2。 69/8

3。 We don't stop the triangle from moving to the left on the right side of the square.

The area ratio of a triangle is equal to the square of the side length ratio (this is self-evident)

Let the midpoint of the bottom of the triangle PQR be O, then there is OP vertical QR, and the area of the triangle PQO is 6.

When t=5, the area of the small triangle on the right side of the square is 27/8.

Area coincident with the square: 69/8

When 5≤t≤8, move to the left, and the square on the left will reveal a small triangle with a base length of (t-5).

Come out, this little triangle is similar to PQO,

Its area: {[(t-5)/4] square} * 6 This is also the area removed from the left.

And the base of the original small triangle on the right is now 3-(t-5)=8-t,

Area: {[(8-t)/4] square} * 6,

The area moved in from the right: [27/8]-{[(8-t)/4] square} * 6.

So the overlapping area = 69/8- the area moved out on the left+the area moved in on the right.

=[69/8]-{[(t-5)/4]square } * 6+[27/8]-{[(8-t)/4]square } * 6

=12-(3/8) * [2 * (t2)-26t+89] (t2) represents the square of t.

When t=6.5, the maximum value is 165/ 16.