If a goat is tied to a pole at Point A with a rope 12 m long such that it…

Concept: When an animal tied by a rope at a vertex is kept outside a polygon, the grazing region equals the major sector swept at the tether point (interior angle excluded) plus, whenever the rope is longer than an adjacent side, an extra wrap-around sector at that neighbouring vertex — with radius reduced by the side length, and angle equal to the exterior angle there.

1. Triangle ABC is isosceles with AB = AC = 10 m and ∠A = 30°, so ∠B = ∠C = (180° − 30°) ÷ 2 = 75°.

2. Sector at A: radius = rope length = 12 m, angle = 360° − 30° = 330° (the interior angle is blocked by the triangle). Area = (330/360) × π × 122 = 132π.

3. Since the rope (12 m) is longer than AB and AC (10 m each), it still has 12 − 10 = 2 m free after wrapping around B and C.

4. Extra sector at B: radius 2 m, angle = 180° − 75° = 105°. Extra sector at C: same, radius 2 m, angle 105°. Each area = (105/360) × π × 22 = 7π/6.

5. Combined extra area from both B and C = 2 × 7π/6 = 7π/3. Total grazing area = 132π + 7π/3 = 403π/3 ≈ 134.33π.

Cross-check: the base sector alone — the 330° sweep at the tether point using the actual 12 m rope as radius — is exactly 132π. The extra wrap-around term is zero only when the rope does not extend past the sides at all; here the rope is 2 m longer than each side, so that term is strictly positive (7π/3), and the total must be strictly more than 132π — confirming 'more than 132π', not 'equal to' or 'less than' it.