https://file.aiquickdraw.com/mathbot/file/172684524683891e3fjqq.jpg

The problem involves a triangular lawn model with a scale of **1 cm representing 10 yards**. The model dimensions are given as:

- Height = **3 cm**
- Base = **8 cm**

### Steps to solve:

1. **Convert the model dimensions to actual dimensions:**
   - Actual height = \( 3 \, \text{cm} \times 10 = 30 \, \text{yards} \)
   - Actual base = \( 8 \, \text{cm} \times 10 = 80 \, \text{yards} \)

2. **Find the area of the actual triangular lawn:**
   The area of a triangle is given by the formula:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Substituting the actual dimensions:
   \[
   \text{Area} = \frac{1}{2} \times 80 \, \text{yards} \times 30 \, \text{yards}
   \]
   \[
   \text{Area} = \frac{1}{2} \times 2400 \, \text{square yards} = 1200 \, \text{square yards}
   \]

Thus, the area of the actual lawn is **1200 square yards**.

Would you like more details on the steps, or do you have any questions?

Here are some related questions for further exploration:
1. How would the area change if the scale was 1 cm = 15 yards?
2. How can you calculate the perimeter of the actual lawn?
3. How would the formula change if the triangle was not a right triangle?
4. If the base was doubled but the height remained the same, what would be the new area?
5. How do you determine the area of irregular shapes like a trapezoid?

**Tip:** When working with scale models, always ensure to correctly convert all measurements to the actual dimensions before performing calculations.

A model of a triangular lawn is scaled to have 1 cm represent 10 yards. The dimensions of the model are a height of 3 cm and a base of 8 cm. What is the area of the actual lawn, in square yards?

Learn how to calculate the area of a triangular lawn from its scaled model dimensions. The model's height is 3 cm and base is 8 cm, where each centimeter represents 10 yards in the actual lawn. Find the area of the real lawn using the triangle area formula and scale conversion. Suitable for middle school and high school students.

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