Question-214228

Question Number 214228 by mr W last updated on 02/Dec/24

Commented by mr W last updated on 02/Dec/24

$${find}\:{area}\:{of}\:{the}\:{square} \\ $$

Answered by aleks041103 last updated on 02/Dec/24

let the point have a name G and let it be at (y,z) let B be at (0,0) let C be at (x,0) let A be at (0,x) then: y^2 +(x−z)^2 =7^2 y^2 +z^2 =4^2 (x−y)^2 +z^2 =9^2 x^2 +y^2 +z^2 −2xz=49 x^2 +y^2 +z^2 −2xy=81 y^2 +z^2 =16 x^2 −33=2xz x^2 −65=2xy y^2 +z^2 =16 ⇒x(z−y)=16 , 4x^2 yz=(x^2 −33)(x^2 −65) ⇒x^2 (z−y)^2 =16^2 = =x^2 (y^2 +z^2 )−2x^2 yz= =16x^2 −(1/2)(x^2 −33)(x^2 −65) ⇒S=x^2 and 2.16^2 =2.16S−(S−33)(S−65) (S−33)(S−65)−32S+512=0 S^2 −130S+33.65+512=0 S^2 −130S+2657=0 ⇒S=65±(√(65^2 −2657))=65±(√(1568)) remains to be checked whether these are valid solutions. (also i may have made a mistake somewhere ;) )

$${let}\:{the}\:{point}\:{have}\:{a}\:{name}\:{G} \\ $$$${and}\:{let}\:{it}\:{be}\:{at}\:\left({y},{z}\right) \\ $$$${let}\:{B}\:{be}\:{at}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$${let}\:{C}\:{be}\:{at}\:\left({x},\mathrm{0}\right) \\ $$$${let}\:{A}\:{be}\:{at}\:\left(\mathrm{0},{x}\right) \\ $$$${then}: \\ $$$${y}^{\mathrm{2}} +\left({x}−{z}\right)^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}=\mathrm{49} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{81} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{16} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{33}=\mathrm{2}{xz} \\ $$$${x}^{\mathrm{2}} −\mathrm{65}=\mathrm{2}{xy} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{16} \\ $$$$ \\ $$$$\Rightarrow{x}\left({z}−{y}\right)=\mathrm{16}\:,\:\mathrm{4}{x}^{\mathrm{2}} {yz}=\left({x}^{\mathrm{2}} −\mathrm{33}\right)\left({x}^{\mathrm{2}} −\mathrm{65}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({z}−{y}\right)^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} = \\ $$$$={x}^{\mathrm{2}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{x}^{\mathrm{2}} {yz}= \\ $$$$=\mathrm{16}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\mathrm{33}\right)\left({x}^{\mathrm{2}} −\mathrm{65}\right) \\ $$$$\Rightarrow{S}={x}^{\mathrm{2}} \:{and} \\ $$$$\mathrm{2}.\mathrm{16}^{\mathrm{2}} =\mathrm{2}.\mathrm{16}{S}−\left({S}−\mathrm{33}\right)\left({S}−\mathrm{65}\right) \\ $$$$\left({S}−\mathrm{33}\right)\left({S}−\mathrm{65}\right)−\mathrm{32}{S}+\mathrm{512}=\mathrm{0} \\ $$$${S}^{\mathrm{2}} −\mathrm{130}{S}+\mathrm{33}.\mathrm{65}+\mathrm{512}=\mathrm{0} \\ $$$${S}^{\mathrm{2}} −\mathrm{130}{S}+\mathrm{2657}=\mathrm{0} \\ $$$$\Rightarrow{S}=\mathrm{65}\pm\sqrt{\mathrm{65}^{\mathrm{2}} −\mathrm{2657}}=\mathrm{65}\pm\sqrt{\mathrm{1568}} \\ $$$$ \\ $$$${remains}\:{to}\:{be}\:{checked}\:{whether}\:{these} \\ $$$${are}\:{valid}\:{solutions}. \\ $$$$\left.\left({also}\:{i}\:{may}\:{have}\:{made}\:{a}\:{mistake}\:{somewhere}\:;\right)\:\right) \\ $$

Commented by mr W last updated on 02/Dec/24

$${S}=\mathrm{65}+\sqrt{\mathrm{1568}}=\mathrm{65}+\mathrm{28}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{correct}! \\ $$

Answered by mr W last updated on 02/Dec/24

Commented by mr W last updated on 02/Dec/24

ΔBEA≡ΔBFG cos β=(((4(√2))^2 +7^2 −9^2 )/(2×4(√2)×7))=0 ⇒β=90° AB^2 =4^2 +7^2 −2×4×7 cos 135° =65+28(√2)=area of square

$$\Delta{BEA}\equiv\Delta{BFG} \\ $$$$\mathrm{cos}\:\beta=\frac{\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }{\mathrm{2}×\mathrm{4}\sqrt{\mathrm{2}}×\mathrm{7}}=\mathrm{0}\:\Rightarrow\beta=\mathrm{90}° \\ $$$${AB}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\mathrm{7}\:\mathrm{cos}\:\mathrm{135}° \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{65}+\mathrm{28}\sqrt{\mathrm{2}}={area}\:{of}\:{square} \\ $$

Answered by TonyCWX08 last updated on 02/Dec/24

Answered by TonyCWX08 last updated on 02/Dec/24

BF=2(√2) BF^2 +FC^2 =BC^2 (2(√2))^2 +(2(√2)+7)^2 =BC^2 BC^2 =8+8+28(√2)+49=65+28(√2)=Area

$${BF}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${BF}^{\mathrm{2}} +{FC}^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{7}\right)^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$${BC}^{\mathrm{2}} =\mathrm{8}+\mathrm{8}+\mathrm{28}\sqrt{\mathrm{2}}+\mathrm{49}=\mathrm{65}+\mathrm{28}\sqrt{\mathrm{2}}={Area} \\ $$

Commented by TonyCWX08 last updated on 02/Dec/24

PE=(√(4^2 +4^2 ))=4(√2) PE^2 +EC^2 =(4(√2))^2 +(7)^2 =81 =PC^2 Triangle_(PEC) is a right angle triangle. EC is collinear with FC.

$${PE}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${PE}^{\mathrm{2}} +{EC}^{\mathrm{2}} \\ $$$$=\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{7}\right)^{\mathrm{2}} \\ $$$$=\mathrm{81} \\ $$$$={PC}^{\mathrm{2}} \\ $$$${Triangle}_{{PEC}} \:{is}\:{a}\:{right}\:{angle}\:{triangle}. \\ $$$${EC}\:{is}\:{collinear}\:{with}\:{FC}. \\ $$

Commented by mr W last updated on 02/Dec/24

you must prove at first that FE and EC are collinear! otherwise you can not take FC=FE+EC=2(√2)+7.

$${you}\:{must}\:{prove}\:{at}\:{first}\:{that}\:{FE}\:{and} \\ $$$${EC}\:{are}\:{collinear}!\:{otherwise}\:{you} \\ $$$${can}\:{not}\:{take}\:{FC}={FE}+{EC}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{7}. \\ $$

Commented by TonyCWX08 last updated on 02/Dec/24

$${My}\:{diagram}\:{is}\:{a}\:{bit}\:{misleading}. \\ $$

Commented by mr W last updated on 02/Dec/24

$${yes}!\:{good}\:{approach}! \\ $$

Commented by TonyCWX08 last updated on 02/Dec/24

$${Fixed}. \\ $$

Answered by ajfour last updated on 03/Dec/24

p^2 +q^2 =16 s=q+(√(49−p^2 ))=p+(√(81−q^2 )) ⇒ s^2 −2qs=33 & s^2 −2ps=65 ⇒ (s^2 −33)^2 +(s^2 −65)^2 =64s^2 2s^2 −260s^2 +1089+4225=0 s=65±(√(4225−((1089)/2)−((4225)/2))) s=65±(√(1568)) s=65±28(√2)

$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{16} \\ $$$${s}={q}+\sqrt{\mathrm{49}−{p}^{\mathrm{2}} }={p}+\sqrt{\mathrm{81}−{q}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{s}^{\mathrm{2}} −\mathrm{2}{qs}=\mathrm{33} \\ $$$$\&\:\:\:{s}^{\mathrm{2}} −\mathrm{2}{ps}=\mathrm{65} \\ $$$$\Rightarrow\:\:\left({s}^{\mathrm{2}} −\mathrm{33}\right)^{\mathrm{2}} +\left({s}^{\mathrm{2}} −\mathrm{65}\right)^{\mathrm{2}} =\mathrm{64}{s}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} −\mathrm{260}{s}^{\mathrm{2}} +\mathrm{1089}+\mathrm{4225}=\mathrm{0} \\ $$$${s}=\mathrm{65}\pm\sqrt{\mathrm{4225}−\frac{\mathrm{1089}}{\mathrm{2}}−\frac{\mathrm{4225}}{\mathrm{2}}} \\ $$$$\:\:{s}=\mathrm{65}\pm\sqrt{\mathrm{1568}} \\ $$$$\:\:{s}=\mathrm{65}\pm\mathrm{28}\sqrt{\mathrm{2}} \\ $$

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