The shape after joining the midpoints of sides of a rhombus will form :

The correct option is B

Rectangle

Let ABCD to be rhombus of side a since E,F,G,H one midpoint of side, they will divide the side equally of length $\frac{a}{2}$

using midpoint theorem in triangle DCB we can say HG||DB ...(i)

using midpoint theorem in triangle DAB we can say EF||DB ...(i)

from (i) and (ii) we get HG||EF

similarly we can prove EH||FG so opposite side of figure EHGF are parallel

Now in $\Delta$ ADC

O is the midpoint of AC (Diagonal bisect each other)

H is the midpoint of DC (Given)

From midpoint theorem DA||HO $\Rightarrow$DA||HF

similarly we can prove AB||EG

$\therefore$ AFHD is a parallelogram and AD=FH ...(iii)

also ABGE is a parallelogram AB= EG ....(iv)

from (iii) and (iv) we have FH=EG (AS AD=AB side of rhombus)

The diagonal FH & EG are equal

A parallelogram in which diagonals are equal is a rectangle