14. (Hengyang, Hunan, 20 10) It is known that the side length of equilateral triangle ABC is 4cm, and the line segment MN with the length of 1cm moves to point B at the speed of 1cm/s on the side AB of △ABC (at the beginning of the movement, the point? Take some? Coincidence, point n to point? At the end of the exercise), do M and N points respectively? The perpendicular of one side intersects the other sides of △ABC at P and Q. What is the movement time of segment MN? second
(1) Line segment MN is moving. Why is the quadrilateral MNQP just a rectangle when the value is 0? And calculating the area of the rectangle;
(2) In the process of moving line segment MN, the area of quadrilateral mnqp is S, and the moving time is T. Use the moving time to find the area s of quadrilateral MNQP? Change the functional relationship and write the range of the independent variable t.
Answer (1) If the quadrilateral MNQP is a rectangle, there is MP=QN. At this time, because ∠ PMA = ∠ QNB = 90, ∠ A = ∠ B = 60, Rt△PMA≌Rt△QNB, am. Understand? t= 1.5。 ? In Rt△AMP, AM= 1.5 and ∠ A = 60, so MP= and MN= 1, so the rectangular area is.
(2) According to the thinking of the above question, if M and N are divided into two ends of the AB midline at the bottom of the triangle, then AM=t, MP=? T, NQ=(3-t) Because BN=4-t- 1=3-t, what is the area of trapezoid MNQP? MN？ (MP+QN)=？ ×(? T+(3-t))= is a constant value (that is, it does not change with time). At this time? 1 & lt； t & lt2.
What if? t & lt= 1? Or? t≥2？ Then m and n are both on the same side of the bottom centerline, as shown in the second and third figures. The second diagram, BM=t, BN= 1+t, then the trapezoidal area is S=? × 1×[? t+(3-t))]=？ (2t+ 1), where 0≤t≤ 1.
Can be obtained similarly? 2≤t≤=3？ What was the situation in the area at that time? S=？ (7-2t)。