<Question>
The lengths of two opposite sides of a square are decreased by $40\%$ while the lengths of the other two sides are increased by $50\%$ to form a rectangle. By what percent does the square's area decrease?
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this problem step by step. So, we have a square, and we're supposed to decrease the length of two opposite sides by 40% and increase the other two opposite sides by 50%. Then, this forms a rectangle, and we need to find by what percent the square's area decreases. Hmm. Let me start by recalling some basic formulas.

<prompt_1.1> Recall the basic formula for the area of a square and understand how modifying the side lengths can form a rectangle. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, the area of a square is side length squared, right? So if the original square has a side length of, let's say, 's', then the original area is s². Now, the problem says we decrease two opposite sides by 40%. Wait, does that mean we're modifying the length of those sides by 40% of their original length, or reducing them to 40% of the original length? Hmm. The wording says "decreased by 40%", which usually means subtracting 40% of the original from the original. So if the original length is 's', decreasing by 40% would make the new length s - 0.4s = 0.6s. Similarly, increasing the other two sides by 50% would mean adding 50% of the original to the original, so that's s + 0.5s = 1.5s. 

Wait, but in a square, all sides are equal. So if we are modifying two opposite sides, does that mean those sides are being changed, and the other two opposite sides are changed? So the resulting figure is a rectangle with length and width different from the original square. So, the original square has all sides equal to 's', and the new rectangle has two sides of 0.6s and two sides of 1.5s. Therefore, the area of the rectangle would be length times width, which is 0.6s * 1.5s. Let me compute that.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
0.6 times 1.5 is... 0.6*1.5. Let me think.
0.6 is 3/5, and 1.5 is 3/2. So 3/5 * 3/2 = 9/10, which is 0.9. So 0.6s * 1.5s = 0.9s². 

Wait, so the original area was s², and the new area is 0.9s².
So the area has decreased by 0.1s², which is a 10% decrease. Therefore, the answer is 10%? Hmm. That seems straightforward, but let me verify again to make sure I didn't make a mistake.

<prompt_2.1> Verify the calculations to ensure the area decrease is 10%. </prompt_2.1>

But wait, let me think again. The problem says "the lengths of two opposite sides are decreased by 40%". So if the original side is 's', decreasing by 40% would be s - 0.4s = 0.6s, as I did before.

<prompt_2.2> Re-evaluate the problem statement to confirm the interpretation of the percentage changes. </prompt_2.2>

But maybe there's a trick here. Let me make sure.

<prompt_2.3> Consider if there is a trick in the problem statement that could affect the solution. </prompt_2.3>

But let me check with actual numbers to be sure. Suppose the original square has sides of length 10 units.

<prompt_2.4> Use specific numerical examples to verify the solution and conclude the area decrease. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, but let me confirm the calculations. 0.6 times 1.5. 0.6 multiplied by 1 is 0.6, and 0.6 multiplied by 0.5 is 0.3, so adding them together gives 0.6 + 0.3 = 0.9. Yep, that's correct. So 0.9 times s squared. So the new area is 90% of the original area. Therefore, the area has decreased by 10%.
</execution_2.1>

<execution_2.2>
Then, the other two opposite sides are increased by 50%, so that's s + 0.5s = 1.5s. So the rectangle formed has sides 0.6s and 1.5s. Multiplying those gives 0.9s², which is indeed a 10% decrease from the original area of s². So 10% decrease.
</execution_2.2>

<execution_2.3>
Sometimes, when you modify sides of a square to form a rectangle, the sides could be rearranged or something. But in this case, it's two opposite sides decreased by 40% and the other two increased by 50%. So, for example, if the square is ABCD with AB, BC, CD, DA as sides, then AB and CD are decreased by 40%, making them 0.6s each, and BC and DA are increased by 50%, making them 1.5s each. Therefore, the rectangle formed has sides 0.6s and 1.5s. 

Calculating the area of the rectangle: 0.6s * 1.5s = 0.9s². Original area was s², so the difference is 0.1s², which is a 10% decrease. So, 10% is the answer.
</execution_2.3>

<execution_2.4>
So the original area is 10*10 = 100 square units. 

If we decrease two opposite sides by 40%, those sides become 10 - 40% of 10 = 10 - 4 = 6 units. Then, the other two sides are increased by 50%, so 10 + 50% of 10 = 15 units. So the rectangle formed has sides 6 and 15. The area is 6*15 = 90. The original area was 100, so the decrease is 10, which is 10/100 = 0.1, so 10%. 

Hmm, that seems to confirm it. So the area decreases by 10%.
</execution_2.4>
</Execution_2>

<Final_answer>
Therefore, the square's area decreases by $\boxed{10\%}$.
</Final_answer>
</think>