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.

   • \(\text{Angles: }90^\circ,\ \arctan\frac{8}{15} = 28.07^\circ,\ 90^\circ - 28.07^\circ = 61.93^\circ\)
   • \(A = \frac{\theta}{360}\times\pi\times 3.5^2\) 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 180°, so the sectors together make a semicircle: \(\frac{180}{360}\times\pi\times 4.5^2 = 31.8\) 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 (degrees): \(\theta = \arcsin\frac{5}{6.5} = 50.28^\circ,\ \frac{50.28}{360}\times\pi\times 6.5^2 = 18.5\) 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: \(\frac{180}{360}\times\pi\times 4.00^2 = 25.1\) m²
   • Angle at the end of the 8 m side: \(\arctan\frac{6}{8} = 36.87^\circ\); 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 \(180^\circ - 2\times 36.87^\circ = 106.3^\circ\)
   • Segment: \(\frac{106.3}{360}\times\pi\times 4.00^2 - \frac12\times 4.00^2\sin 106.3^\circ = 7.16\) 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 \(\frac{\theta}{360}\times\pi\times 8^2\) with \(\theta=\arcsin\frac{6}{8} = 48.59^\circ\) gives 27.1 m²
• \(\frac{15.9 + 27.1}{90}\times 100 = 47.8\% \approx 48\%\)