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.