) ⇒21(a−b)=ab ...(II) 14(2b−a)=21(a−b) ⇒(b/a)=(5/7) 21((a/b)−1)=a ⇒a=21((7/5)−1)=((42)/5) ⇒((d+c)/d)=((28)/a)=((28×5)/(42))=((10)/3) A_(blue) =((2b(d+c))/2) A_(gray) =((ad)/2) (=21) (A_(blue) /A_(gray) )=((2b)/a)×((d+c)/d)=((2×5)/7)×((10)/3)=((100)/(21)) A_(blue) =((100)/(21))A_(gray) =((100)/(21))×21=100 ✓

$$\frac{{d}+{c}}{{d}}=\frac{\mathrm{28}}{{a}}\:\Rightarrow\frac{{c}}{{d}}=\frac{\mathrm{28}}{{a}}−\mathrm{1}\:\:\:...\left({i}\right) \\ $$$$\frac{{d}+{c}}{{c}}=\frac{\mathrm{2}{b}}{{a}}\:\Rightarrow\frac{{d}}{{c}}=\frac{\mathrm{2}{b}}{{a}}−\mathrm{1}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\mathrm{1}=\left(\frac{\mathrm{28}}{{a}}−\mathrm{1}\right)\left(\frac{\mathrm{2}{b}}{{a}}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{14}\left(\mathrm{2}{b}−{a}\right)={ab}\:\:\:...\left({I}\right) \\ $$$$ \\ $$$$\frac{{d}+{e}}{{d}}=\frac{\mathrm{21}}{{b}}\:\Rightarrow\frac{{e}}{{d}}=\frac{\mathrm{21}}{{b}}−\mathrm{1}\:\:\:...\left({iii}\right) \\ $$$$\frac{{d}+{e}}{{e}}=\frac{{a}}{{b}}\:\Rightarrow\frac{{d}}{{e}}=\frac{{a}}{{b}}−\mathrm{1}\:\:\:...\left({iv}\right) \\ $$$$\left({iii}\right)×\left({iv}\right): \\ $$$$\mathrm{1}=\left(\frac{\mathrm{21}}{{b}}−\mathrm{1}\right)\left(\frac{{a}}{{b}}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{21}\left({a}−{b}\right)={ab}\:\:\:...\left({II}\right) \\ $$$$ \\ $$$$\mathrm{14}\left(\mathrm{2}{b}−{a}\right)=\mathrm{21}\left({a}−{b}\right) \\ $$$$\Rightarrow\frac{{b}}{{a}}=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\mathrm{21}\left(\frac{{a}}{{b}}−\mathrm{1}\right)={a} \\ $$$$\Rightarrow{a}=\mathrm{21}\left(\frac{\mathrm{7}}{\mathrm{5}}−\mathrm{1}\right)=\frac{\mathrm{42}}{\mathrm{5}} \\ $$$$\Rightarrow\frac{{d}+{c}}{{d}}=\frac{\mathrm{28}}{{a}}=\frac{\mathrm{28}×\mathrm{5}}{\mathrm{42}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$${A}_{{blue}} =\frac{\mathrm{2}{b}\left({d}+{c}\right)}{\mathrm{2}} \\ $$$${A}_{{gray}} =\frac{{ad}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left(=\mathrm{21}\right) \\ $$$$\frac{{A}_{{blue}} }{{A}_{{gray}} }=\frac{\mathrm{2}{b}}{{a}}×\frac{{d}+{c}}{{d}}=\frac{\mathrm{2}×\mathrm{5}}{\mathrm{7}}×\frac{\mathrm{10}}{\mathrm{3}}=\frac{\mathrm{100}}{\mathrm{21}} \\ $$$${A}_{{blue}} =\frac{\mathrm{100}}{\mathrm{21}}{A}_{{gray}} =\frac{\mathrm{100}}{\mathrm{21}}×\mathrm{21}=\mathrm{100}\:\checkmark \\ $$

Commented by normans last updated on 25/Feb/23

yes, this is the right solution I mean. Thanks Prof..

$$\boldsymbol{{yes}},\:\boldsymbol{{this}}\:\boldsymbol{{is}}\:\:\boldsymbol{{the}}\:\boldsymbol{{right}}\:\boldsymbol{{solution}}\:\boldsymbol{{I}}\:\boldsymbol{{mean}}. \\ $$$$\boldsymbol{{Thanks}}\:\boldsymbol{{Prof}}.. \\ $$

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