Let m_{1}, m_{2} be the slopes of two adjacen | vedclass

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(B) Since $m_{1}, m_{2}$ are slopes of adjacent sides of a square,$m_{1} m_{2} = -1$,so $m_{2} = -\frac{1}{m_{1}}$.Given $a^{2}+11 a+3(m_{1}^{2}+m_{2}

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$128$

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(B) Since $m_{1}, m_{2}$ are slopes of adjacent sides of a square,$m_{1} m_{2} = -1$,so $m_{2} = -\frac{1}{m_{1}}$.Given $a^{2}+11 a+3(m_{1}^{2}+m_{2}^{2})=220$,which becomes $a^{2}+11 a+3(m_{1}^{2}+\frac{1}{m_{1}^{2}})=220$.The diagonal equation is $(\cos \alpha-\sin \alpha) x +(\sin \alpha+\cos \alpha) y = 10$. The slope of this diagonal is $M = -\frac{\cos \alpha-\sin \alpha}{\sin \alpha+\cos \alpha} = \frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} = 	an(\alpha - \frac{\pi}{4})$.The sides of the square make an angle of $\frac{\pi}{4}$ with the diagonal. Thus,the slopes $m$ of the sides satisfy $	an(\frac{\pi}{4}) = |\frac{m-M}{1+mM}|$,which leads to $m = 	an \alpha$ or $m = \cot \alpha$.Thus,$m_{1}^{2}+m_{2}^{2} = 	an^{2} \alpha + \cot^{2} \alpha$.Given the vertex $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$,the distance from the origin to the vertex is $10\sqrt{2}$. The diagonal length is $a\sqrt{2}$. By analyzing the geometry,we find $a=10$.Substituting $a=10$ into the equation: $100 + 110 + 3(	an^{2} \alpha + \cot^{2} \alpha) = 220$ $\Rightarrow 3(	an^{2} \alpha + \cot^{2} \alpha) = 10$ $\Rightarrow 	an^{2} \alpha + \cot^{2} \alpha = \frac{10}{3}$.This implies $	an^{2} \alpha = 3$ or $\frac{1}{3}$.Then $\sin^{4} \alpha + \cos^{4} \alpha = (\sin^{2} \alpha + \cos^{2} \alpha)^{2} - 2\sin^{2} \alpha \cos^{2} \alpha = 1 - \frac{1}{2}\sin^{2}(2\alpha)$.Since $	an^{2} \alpha = 3$,$\sin^{2} \alpha = \frac{3}{4}, \cos^{2} \alpha = \frac{1}{4}$,so $\sin^{4} \alpha + \cos^{4} \alpha = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$.Finally,$72(\frac{5}{8}) + 100 - 30 + 13 = 45 + 83 = 128$.

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