Question-195892

Question Number 195892 by AROUNAMoussa last updated on 12/Aug/23

Answered by mr W last updated on 13/Aug/23

1+(√((4+1)^2 −(4−1)^2 ))+(√((4+2)^2 −(4−2)^2 ))+2 =7+4(√2)

$$\mathrm{1}+\sqrt{\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{2} \\ $$$$=\mathrm{7}+\mathrm{4}\sqrt{\mathrm{2}} \\ $$

Commented by AROUNAMoussa last updated on 13/Aug/23

$${Merci}\:{mais}\:{je}\:{n}'{ai}\:{rien}\:{compris},{veillez}\:{donner}\:{des}\:{detailles}\:! \\ $$

Commented by mr W last updated on 13/Aug/23

Commented by mr W last updated on 13/Aug/23

length of rectangle =1+a+b+2 a^2 +(4−1)^2 =(4+1)^2 ⇒a=(√((4+1)^2 −(4−1)^2 ))=4 b^2 +(4−2)^2 =(4+2)^2 ⇒b=(√((4+2)^2 −(4−2)^2 ))=4(√2)

$${length}\:{of}\:{rectangle}\:=\mathrm{1}+{a}+{b}+\mathrm{2} \\ $$$${a}^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow{a}=\sqrt{\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4} \\ $$$${b}^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow{b}=\sqrt{\left(\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{2}} \\ $$

Commented by AROUNAMoussa last updated on 14/Aug/23

$${Merci}\:{bcp}\:{Mr}\:{W}\:! \\ $$

Facebook Tweet Pin

Question-195892

Leave a Reply Cancel reply