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Question-171528




Question Number 171528 by cortano1 last updated on 17/Jun/22
Answered by som(math1967) last updated on 17/Jun/22
(x+y+z)^2 =162  ⇒(x+y+z)=(√(162))=9(√2)   x^2 +y^2 +z^2 =58  2(x^2 y^2 +y^2 z^2 +z^2 x^2 )=58^2 −1412=1952  ar. of△     =(√(s(s−x)(s−y)(s−z)))  =(1/4)(√(2(x^2 y^2 +y^2 z^2 +x^2 z^2 )−x^4 −y^4 −z^4 ))  =(1/4)(√(1952−1412))  =(1/4)(√(540))=((3(√(15)))/2)   rad. of circle=((2×3(√(15)))/(2×9(√2)))=((√(15))/(3(√2)))  area=π×((15)/(18))=((5π)/6) sq unit
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{162} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)=\sqrt{\mathrm{162}}=\mathrm{9}\sqrt{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{58} \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)=\mathrm{58}^{\mathrm{2}} −\mathrm{1412}=\mathrm{1952} \\ $$$${ar}.\:{of}\bigtriangleup \\ $$$$\:\:\:=\sqrt{{s}\left({s}−{x}\right)\left({s}−{y}\right)\left({s}−{z}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{2}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{x}^{\mathrm{2}} {z}^{\mathrm{2}} \right)−{x}^{\mathrm{4}} −{y}^{\mathrm{4}} −{z}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{1952}−\mathrm{1412}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{540}}=\frac{\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$$\:{rad}.\:{of}\:{circle}=\frac{\mathrm{2}×\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{2}×\mathrm{9}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{3}\sqrt{\mathrm{2}}} \\ $$$${area}=\pi×\frac{\mathrm{15}}{\mathrm{18}}=\frac{\mathrm{5}\pi}{\mathrm{6}}\:{sq}\:{unit} \\ $$
Commented by Tawa11 last updated on 17/Jun/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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