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(B) The height $h$ of the equilateral triangle is the perpendicular distance from the vertex $(2,1)$ to the line $x+y-2=0$. $h = \frac{|2+1-2|}{\sqrt{

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$\frac{8}{\sqrt{3}}$

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(B) The height $h$ of the equilateral triangle is the perpendicular distance from the vertex $(2,1)$ to the line $x+y-2=0$. $h = \frac{|2+1-2|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$. Since $h = \frac{\sqrt{3}}{2} a$,we have $a = \frac{2h}{\sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$. Thus,$a \sqrt{2} = \sqrt{\frac{2}{3}} 	imes \sqrt{2} = \frac{2}{\sqrt{3}}$. The slope of the base is $m = -1$. Let the slopes of the other two sides be $m_1$ and $m_2$. The angle between the base and the sides is $60^\circ$. Using $	an 60^\circ = |\frac{m_1 - (-1)}{1 + m_1(-1)}| = |\frac{m_1+1}{1-m_1}| = \sqrt{3}$. Solving $\frac{m_1+1}{1-m_1} = \sqrt{3}$ gives $m_1 = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$. Solving $\frac{m_1+1}{1-m_1} = -\sqrt{3}$ gives $m_2 = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}$. Then $|m_1-m_2| = |(2-\sqrt{3}) - (2+\sqrt{3})| = |-2\sqrt{3}| = 2\sqrt{3}$. Finally,$|m_1-m_2| + a\sqrt{2} = 2\sqrt{3} + \frac{2}{\sqrt{3}} = \frac{6+2}{\sqrt{3}} = \frac{8}{\sqrt{3}}$.

The equation of the base of an equilateral triangle is $x+y=2$ and its opposite vertex is $(2,1)$. If $m_1, m_2$ are the slopes of the other two sides and the length of its side is $a$,then $|m_1-m_2|+a \sqrt{2}=$

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