$ABCD$ is a square, $E$ and $F$ are midpoints of sides $AD$ and $BC$ respectively,

and $DI= IJ=JC$ ,

and $AB=1$ ,

If area of shaded region is $\dfrac{a}{b}$ , where $a$ and $b$ are positive co-prime Integers,

Then find the value of $a+b$ .

The answer is 17.

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Considering that E is the midpoint of AD and F is the midpoint of BC, the length of GH is the average of the length of IJ and AB which is $\dfrac{\frac{1}{3}+1}{2}=\dfrac{2}{3}$ . Note that the perpendicular height of the trapezium is $\dfrac{1}{2}$ . Then the area of the trapezium is as follows:

$\dfrac{1}{2} \times \dfrac{1}{2} \times (\dfrac{2}{3}+1)=\dfrac{1}{4} \times \dfrac{5}{3}= \dfrac{5}{4\times 3}=\dfrac{5}{12}$

Therefore as required, $a+b=5+12=\boxed{17}$ .