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 denote:

- The length of the rectangular sheet metal as \( L \) (in cm),
- The width of the rectangular sheet metal as \( W \) (in cm).

From the problem, we are given that:

1. The width is 50 cm shorter than the length:
   \[
   W = L - 50
   \]
   
2. After cutting out \( 10 \, \text{cm} \times 10 \, \text{cm} \) squares from each corner and folding the flaps, the resulting box has a height of 10 cm. The length and width of the base of the box are reduced by 20 cm (since 10 cm is cut from each side).

   So, the dimensions of the base of the box are:
   - Length of the base = \( L - 20 \) cm,
   - Width of the base = \( W - 20 \) cm.

3. The volume of the box is given as \( 104,000 \, \text{cm}^3 \). The volume of the box can be expressed as:
   \[
   \text{Volume} = \text{Base length} \times \text{Base width} \times \text{Height}
   \]
   Substituting the known values:
   \[
   104,000 = (L - 20)(W - 20)(10)
   \]

4. Simplifying this equation:
   \[
   104,000 = 10(L - 20)(W - 20)
   \]
   \[
   10,400 = (L - 20)(W - 20)
   \]

5. Now, substitute \( W = L - 50 \) into the equation:
   \[
   10,400 = (L - 20)((L - 50) - 20)
   \]
   \[
   10,400 = (L - 20)(L - 70)
   \]

6. Expand the equation:
   \[
   10,400 = L^2 - 70L - 20L + 1400
   \]
   \[
   10,400 = L^2 - 90L + 1400
   \]

7. Rearranging the equation:
   \[
   L^2 - 90L - 9,000 = 0
   \]

8. Now, solve this quadratic equation using the quadratic formula:
   \[
   L = \frac{-(-90) \pm \sqrt{(-90)^2 - 4(1)(-9,000)}}{2(1)}
   \]
   \[
   L = \frac{90 \pm \sqrt{8100 + 36,000}}{2}
   \]
   \[
   L = \frac{90 \pm \sqrt{44,100}}{2}
   \]
   \[
   L = \frac{90 \pm 210}{2}
   \]

   This gives two possible solutions:
   \[
   L = \frac{90 + 210}{2} = \frac{300}{2} = 150
   \]
   or
   \[
   L = \frac{90 - 210}{2} = \frac{-120}{2} = -60
   \]

Since the length cannot be negative, we discard \( L = -60 \) and take \( L = 150 \, \text{cm} \).

9. Using \( W = L - 50 \), we find the width:
   \[
   W = 150 - 50 = 100 \, \text{cm}
   \]

Thus, the dimensions of the original sheet metal are:
- Length \( L = 150 \, \text{cm} \),
- Width \( W = 100 \, \text{cm} \).

### Summary:
- Length: 150 cm
- Width: 100 cm

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How would the dimensions change if the height of the box was different?
2. Can the same process be applied if the squares cut from each corner are not equal in size?
3. How would the volume change if the width was 30 cm shorter instead of 50 cm shorter?
4. What if the sheet metal had a length of 200 cm, how would the width be affected to maintain the same volume?
5. How can this approach be adapted for a non-rectangular sheet of metal?

**Tip:** When working with quadratic equations, always check the physical constraints of the problem to discard unrealistic solutions.

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.

Explore how to find the length and width of a rectangular sheet metal with width 50 cm shorter than its length. Learn step-by-step how cutting and folding squares forms a box with a given volume. This problem involves algebra, geometry, and solving quadratic equations, ideal for Grades 9-11.

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