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In the figure AHKF, FKDE and HBCK are unit squares ; AD and BF intersect in $\mathrm{X}$. Then the ratio of the areas of triangles AXF and ABF is -
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Verified Answer
The correct answer is:
$1 / 5$
Equation of $\mathrm{BF}$
$y=2 x$
equation of AD
$\begin{array}{l}
x+2 y=4 \\
x\left(\frac{4}{5}, \frac{8}{5}\right) \\
\frac{\operatorname{ar}(\triangle \mathrm{AFX})}{\operatorname{ar}(\Delta \mathrm{ABF})}=\frac{\frac{1}{2} \times 1 \times\left(2-\frac{8}{5}\right)}{\frac{1}{2} \times 1 \times 2}=\frac{1}{5}
\end{array}$
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