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$ \left( a \right){\text{ Quadrilateral with unequal sides}} $

$ \left( b \right){\text{ Parallelogram}} $

$ \left( c \right){\text{ Rhombus}} $

$ \left( d \right){\text{ Rectangle}} $

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Since it is given that the $ ABCD $ is a quadrilateral with unequal sides and $ E,F,G,H $ are the middle points of the four sides. So on joining the diagonals of the quadrilateral which will be named as $ BD $ and $ AC $ .

Now we will take a triangle $ ABD $ , for the reference we take it from the figure. So in this triangle $ H $ and $ E $ will be the midpoints of the sides $ AD $ and $ AB $ respectively.

Hence, from this we have $ HE||BD $ .

Similarly, we will take a triangle $ BCD $ , in this triangle $ FG||BD $ and in the same way we will get $ HE||GF $ .

Similarly, on considering the triangle $ ABC $ and the triangle $ ACD $ , we will get $ EF||AC $ and $ GH||CA $ . So from this, we can see that the quadrilateral $ EFGH $ , in this the opposite sides are parallel. And we know that this is the property of parallelogram.

Hence, it makes it a parallelogram.

Therefore, the option $ \left( b \right) $ is correct.