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.

   • \(\text{Angles: }\frac{\pi}{2},\ \arctan\frac{7}{24} = 0.284,\ \frac{\pi}{2} - 0.284 = 1.29\)
   • \(A = \frac12\times 3^2\times\theta\) 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 π, so the sectors together make a semicircle: \(\frac12\pi\times 5^2 = 39.3\) 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 (radians): \(\theta = \arcsin\frac{4}{5.5} = 0.814,\ \frac12\times 5.5^2\times0.814 = 12.3\) 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: \(\frac12\pi\times 10.0^2 = 157\) m²
   • Angle at the end of the 20 m side: \(\arctan\frac{15}{20} = 0.644\); 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 \(\pi - 2\times 0.644 = 1.85\)
   • Segment: \(\frac12\times 10.0^2(1.85 - \sin 1.85) = 44.7\) 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 \(\frac12\times 7^2\times\arcsin\frac{5}{7} = 19.5\) m²
• \(\frac{12.2 + 19.5}{70}\times 100 = 45.3\% \approx 45\%\)

A fenced triangular pen CDE stands in a large field. A ram is tied by a 6.5 m rope at a point A on the outside of the fence CD. A is 3.5 m from the corner C, and CD is 11 m long. The angle of the pen at C is θ radians. The ram can graze 78 m² of grass. Find θ.

• Semicircle: \(\frac12\pi\times 6.5^2 = 66.4\) m²
• Rope left at C: 6.5 − 3.5 = 3 m. It sweeps the angle π − θ.
• \(66.4 + \frac12\times 3^2(\pi - \theta) = 78\)
• \(\pi - \theta = 2.59,\quad \theta = 0.556\) radians

A rectangular paddock is 13 m by 3 m. A horse is tied to a post at one corner by a rope of length a metres, with \(3 < a < 13\), so that it can graze exactly half of the paddock.

1. (a) Show that \(a^2\arcsin\frac{3}{a} + 3\sqrt{a^2-9} = 39\).

   • Triangle: base \(\sqrt{a^2 - 3^2}\), height 3: \(\frac12\times 3\sqrt{a^2-9}\)
   • Sector: \(\theta = \arcsin\frac{3}{a},\ \frac12 a^2\arcsin\frac{3}{a}\)
   • Half the field: \(\frac32\sqrt{a^2-9} + \frac12 a^2\arcsin\frac{3}{a} = \frac12\times 39\)
   • Multiply by 2: \(a^2\arcsin\frac{3}{a} + 3\sqrt{a^2-9} = 39\)

2. (b) Find the value of a.

   • From the calculator, in radian mode: a = 6.73 m on 3 < a < 13

A rectangular paddock is 13 m by 5 m. A horse is tied to a post at one corner by a rope of length a metres, with \(5 < a < 13\). The horse can graze exactly 45% of the paddock. Find the value of a.

• Triangle plus sector: \(\frac12\times 5\sqrt{a^2-25} + \frac12 a^2\arcsin\frac{5}{a}\)
• Grazed area = 45% of \(13\times 5 = 65\) = \(29.25\) m²
• Multiply by 2: \(a^2\arcsin\frac{5}{a} + 5\sqrt{a^2-25} = 58.5\)
• From the calculator, in radian mode: a = 6.56 m