In the figure,AHKF,FKDE and HBCK are unit squ | vedclass

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(B) Given that $AHKF$,$FKDE$,and $HBCK$ are unit squares. Let the side length of each square be $1$. We can set up a coordinate system with $K$ at the

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$\frac{1}{5}$

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(B) Given that $AHKF$,$FKDE$,and $HBCK$ are unit squares. Let the side length of each square be $1$. We can set up a coordinate system with $K$ at the origin $(0,0)$. Then the coordinates are: $K(0,0)$,$H(-1,0)$,$F(0,1)$,$A(-1,1)$,$B(-1,-1)$,$D(1,0)$. The line $AD$ passes through $A(-1,1)$ and $D(1,0)$. The slope $m_1 = \frac{0-1}{1-(-1)} = -\frac{1}{2}$. The equation is $y - 0 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{1}{2}$. The line $BF$ passes through $B(-1,-1)$ and $F(0,1)$. The slope $m_2 = \frac{1-(-1)}{0-(-1)} = 2$. The equation is $y - 1 = 2(x - 0) \implies y = 2x + 1$. To find $X$,set $- \frac{1}{2}x + \frac{1}{2} = 2x + 1 \implies -x + 1 = 4x + 2 \implies 5x = -1 \implies x = -\frac{1}{5}$. Then $y = 2(-\frac{1}{5}) + 1 = \frac{3}{5}$. So $X(-\frac{1}{5}, \frac{3}{5})$. In $	riangle ABF$,the base $AF = 1$ and height $AB = 2$. Area $= \frac{1}{2} 	imes 1 	imes 2 = 1$. In $	riangle AXF$,the base $AF = 1$ and the height is the perpendicular distance from $X$ to $AF$. Since $AF$ lies on the line $y=1$,the height is $|1 - \frac{3}{5}| = \frac{2}{5}$. Area of $	riangle AXF = \frac{1}{2} 	imes 1 	imes \frac{2}{5} = \frac{1}{5}$. The ratio of the areas is $\frac{	ext{Area}(	riangle AXF)}{	ext{Area}(	riangle ABF)} = \frac{1/5}{1} = \frac{1}{5}$.

In the figure,$AHKF$,$FKDE$ and $HBCK$ are unit squares. $AD$ and $BF$ intersect at $X$. Then,the ratio of the areas of triangles $AXF$ and $ABF$ is

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