Grazing goat problems: sector and triangle area Problem set A Worked solutions

A field is a right-angled triangle with sides 8 m, 15 m and 17 m. A goat, a sheep and a pony are each tied to a different corner by a 3.5 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.

   • \(\text{Angles: }\frac{\pi}{2},\ \arctan\frac{8}{15} = 0.490,\ \frac{\pi}{2} - 0.490 = 1.08\)
   • \(A = \frac12\times 3.5^2\times\theta\) gives 9.62 m², 3.00 m², 6.62 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 10 m, 24 m and 26 m. Three goats are each tied to a different corner by a 4.5 m rope. The ropes are too short to meet, and the goats cannot leave the field. Find the percentage of the field the goats graze altogether.

• The three angles of the triangle add to π, so the sectors together make a semicircle: \(\frac12\pi\times 4.5^2 = 31.8\) m²
• Area of the field: \(\frac12\times 24\times 10 = 120\) m²
• \(\frac{31.8}{120}\times 100 = 26.5\%\)

A field is a rectangle 14 m by 5 m. A goat is tied by a 6.5 m rope to a post at one corner. Find the area of the field the goat can graze.

• \(\sqrt{6.5^2 - 5^2} = 4.15\) m: the rope meets the far side 4.15 m along
• Triangle: \(\frac12\times 5\times 4.15 = 10.4\) m²
• Sector (radians): \(\theta = \arcsin\frac{5}{6.5} = 0.878,\ \frac12\times 6.5^2\times0.878 = 18.5\) m²
• \(A = 10.4 + 18.5 = 28.9\) m²

A paddock is a right-angled triangle with sides 6 m, 8 m and 10 m. A donkey is tied to the midpoint of the hypotenuse by a rope just long enough to reach both ends of it. A second donkey is tied to the midpoint of the 8 m side by a 4 m rope. Neither donkey can leave the paddock.

1. (a) Find the percentage of the paddock the first donkey can graze.

   • The rope is 10.0/2 = 5.00 m. The midpoint of the hypotenuse is the centre of the circle through all three corners, so every corner is 5.00 m from the donkey.
   • The whole paddock is within reach: 100%.

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

   • Half of the circle of radius 4.00: \(\frac12\pi\times 4.00^2 = 25.1\) m²
   • Angle at the end of the 8 m side: \(\arctan\frac{6}{8} = 0.644\); the triangle formed by the donkey and the two points on the hypotenuse has base angles equal to it, so the angle at the donkey is \(\pi - 2\times 0.644 = 1.85\)
   • Segment: \(\frac12\times 4.00^2(1.85 - \sin 1.85) = 7.16\) m²
   • \(25.1 - 7.16 = 18.0\) m²

A rectangular lawn measures 15 m by 6 m. A goat is tied by an 8 m rope to a post at one corner of the lawn. Find the percentage of the lawn the goat can graze, to the nearest integer.

• \(\sqrt{8^2 - 6^2} = 5.29\)
• Triangle \(\frac12\times 6\times 5.29 = 15.9\) m², sector \(\frac12\times 8^2\times\arcsin\frac{6}{8} = 27.1\) m²
• \(\frac{15.9 + 27.1}{90}\times 100 = 47.8\% \approx 48\%\)

A fenced triangular pond CDE stands in a large field. A goat is tied by an 8 m rope at a point A on the outside of the fence CD. A is 5 m from the corner C, and CD is 14 m long. The angle of the pond at C is θ radians. The goat can graze 110 m² of grass. Find θ.

• Semicircle: \(\frac12\pi\times 8^2 = 100.5\) m²
• Rope left at C: 8 − 5 = 3 m. It sweeps the angle π − θ.
• \(100.5 + \frac12\times 3^2(\pi - \theta) = 110\)
• \(\pi - \theta = 2.104,\quad \theta = 1.04\) radians

A rectangular field is 12 m by 5 m. A goat is tied to a post at one corner by a rope of length a metres, with \(5 < a < 12\), so that it can graze exactly half of the field.

1. (a) Show that \(a^2\arcsin\frac{5}{a} + 5\sqrt{a^2-25} = 60\).

   • Triangle: base \(\sqrt{a^2 - 5^2}\), height 5: \(\frac12\times 5\sqrt{a^2-25}\)
   • Sector: \(\theta = \arcsin\frac{5}{a},\ \frac12 a^2\arcsin\frac{5}{a}\)
   • Half the field: \(\frac52\sqrt{a^2-25} + \frac12 a^2\arcsin\frac{5}{a} = \frac12\times 60\)
   • Multiply by 2: \(a^2\arcsin\frac{5}{a} + 5\sqrt{a^2-25} = 60\)

2. (b) Find the value of a.

   • From the calculator, in radian mode: a = 6.69 m on 5 < a < 12

A rectangular field is 14 m by 6 m. A goat is tied to a post at one corner by a rope of length a metres, with \(6 < a < 14\). The goat can graze exactly 55% of the field. Find the value of a.

• Triangle plus sector: \(\frac12\times 6\sqrt{a^2-36} + \frac12 a^2\arcsin\frac{6}{a}\)
• Grazed area = 55% of \(14\times 6 = 84\) = \(46.2\) m²
• Multiply by 2: \(a^2\arcsin\frac{6}{a} + 6\sqrt{a^2-36} = 92.4\)
• From the calculator, in radian mode: a = 8.47 m