用中文视频解答一下---**Question Stem:**
(2025·天津) In quadrilateral ABCD, AD∥BC, ∠B=90°, AB=8cm, AD=10cm, BC=16cm. Moving point M starts from point A and moves along side BA, then AD, towards endpoint D at a speed of 2cm/s; moving point N starts from point C at the same time and moves along side CB towards endpoint B at a speed of 1cm/s. It is stipulated that when one moving point reaches its endpoint, the other moving point stops. Let the time of motion be t s. When t=2s, the positions of points M and N are shown in the figure. The following conclusions are given:
① When t=6s, CN=DM;
② When 1≤t≤2, the maximum area of △BMN is 26cm²;
③ There are two different values of t that satisfy the area of △BMN is 39cm².
Among them, the number of correct conclusions is ( )
**Options:**
A. 0
B. 1
C. 2
D. 3
**Chart/Diagram Description:**
* **Type:** Geometric figure, specifically a right trapezoid.
* **Main Elements:**
* **Shape:** Trapezoid ABCD.
* **Points:** Vertices A, B, C, D are labeled. Point M is located on side AB. Point N is located on side BC.
* **Lines:** The sides of the trapezoid (AB, BC, CD, DA) are shown. A line segment MN connects points M and N, forming triangle BMN.
* **Angles:** Angle B (∠ABC) is marked as a right angle (90°).
* **Relative Position:** Point M is between A and B. Point N is between B and C.
**Other Relevant Text:**
**Knowledge Points (考点):**
Maximum value of a quadratic function; Distance between parallel lines; Application of a quadratic equation. Explanations for quadratic function maximum/minimum value problems.
**Topic (专题):**
Graph and properties of quadratic functions; Deductive reasoning ability.
**Answer (答案):**
C
**Analysis (分析):**
When t=6s, point M is on AD, calculate DM and CN to evaluate ①; When 1≤t≤2, point M is on AB, use the triangle area formula to find the area of △BMN, and use the properties of quadratic functions to evaluate ②; Consider two cases: when point M is on AB, and when point M is on AD, combine the condition that the area of △BMN is 39cm², set up equations to evaluate ③.
**Solution (解答):**
Solution: According to the problem statement:
The time for point M to move on AB is 8/2 = 4s.
The time for point M to move on AD is 10/2 = 5s.
The time for point N to move on CB is 16s.
① When t=6s, point M is on AD.
At this time, AM = 2×6 - 8 = 4cm, CN = 6cm.
Therefore, DM = AD - AM = 10 - 4 = 6cm.
Therefore, CN = DM, so ① is correct.
② When 1≤t≤2, point M is on AB.
At this time, BM = 2t cm, CN = t cm.
Therefore, BN = (16-t) cm.
Therefore, S△BMN = (1/2) * BM * BN = (1/2) * 2t * (16-t) = -t² + 16t = -(t-8)² + 64.
Since -1 < 0,
Therefore, when t < 8, S△BMN increases as t increases.
Therefore, when t=2, S△BMN obtains its maximum value, the maximum value is -(2-8)² + 64 = 28.
That is, when 1≤t≤2, the maximum area of △BMN is 28cm², so ② is incorrect.
③ When point M is on AB,
The area of △BMN is 39cm².
S△BMN = (1/2) * BM * BN = (1/2) * 2t * (16-t) = -t² + 16t = 39.
Solving this, we get: t₁=3, t₂=13 (discard t₂ because t≤4s when M is on AB).
Therefore, when t=3, the area of △BMN is 39cm².
When point M is on AD,
Since AD∥BC, ∠B=90°, which means AB⊥AD.
At this time, S△BMN = (1/2) * AB * BN = (1/2) * 8 * (16-t) = 64-4t = 39.
Solving this, we get: t = 25/4.
Therefore, when t=25/4, the area of △BMN is 39cm².
Therefore, there are two different values of t that satisfy the area of △BMN is 39cm², so ③ is correct.
Hence, choose C.
**Review/Comment (点评):**
This problem mainly examines the properties of quadratic functions and the application of quadratic equations. Mastering the properties of quadratic functions is key to solving this problem.
**Supplementary Information:**
Statement: The copyright of this problem solution belongs to Youwang. No reproduction or distribution without written permission.
Answered by: zj Dajiaozi Teacher 2025/6/25