https://file.aiquickdraw.com/mathbot/file/1730269097972hb39elu9.jpg

Let's solve each part of the problem step-by-step.

### Part 1
A rectangle with width \(6 \, \text{cm}\) and length \(8 \, \text{cm}\) has the same area as a trapezium with a height of \(4 \, \text{cm}\). We need to find the sum of the lengths of the parallel sides of the trapezium.

1. **Area of the Rectangle**:
   \[
   \text{Area} = \text{width} \times \text{length} = 6 \times 8 = 48 \, \text{cm}^2
   \]

2. **Using the Trapezium Area Formula**:
   The area of a trapezium is given by:
   \[
   \text{Area} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height}
   \]
   where \(\text{height} = 4 \, \text{cm}\) and \(\text{Area} = 48 \, \text{cm}^2\).

3. **Setting up the Equation**:
   \[
   48 = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times 4
   \]
   Simplify:
   \[
   48 = 2 \times (\text{base}_1 + \text{base}_2)
   \]
   \[
   \text{base}_1 + \text{base}_2 = \frac{48}{2} = 24 \, \text{cm}
   \]

So, the lengths of the parallel sides of the trapezium add up to **24 cm**.

---

### Part 2
The areas of a rectangle, a triangle, and a trapezium are all \(128 \, \text{cm}^2\), and each shape has a height of \(8 \, \text{cm}\). We need to find the lengths of each shape.

1. **Rectangle**:
   \[
   \text{Area of Rectangle} = \text{height} \times \text{length}
   \]
   \[
   128 = 8 \times \text{length}
   \]
   \[
   \text{length} = \frac{128}{8} = 16 \, \text{cm}
   \]

2. **Triangle**:
   The area of a triangle is given by:
   \[
   \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   \[
   128 = \frac{1}{2} \times \text{base} \times 8
   \]
   Simplify:
   \[
   128 = 4 \times \text{base}
   \]
   \[
   \text{base} = \frac{128}{4} = 32 \, \text{cm}
   \]

3. **Trapezium**:
   The area of a trapezium is:
   \[
   \text{Area of Trapezium} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height}
   \]
   \[
   128 = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times 8
   \]
   Simplify:
   \[
   128 = 4 \times (\text{base}_1 + \text{base}_2)
   \]
   \[
   \text{base}_1 + \text{base}_2 = \frac{128}{4} = 32 \, \text{cm}
   \]

Thus, the lengths are:
- Rectangle length: **16 cm**
- Triangle base: **32 cm**
- Sum of trapezium's parallel sides: **32 cm**

Would you like more details on any step, or do you have further questions?

---

### Related Questions
1. What would happen if the heights of the shapes were different?
2. How does the area formula change for shapes with more complex geometries?
3. What if the trapezium's parallel sides were not equal—how would that affect the solution?
4. Could we calculate the perimeters given only the areas and heights?
5. How can these formulas be derived from first principles in geometry?

### Tip
Always verify unit consistency, especially when working with area and length measurements.

A rectangle of width 6 cm and length 8 cm has the same area as a trapezium of height 4 cm. What do the lengths of the parallel sides of the trapezium add up to?

The areas of a rectangle, a triangle, and a trapezium are all 128 cm². If the heights of all three shapes are 8 cm, what are the lengths of each shape?

This problem explores the calculation of side lengths of a rectangle, triangle, and trapezium with equal areas and specified heights. Learn how to apply area formulas for each shape, with solutions suitable for Grades 7-9.

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