16. (15 points) As shown in the figure, points $M$ and $N$ are the midpoints of the opposite sides $AD$ and $BC$ of a square $ABCD$ with side length 4 decimeters. Points $P$ and $Q$ start moving from $M$ simultaneously. $P$ moves counterclockwise along the edges of the square at a speed of 1 meter/second, and $Q$ moves clockwise along the edges of the square at a speed of 2 meters/second. Find:
(1) The area of $\triangle N P Q$ after 1 second;
(2) The area of $\triangle N P Q$ after 15 seconds;
(3) The area of $\triangle N P Q$ after 2015 seconds.
16. (15 points) As shown in the figure, points $M$ and $N$ are the midpoints of the sides $AD$ and $BC$ of a square $ABCD$ with side length 4 decimeters. Points $P$ and $Q$ start from $M$ at the same time. Point $P$ moves counterclockwise along the edges of the square at a speed of 1 meter/second, and point $Q$ moves clockwise along the edges of the square at a speed of 2 meters/second. Find:
(1) The area of $\triangle N P Q$ after 1 second;
(2) The area of $\triangle N P Q$ after 15 seconds;
(3) The area of $\triangle N P Q$ after 2015 seconds.
【Solution】Solution: (1) After 1 second, as shown in the figure,
The area of $\triangle N P Q$: $(1+2) \times 4 \div 2$
$$
=3 \times 4 \div 2
$$
$=6$ (square decimeters);
(2) After 15 seconds, as shown in the figure,
The area of $\triangle N P Q$: $(2+1) \times 4 \div 2$
$$
=3 \times 4 \div 2
$$
$=6$ (square decimeters);
(3) Since $16 \div 1=16,16 \div 2=8$,
Therefore, after 16 seconds, points $P$ and $Q$ both return to the starting point $M$,
$$
2015 \div 16=125 \cdots 15 \text { (seconds) }
$$

So, after 2015 seconds, the positions of points $P$ and $Q$ are the same as after 15 seconds,
The area is also 6 square decimeters.