Grazing goat problems: sector and triangle area Problem set B One question at a time

A field is a right-angled triangle with sides 7 m, 24 m and 25 m. A cow, a horse and a donkey are each tied to a different corner by a 3 m rope. The ropes are too short to meet, and the animals cannot leave the field.

1. (a) Find the area each animal can graze.

   • \(\text{Angles: }90^\circ,\ \arctan\frac{7}{24} = 16.26^\circ,\ 90^\circ - 16.26^\circ = 73.74^\circ\)
   • \(A = \frac{\theta}{360}\times\pi\times 3^2\) gives 7.07 m², 1.28 m², 5.79 m²

2. (b) Which animal grazes the most?

   • The area depends on the angle because every rope is the same length. The right angle is the largest angle.

A field is a right-angled triangle with sides 12 m, 16 m and 20 m. Three llamas are each tied to a different corner by a 5 m rope. The ropes are too short to meet, and the llamas cannot leave the field. Find the percentage of the field the llamas graze altogether.

• The three angles of the triangle add to 180°, so the sectors together make a semicircle: \(\frac{180}{360}\times\pi\times 5^2 = 39.3\) m²
• Area of the field: \(\frac12\times 16\times 12 = 96.0\) m²
• \(\frac{39.3}{96.0}\times 100 = 40.9\%\)

A field is a rectangle 11 m by 4 m. A pony is tied by a 5.5 m rope to a post at one corner. Find the area of the field the pony can graze.

• \(\sqrt{5.5^2 - 4^2} = 3.77\) m: the rope meets the far side 3.77 m along
• Triangle: \(\frac12\times 4\times 3.77 = 7.55\) m²
• Sector (degrees): \(\theta = \arcsin\frac{4}{5.5} = 46.66^\circ,\ \frac{46.66}{360}\times\pi\times 5.5^2 = 12.3\) m²
• \(A = 7.55 + 12.3 = 19.9\) m²

A garden is a right-angled triangle with sides 15 m, 20 m and 25 m. A llama is tied to the midpoint of the hypotenuse by a rope just long enough to reach both ends of it. A second llama is tied to the midpoint of the 20 m side by a 10 m rope. Neither llama can leave the garden.

1. (a) Find the percentage of the garden the first llama can graze.

   • The rope is 25.0/2 = 12.5 m. The midpoint of the hypotenuse is the centre of the circle through all three corners, so every corner is 12.5 m from the llama.
   • The whole garden is within reach: 100%.

2. (b) The hypotenuse cuts a segment off the circle of the second llama. Find the area the second llama can graze.

   • Half of the circle of radius 10.0: \(\frac{180}{360}\times\pi\times 10.0^2 = 157\) m²
   • Angle at the end of the 20 m side: \(\arctan\frac{15}{20} = 36.87^\circ\); the triangle formed by the llama and the two points on the hypotenuse has base angles equal to it, so the angle at the llama is \(180^\circ - 2\times 36.87^\circ = 106.3^\circ\)
   • Segment: \(\frac{106.3}{360}\times\pi\times 10.0^2 - \frac12\times 10.0^2\sin 106.3^\circ = 44.7\) m²
   • \(157 - 44.7 = 112\) m²

A rectangular school field measures 14 m by 5 m. A tethered ram is tied by a 7 m rope to a post at one corner of the field. Find the percentage of the field the ram can graze, to the nearest integer.

• \(\sqrt{7^2 - 5^2} = 4.90\)
• Triangle \(\frac12\times 5\times 4.90 = 12.2\) m², sector \(\frac{\theta}{360}\times\pi\times 7^2\) with \(\theta=\arcsin\frac{5}{7} = 45.58^\circ\) gives 19.5 m²
• \(\frac{12.2 + 19.5}{70}\times 100 = 45.3\% \approx 45\%\)