From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm,

We know that the opposite sides of rectangle are equal
∴ AD = BC = 28 cm

Now, Radius of semicircular portion $=\frac{1}{2} \mathrm{BC}=14 \mathrm{~cm}$

∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion

$=40 \times 28-\frac{1}{2}\left(\frac{22}{7} \times 14 \times 14\right)$

$=1120-308$

$=812 \mathrm{~cm}^{2}$

Hence, the area of the remaining paper is 812 cm2