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Solve all three task in a good way using proper symbols which are used in notebooks like a grade 5 student

Let's solve the three tasks step by step:

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### **TASK 1**

Given:
- The area of the trapezium-shaped field is \( 10500 \, \text{sq metres} \).
- The perpendicular distance (height) between the two parallel sides is \( 100 \, \text{m} \).
- One parallel side (along the river) is twice the length of the other parallel side (along the road).

Let:
- The length of the side along the road = \( x \, \text{m} \).
- The length of the side along the river = \( 2x \, \text{m} \).

We know the formula for the area of a trapezium is:

\[
\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}
\]

Substitute the known values:

\[
10500 = \frac{1}{2} \times (x + 2x) \times 100
\]
\[
10500 = \frac{1}{2} \times 3x \times 100
\]
\[
10500 = 150x
\]

Solve for \( x \):

\[
x = \frac{10500}{150} = 70 \, \text{m}
\]

Thus:
- The length of the side along the road = \( 70 \, \text{m} \).
- The length of the side along the river = \( 2 \times 70 = 140 \, \text{m} \).

---

### **TASK 2**

Given:
- The trapezium-shaped field \( ABCD \) has:
  - \( AB = 100 \, \text{m} \) (perpendicular distance),
  - \( BC = 48 \, \text{m} \),
  - \( CD = 17 \, \text{m} \),
  - \( AD = 40 \, \text{m} \),
  - and the total length of the fence \( ABCD = 120 \, \text{m} \).

To find:
- The area of the field.

The formula for the area of a trapezium is:

\[
\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}
\]

Here:
- The parallel sides are \( AD \) and \( BC \).
- The height \( AB = 100 \, \text{m} \).

Substitute the values:

\[
\text{Area} = \frac{1}{2} \times (40 + 48) \times 100
\]
\[
\text{Area} = \frac{1}{2} \times 88 \times 100
\]
\[
\text{Area} = 4400 \, \text{sq metres}
\]

Thus, the area of the trapezium-shaped field is \( 4400 \, \text{sq metres} \).

---

### **TASK 3**

Given:
- The dimensions of the field in the figure are in cm.

The figure is composed of several rectangles. We will break the figure down into parts:

- **Rectangle A**: \( 4 \, \text{cm} \times 4 \, \text{cm} \)
  - Area \( = 4 \times 4 = 16 \, \text{sq cm} \).

- **Rectangle B**: \( 12 \, \text{cm} \times 4 \, \text{cm} \)
  - Area \( = 12 \times 4 = 48 \, \text{sq cm} \).

- **Rectangle C**: \( 4 \, \text{cm} \times 4 \, \text{cm} \)
  - Area \( = 4 \times 4 = 16 \, \text{sq cm} \).

- **Right triangle D**: \( \frac{1}{2} \times 4 \, \text{cm} \times 3 \, \text{cm} \)
  - Area \( = \frac{1}{2} \times 4 \times 3 = 6 \, \text{sq cm} \).

Now, the total area of the field is:

\[
\text{Total area} = 16 + 48 + 16 + 6 = 86 \, \text{sq cm}
\]

Thus, the area of the field shown in Task 3 is \( 86 \, \text{sq cm} \).

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Would you like further details on any of these tasks?

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### **Related Questions:**
1. How can the area of a triangle be calculated in other cases?
2. What is the difference between the area formulas of trapezium and parallelogram?
3. How does scaling affect the area of geometrical figures?
4. What are some real-life examples where trapezium-shaped fields are used?
5. How would the answer change if the height of the trapezium in Task 1 was different?

**Tip**: For composite shapes, break them down into simpler shapes (like rectangles and triangles) to calculate the area easily.

Solve all three tasks in a good way using proper symbols like a grade 5 student.

Step-by-step solutions to problems involving trapezium-shaped fields and composite shapes. These tasks include the area of a trapezium, decomposing shapes into rectangles and triangles, and applying basic algebra to solve problems. Suitable for Grade 5 students.

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