Given: Triangle \(ABC\) with side lengths \(35\,\text{m}, 40\,\text{m}, 50\,\text{m}\). A horse, cow, and goat are tethered at vertices \(A, B, C\) respectively, with equal rope lengths \(r=14\,\text{m}\). The animals graze exclusively within the triangle's boundaries.
At each vertex, the animal grazes a circular sector with radius \(r\), defined by the triangle's interior angle at that vertex. The total grazed area is the sum of these sectors. Mathematically, the total grazed area is given by: \[ \mathcal{A} \;=\; \frac{1}{2}r^2(\angle A+\angle B+\angle C). \] Since the sum of interior angles in any triangle is \(\pi\) radians (\(\angle A+\angle B+\angle C=\pi\)), the formula simplifies to: \[ \mathcal{A} \;=\; \frac{1}{2}\,r^2\,\pi. \] In this scenario, \(r=14\,\text{m}\), and crucially, each side adjacent to a vertex is longer than \(14\,\text{m}\). This ensures that no grazed sector is limited or cut off by reaching the opposite side of the triangle.
The total grazed area is calculated as: \[ \mathcal{A}=\frac{1}{2}\times 14^2 \times \pi \;=\; \frac{1}{2}\times 196 \times \pi \;=\; 98\pi\ \text{m}^2. \] Numerically, this is approximately \(307.88\ \text{m}^2\).
Total grazed area: \(\boxed{98\pi\ \text{m}^2 \;(\approx 307.9\ \text{m}^2)}\).
Let \( A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix} \) and \( P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0. \) If \( B = P A P^T \), \( C = P^T B P \), and the sum of the diagonal elements of \( C \) is \( \frac{m}{n} \), where gcd(m, n) = 1, then \( m + n \) is: