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Question Number 183227 by mr W last updated on 23/Dec/22

Commented by mr W last updated on 23/Dec/22

find the length of other diagonal.

$${find}\:{the}\:{length}\:{of}\:{other}\:{diagonal}. \\ $$

Answered by mr W last updated on 24/Dec/22

Commented by mr W last updated on 24/Dec/22

BD=(√((6+6)^2 −7^2 ))=(√(95))

$${BD}=\sqrt{\left(\mathrm{6}+\mathrm{6}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\sqrt{\mathrm{95}} \\ $$

Commented by manxsol last updated on 24/Dec/22

We did not have the vision but it is a special case. I send another case

$${We}\:{did}\:{not}\:{have}\:{the} \\ $$$$\:{vision}\:{but}\:{it}\:{is}\:{a}\: \\ $$$${special}\:{case}.\:{I}\:{send}\:{another} \\ $$$$\:{case} \\ $$

Commented by manxsol last updated on 24/Dec/22

Commented by mr W last updated on 24/Dec/22

yes it′s a special case. but the question only requests you to solve this special case, nothing else.

$${yes}\:{it}'{s}\:{a}\:{special}\:{case}.\:{but}\:{the}\:{question} \\ $$$${only}\:{requests}\:{you}\:{to}\:{solve}\:{this}\:{special} \\ $$$${case},\:{nothing}\:{else}. \\ $$

Commented by mr W last updated on 24/Dec/22

as for the case you sent, see Q183281

$${as}\:{for}\:{the}\:{case}\:{you}\:{sent},\:{see}\:{Q}\mathrm{183281} \\ $$

Commented by manxsol last updated on 24/Dec/22

Oh,Sir W. but my case has a surprise

$${Oh},{Sir}\:{W}.\:{but}\:{my}\:{case}\:\:{has} \\ $$$${a}\:{surprise} \\ $$

Answered by Frix last updated on 23/Dec/22

19×5×7^2 =(12+x)(12−x)x^2 x≠7 ⇒ x=(√(95))

$$\mathrm{19}×\mathrm{5}×\mathrm{7}^{\mathrm{2}} =\left(\mathrm{12}+{x}\right)\left(\mathrm{12}−{x}\right){x}^{\mathrm{2}} \\ $$$${x}\neq\mathrm{7}\:\Rightarrow\:{x}=\sqrt{\mathrm{95}} \\ $$

Answered by a.lgnaoui last updated on 23/Dec/22

{ ((x^2 +y^2 =36 (1))),((x^2 +z^2 =49 (2))) :} (2)−(1)⇒z^2 −y^2 =13 (z−y)(z+y)=13 6(z−y)=13 ⇒z−y=((13)/6) { ((z+y=6 z=((49)/(12)))),((z−y=((13)/6) y=((23)/(12)))) :} x^2 =6^2 −y^2 =36−(((23)/(12)))^2 BD^2 =DE^2 +BE^2 =x^2 +(2y+z)^2 BD=(√((36−y^2 )+(2y+z)^2 )) BD =(√(95 )) =9,746794

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{36}\:\:\:\:\:\left(\mathrm{1}\right)}\\{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{49}\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\:\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\Rightarrow{z}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{13} \\ $$$$\:\:\left({z}−{y}\right)\left({z}+{y}\right)=\mathrm{13} \\ $$$$\:\:\:\mathrm{6}\left({z}−{y}\right)=\mathrm{13}\:\Rightarrow{z}−{y}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\begin{cases}{{z}+{y}=\mathrm{6}\:\:\:\:\:\:\:\:{z}=\frac{\mathrm{49}}{\mathrm{12}}}\\{{z}−{y}=\frac{\mathrm{13}}{\mathrm{6}}\:\:\:\:\:\:{y}=\frac{\mathrm{23}}{\mathrm{12}}}\end{cases} \\ $$$${x}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{36}−\left(\frac{\mathrm{23}}{\mathrm{12}}\right)^{\mathrm{2}} \\ $$$$\mathrm{BD}^{\mathrm{2}} =\mathrm{DE}^{\mathrm{2}} +\mathrm{BE}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} +\left(\mathrm{2}{y}+{z}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{BD}=\sqrt{\left(\mathrm{36}−{y}^{\mathrm{2}} \right)+\left(\mathrm{2}{y}+{z}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\mathrm{BD}\:\:\:=\sqrt{\mathrm{95}\:}\:=\mathrm{9},\mathrm{746794} \\ $$

Commented by a.lgnaoui last updated on 23/Dec/22

Commented by a.lgnaoui last updated on 23/Dec/22

Answered by manxsol last updated on 23/Dec/22

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