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Question Number 171528 by cortano1 last updated on 17/Jun/22

Answered by som(math1967) last updated on 17/Jun/22

$${(x+y+z)}^2=162$$$$\Rightarrow (x+y+z)=\sqrt{162}=9\sqrt{2}$$$$x^2+y^2+z^2=58$$$$2(x^2{y}^2+y^2{z}^2+z^2{x}^2)={58}^2-1412=1952$$$$ar.of△$$$$=\sqrt{s(s-x)(s-y)(s-z)}$$$$=\frac{1}{4}\sqrt{2(x^2{y}^2+y^2{z}^2+x^2{z}^2)-x^4-y^4-z^4}$$$$=\frac{1}{4}\sqrt{1952-1412}$$$$=\frac{1}{4}\sqrt{540}=\frac{3\sqrt{15}}{2}$$$$rad.ofcircle=\frac{2\times 3\sqrt{15}}{2\times 9\sqrt{2}}=\frac{\sqrt{15}}{3\sqrt{2}}$$$$area=\pi \times \frac{15}{18}=\frac{5\pi }{6}squnit$$

Commented by Tawa11 last updated on 17/Jun/22

$$\mathrm{Great}\mathrm{sir}.$$

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