AI HL · 12.10
0 of 79 skills done
Grazing goat problems: sector and triangle area
12.10.15
Screenshot and video to come
Screenshot and video to come
A goat is tied to a 12 m by 8 m shed, seen from above. Change the length of the rope. The grass it can reach is made of sectors, and the area is their sum.
On the long wall, drag the hook X along the wall, or click it and use the arrow keys. Angles are in radians.
Move the rope slider to change the grass the goat can reach.
Set the radius with the slider and drag the point to set the angle at the centre. The arc length and the area are worked out with the numbers.
Drag the point to change the angle.
A field is a right-angled triangle with sides 9 m, 12 m and 15 m. A goat, a sheep and a pony are each tied to a different corner by a 4 m rope. The ropes are too short to meet, and the animals cannot leave the field. Which animal grazes the largest area? Start by finding the area the animal at the right angle can graze.
Have a go first. Then open the working one line at a time.
\(\text{right angle} = \frac{\pi}{2},\quad \text{other angles: } \arcsin\frac{9}{15} = 0.644,\ \ \frac{\pi}{2} - 0.644 = 0.927\)
Work in radians. The angles of a triangle add to π, so the three angles are π/2, 0.644 and 0.927.
\(A = \frac12 r^2\theta = 8\theta\)
All three ropes are 4 m, so \(\frac12\times 4^2 = 8\) is the same each time.
\(\frac{\pi}{2}: 12.6\text{ m}^2,\quad 0.644: 5.15\text{ m}^2,\quad 0.927: 7.42\text{ m}^2\)
The area depends only on the angle, because the rope is the same.
The animal at the right angle grazes the most.
The largest angle gives the largest sector. A right angle is bigger than either of the other two.
In the same field (9 m, 12 m, 15 m) all three animals graze at once. What percentage of the field do they graze altogether?
Have a go first. Then open the working one line at a time.
\(\frac12 r^2\left(\frac{\pi}{2} + 0.644 + 0.927\right) = 8\times\pi = 25.1\text{ m}^2\)
The three angles add to π, so the three sectors together make a semicircle of radius 4: \(\frac12\pi\times 4^2 = 8\pi\).
\(\frac12\times 9\times 12 = 54\text{ m}^2\)
The two short sides are the base and the height.
\(\frac{25.1}{54}\times 100 = 46.5\%\)
Divide the grazed area by the area of the field.
A field is a right-angled triangle with sides 5 m, 12 m and 13 m. Horace the goat is tied to the midpoint M of the hypotenuse by a rope just long enough to reach both ends of the hypotenuse. Two sheep are tied to the midpoints of the other two sides, each on a rope just long enough to reach both ends of that side. The animals cannot leave the field. Find the percentage of the field the goat can graze.
Have a go first. Then open the working one line at a time.
\(\frac{13}{2} = 6.5\text{ m}\)
The rope reaches both ends of the hypotenuse, so it is half of 13.
\(MA = MB = MC = 6.5\)
The angle in a semicircle is a right angle, so the right-angled corner lies on the circle with the hypotenuse as diameter. The midpoint of the hypotenuse is the centre of that circle.
Goat: 100% of the field = 30.0 m²
All three corners are 6.5 m from M. The whole triangle is inside the circle, so the goat grazes the whole field.
\(12\text{ m side: }56.5 - 29.6 = 27.0\text{ m}^2,\quad 5\text{ m side: }9.57\text{ m}^2\)
Each sheep's semicircle is cut by the hypotenuse, so each grazes less than the whole field. The goat grazes the largest area.
A field is a rectangle 9 m by 3 m. A goat is tied by a 4 m rope to a post at one corner. Find the percentage of the field the goat can graze, to the nearest integer.
Have a go first. Then open the working one line at a time.
\(\sqrt{4^2 - 3^2} = \sqrt{7} = 2.65\text{ m}\)
The right-angled triangle has hypotenuse 4 (the rope) and height 3 (the width of the field).
\(\frac12\times 3\times\sqrt7 = 3.97\text{ m}^2\)
This is the part of the field next to the short side, up to the end of the rope.
\(\theta = \arcsin\frac34 = 0.848,\quad \frac12\times 4^2\times 0.848 = 6.78\text{ m}^2\)
Put the calculator in radian mode. The sector lies along the long side, between the long side and the line to where the rope meets the far side.
\(\frac{3.97 + 6.78}{27}\times 100 = 39.8\% \approx 40\%\)
Field area \(9\times 3 = 27\) m².
The goat's rope is replaced by one of length a metres, with \(3 < a < 9\), so that the goat can graze exactly half of the 9 m by 3 m field. Show that \(a^2\arcsin\frac{3}{a} + 3\sqrt{a^2 - 9} = 27\). Before you do, find the value of the left-hand side when a = 4.
Have a go first. Then open the working one line at a time.
\(\frac12\times 3\times\sqrt{a^2 - 9}\)
The rope of length a reaches the far side where the horizontal distance is \(\sqrt{a^2-9}\).
\(\theta = \arcsin\frac{3}{a},\quad \frac12 a^2\arcsin\frac{3}{a}\)
Same angle as in the last example, now with the unknown rope a.
\(\frac32\sqrt{a^2-9} + \frac12 a^2\arcsin\frac3a = \frac12\times 27\)
The field has area \(9\times 3 = 27\). The grazed area is one half of it.
\(3\sqrt{a^2-9} + a^2\arcsin\frac{3}{a} = 27\)
Multiply every term by 2. This is the result to show: write each line.
Find the value of a in the last example.
Have a go first. Then open the working one line at a time.
Check the calculator is in radian mode.
The arcsin is an angle in radians.
From the calculator: \(a^2\arcsin\frac3a + 3\sqrt{a^2-9} = 27\) on 3 < a < 9 gives a = 4.83
Use Solve with a starting guess in the range, or graph both sides and find the intersection.
3 < 4.83 < 9
The rope is longer than the width of the field and shorter than its length.
A fenced triangular pond CDE stands in a large field. A goat is tied by a 7 m rope at a point A on the outside of the fence CD. A is 4.5 m from the corner C, and CD is 12 m long. The angle of the pond at C is θ radians. The goat can graze 85 m² of grass. Find θ.
Have a go first. Then open the working one line at a time.
\(\frac12\times\pi\times 7^2 = 77.0\text{ m}^2\)
The fence CD is straight at A, and the goat is on one side of it. D is 7.5 m from A, so the rope does not reach it.
\(\text{rope left} = 7 - 4.5 = 2.5\text{ m},\quad \text{angle} = \pi - \theta\)
The rope wraps round C and sweeps from the line DC to the fence CE. That is the angle outside the pond at C.
\(\frac12\pi\times 7^2 + \frac12\times 2.5^2(\pi - \theta) = 85\)
Semicircle plus sector equals the area grazed.
\(\pi - \theta = 2.57,\quad \theta = 0.572\) radians
Rearrange, or solve on the calculator. θ is between 0 and π, so the answer is possible.