If one side of an isosceles triangle is given | vedclass

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(C) The base of the triangle is the line segment connecting $(2,0)$ and $(0,2)$. The length of the base is $\sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4+4} = 2\

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$2$

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(C) The base of the triangle is the line segment connecting $(2,0)$ and $(0,2)$. The length of the base is $\sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4+4} = 2\sqrt{2}$.The third vertex lies on the line $y=2$. Let the vertex be $(x, 2)$.Since the triangle is isosceles,the distance from $(x, 2)$ to $(2,0)$ must equal the distance from $(x, 2)$ to $(0,2)$,or one of these must equal the base length $2\sqrt{2}$.From the figure,the vertex is $(2,2)$. The height of the triangle from the base (line $x+y=2$) to the vertex $(2,2)$ is the perpendicular distance from $(2,2)$ to $x+y-2=0$,which is $h = \frac{|2+2-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2\sqrt{2}) 	imes \sqrt{2} = 2$ sq. units.

If one side of an isosceles triangle is given by the line $y=2$ and the base is provided by the points $(2,0)$ and $(0,2)$,then its area (in sq. units) is

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