Rectangle ABCD has AB=5 and BC=4, point E lies on AB so that EB=1 point G lies on BC so that CG=1 and point F lies on CD so that DF=2, segment AG and AC intersect EF at Q and P respectively then ratio $\frac{P Q}{E F}$ is

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$\frac{9}{82}$

$\frac{10}{91}$

$\frac{\sqrt{2}}{13}$

$\frac{\sqrt{13}}{16}$

answer is D.

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Detailed Solution

Equation of AG
$y - 4 = \frac{4 - 1}{0 - 5} (x - 0); y - 4 = - \frac{3}{5} (x)$
$5 y - 20 = - 3 x; y = - \frac{3}{5} x + 4$
Question Image
Equation AC
$y = - \frac{4}{5} x + 4$
Equation EF, $y = 2 x - 7 \Rightarrow \frac{P Q}{E F} = \frac{10}{91}$

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