Let's break down the problem step by step to find the dimensions of the original piece of sheet metal.

Step 1: Define variables

• Let the length of the original piece of sheet metal be $L$ cm.
• Let the width of the original piece of sheet metal be $W$ cm.

According to the problem, the width is 50 cm shorter than the length: $W = L - 50$

Step 2: Dimensions of the box

When a 10 cm × 10 cm square is cut from each corner, the dimensions of the base of the resulting open box will be reduced by 20 cm (since 10 cm is cut from both sides of the length and width).

• The new length of the box base will be $L - 20$ cm.
• The new width of the box base will be $W - 20 = (L - 50) - 20 = L - 70$ cm.
• The height of the box is the size of the square cut from the corners, which is 10 cm.

Step 3: Volume of the box

The volume of the box is given as 104,000 cm³. The formula for the volume of a box is: $\text{Volume} = \text{length} \times \text{width} \times \text{height}$

Substitute the values for the length, width, and height of the box: $104000 = (L - 20)(L - 70)(10)$

Step 4: Solve for $L$

We can simplify the equation: $104000 = 10 \times (L - 20)(L - 70)$ $10400 = (L - 20)(L - 70)$ Now expand the right-hand side: $10400 = L^2 - 70L - 20L + 1400$ $10400 = L^2 - 90L + 1400$ Move all terms to one side of the equation: $L^2 - 90L + 1400 - 10400 = 0$ $L^2 - 90L - 9000 = 0$

Step 5: Solve the quadratic equation

We can solve the quadratic equation $L^2 - 90L - 9000 = 0$ using the quadratic formula: $L = \frac{-(-90) \pm \sqrt{(-90)^2 - 4(1)(-9000)}}{2(1)}$ $L = \frac{90 \pm \sqrt{8100 + 36000}}{2}$ $L = \frac{90 \pm \sqrt{44100}}{2}$ $L = \frac{90 \pm 210}{2}$ This gives us two possible solutions: $L = \frac{90 + 210}{2} = \frac{300}{2} = 150$ or $L = \frac{90 - 210}{2} = \frac{-120}{2} = -60$

Since a negative length doesn't make sense, we discard $L = -60$ and keep $L = 150$.

Step 6: Find the width

Using the relation $W = L - 50$: $W = 150 - 50 = 100$

Final Answer:

• The length of the original piece of sheet metal is $150 \, \text{cm}$.
• The width of the original piece of sheet metal is $100 \, \text{cm}$.

Would you like further clarification or more details?

Here are some related questions you might find useful:

1. How do we use the quadratic formula to solve real-world problems?
2. What would happen if different-sized squares were cut from the corners of the sheet?
3. How does the volume formula for a box apply to other geometric shapes?
4. Can the height of the box be changed to achieve the same volume with different dimensions?
5. How do changes in length and width affect the volume of a box?

Tip: Always check the physical feasibility of your solutions when solving equations in real-world problems.