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 -

• A

  $1 / 4$
• Correct answer
  B

  $1 / 5$
• C

  $1 / 6$
• D

  $1 / 8$

Solution

Equation of $\mathrm{BF}$ $y=2 x$ equation of AD $\begin{array}{l} x+2 y=4 \\ x\left(\frac{4}{5}, \frac{8}{5}\right) \\ \frac{\operatorname{ar}(\triangle \mathrm{AFX})}{\operatorname{ar}(\Delta \mathrm{ABF})}=\frac{\frac{1}{2} \times 1 \times\left(2-\frac{8}{5}\right)}{\frac{1}{2} \times 1 \times 2}=\frac{1}{5} \end{array}$

Hence, the correct answer is option: 'B'.