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Question-181432




Question Number 181432 by mr W last updated on 25/Nov/22
Commented by mr W last updated on 25/Nov/22
find the area of circle.
$${find}\:{the}\:{area}\:{of}\:{circle}. \\ $$
Commented by universe last updated on 25/Nov/22
Commented by universe last updated on 25/Nov/22
AE^2 +BE^2  = 4  CE^2 +ED^2  = 16  AE^2 +BE^2 +CE^2 +ED^2  = 20  4r^2  = 20 ⇒ r^2  = 5  area of cricle = πr^2  = 5π
$${AE}^{\mathrm{2}} +{BE}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$${CE}^{\mathrm{2}} +{ED}^{\mathrm{2}} \:=\:\mathrm{16} \\ $$$${AE}^{\mathrm{2}} +{BE}^{\mathrm{2}} +{CE}^{\mathrm{2}} +{ED}^{\mathrm{2}} \:=\:\mathrm{20} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} \:=\:\mathrm{20}\:\Rightarrow\:{r}^{\mathrm{2}} \:=\:\mathrm{5} \\ $$$${area}\:{of}\:{cricle}\:=\:\pi{r}^{\mathrm{2}} \:=\:\mathrm{5}\pi \\ $$
Commented by mr W last updated on 25/Nov/22
thanks sir!  please prove that  AE^2 +BE^2 +CE^2 +ED^2  = 4r^2 .  because it′s not obvious.    btw, please post your answer as  “answer”, not as “comment”!  thank you!
$${thanks}\:{sir}! \\ $$$${please}\:{prove}\:{that} \\ $$$${AE}^{\mathrm{2}} +{BE}^{\mathrm{2}} +{CE}^{\mathrm{2}} +{ED}^{\mathrm{2}} \:=\:\mathrm{4}{r}^{\mathrm{2}} . \\ $$$${because}\:{it}'{s}\:{not}\:{obvious}. \\ $$$$ \\ $$$${btw},\:{please}\:{post}\:{your}\:{answer}\:{as} \\ $$$$“{answer}'',\:{not}\:{as}\:“{comment}''! \\ $$$${thank}\:{you}! \\ $$
Commented by universe last updated on 25/Nov/22
https://youtu.be/1EkoaQJyrfk
Answered by HeferH last updated on 25/Nov/22
Commented by HeferH last updated on 25/Nov/22
(((a + 2b)/2))^2  + (((2a + b)/2) − b)^2 = r^2    ((a^2  + 4b^2  + 4ab + (2a − b)^2 )/4) = r^2      ((a^2  + 4b^2  + 4ab + 4a^2  + b^2  − 4ab)/4) = r^2    a^2 + b^2  + 4(a^2  + b^2 ) = 4r^2    but a^2  + b^2  = 4 ⇒   4 + 4(4) = 4r^2    20 = 4r^2    5 = r^2    A = πr^2  = 5π u^2
$$\left(\frac{{a}\:+\:\mathrm{2}{b}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\left(\frac{\mathrm{2}{a}\:+\:{b}}{\mathrm{2}}\:−\:{b}\right)^{\mathrm{2}} =\:{r}^{\mathrm{2}} \\ $$$$\:\frac{{a}^{\mathrm{2}} \:+\:\mathrm{4}{b}^{\mathrm{2}} \:+\:\mathrm{4}{ab}\:+\:\left(\mathrm{2}{a}\:−\:{b}\right)^{\mathrm{2}} }{\mathrm{4}}\:=\:{r}^{\mathrm{2}} \:\: \\ $$$$\:\frac{{a}^{\mathrm{2}} \:+\:\mathrm{4}{b}^{\mathrm{2}} \:+\:\mathrm{4}{ab}\:+\:\mathrm{4}{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:−\:\mathrm{4}{ab}}{\mathrm{4}}\:=\:{r}^{\mathrm{2}} \\ $$$$\:{a}^{\mathrm{2}} +\:{b}^{\mathrm{2}} \:+\:\mathrm{4}\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \right)\:=\:\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\:{but}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\mathrm{4}\:\Rightarrow \\ $$$$\:\mathrm{4}\:+\:\mathrm{4}\left(\mathrm{4}\right)\:=\:\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\:\mathrm{20}\:=\:\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\:\mathrm{5}\:=\:{r}^{\mathrm{2}} \\ $$$$\:{A}\:=\:\pi{r}^{\mathrm{2}} \:=\:\mathrm{5}\pi\:{u}^{\mathrm{2}} \\ $$

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