A rectangular piece of sheet metal has the width 50 𝑐𝑚 shorter than its length. From each corner, a 10 𝑐𝑚 × 10 𝑐𝑚 square is cut out and the flaps are then folded up to form an open box. If the volume of the box is 104000 𝑐𝑚3, find the length and width of the original piece of sheet metal.

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.

A rectangular piece of sheet metal has the width 50 cm shorter than its length. From each corner, a 10 cm × 10 cm square is cut out and the flaps are then folded up to form an open box. If the volume of the box is 104000 cm³, find the length and width of the original piece of sheet metal.

Find the dimensions of a rectangular sheet metal with width 50 cm shorter than its length. Using quadratic equations and geometry, solve for the original length and width of the metal piece to form a box with a volume of 104000 cm³. Step-by-step solution included, suitable for Grades 9-11.

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