Three horses are tied each with 7 m long rope at three corners of a triangular field having sides 20 m, 30m and 40 m. find the area of the plot which can be grazed by the horses

To solve this problem, we can break it down step-by-step by calculating the total area grazed by the horses. Each horse is tied to a corner of a triangular field, with a rope length of 7 meters. The triangular field has sides of 20 m, 30 m, and 40 m.

### Steps:

#### 1. **Visualizing the Situation**:
Each horse can graze in a circular area, with the horse tied at a vertex of the triangle. Since the rope is 7 meters long, the horse's grazing area forms part of a circle (a sector) at each corner.

The field is a right triangle because it satisfies the Pythagorean theorem:

\[
20^2 + 30^2 = 40^2 \implies 400 + 900 = 1600
\]

So, this triangle has a right angle at the vertex opposite the side of length 40 m.

#### 2. **Finding the Grazing Area**:
Each horse can graze in a sector of a circle. The radius of the sector is 7 meters, and the angle of the sector depends on the internal angles of the triangle.

- **Internal Angles of the Triangle**:
    - For the side opposite to angle \(A\) (side of 20 m):
    \[
    \cos(A) = \frac{30^2 + 40^2 - 20^2}{2 \cdot 30 \cdot 40} = \frac{900 + 1600 - 400}{2400} = \frac{2100}{2400} = 0.875
    \]
    \[
    A = \cos^{-1}(0.875) = 28.96^\circ
    \]

    - For the side opposite to angle \(B\) (side of 30 m):
    \[
    \cos(B) = \frac{20^2 + 40^2 - 30^2}{2 \cdot 20 \cdot 40} = \frac{400 + 1600 - 900}{1600} = \frac{1100}{1600} = 0.6875
    \]
    \[
    B = \cos^{-1}(0.6875) = 46.57^\circ
    \]

    - For the right angle \(C\) (opposite side of 40 m), \(C = 90^\circ\).

#### 3. **Calculating the Area of Each Grazing Sector**:
The formula for the area of a sector is:
\[
\text{Sector Area} = \frac{\theta}{360^\circ} \times \pi r^2
\]
where \( \theta \) is the angle of the sector and \( r = 7 \) meters.

- **Sector Area for Angle \(A\) (28.96°)**:
\[
\text{Sector Area}_A = \frac{28.96}{360} \times \pi \times 7^2 = \frac{28.96}{360} \times 153.94 = 12.38 \, \text{m}^2
\]

- **Sector Area for Angle \(B\) (46.57°)**:
\[
\text{Sector Area}_B = \frac{46.57}{360} \times \pi \times 7^2 = \frac{46.57}{360} \times 153.94 = 19.91 \, \text{m}^2
\]

- **Sector Area for Angle \(C\) (90°)**:
\[
\text{Sector Area}_C = \frac{90}{360} \times \pi \times 7^2 = \frac{90}{360} \times 153.94 = 38.49 \, \text{m}^2
\]

#### 4. **Total Grazed Area**:
The total area grazed by the horses is the sum of the three sectors:
\[
\text{Total Grazed Area} = 12.38 + 19.91 + 38.49 = 70.78 \, \text{m}^2
\]

Thus, the total area that can be grazed by the horses is approximately **70.78 square meters**.

---

Would you like any further details or have any questions? Here are some additional questions related to the problem:

1. How do you calculate the area of a sector for a different rope length?
2. Can you derive the internal angles of any triangle using the cosine rule?
3. How would the area change if the sides of the triangle were different?
4. What if the horses had different rope lengths, how would the grazing area change?
5. Can you apply a similar method to calculate the grazing area for a quadrilateral field?

**Tip:** Always ensure the triangle is correctly identified as right, acute, or obtuse before calculating angles and areas.

Three horses are tied each with 7 m long rope at three corners of a triangular field having sides 20 m, 30 m, and 40 m. Find the area of the plot which can be grazed by the horses.

This problem involves calculating the grazing area of three horses tied with a 7 m rope at the corners of a right triangle field with sides 20 m, 30 m, and 40 m. Using the cosine rule and sector area formula, the total grazed area is approximately 70.78 square meters. Suitable for high school students (Grades 9-12).

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