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AI SL · 12.10

Geometry & Trigonometry

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12.10.11

Grazing goat problems: sectors round a shed

12.10.11

Grazing goat problems: sectors round a shed

Target: I can find the area of a sector with a reflex angle, such as a rope at the corner of two walls. (+2 more)

By the end of this section

Key points

Sectors and angles

  • Sector area: \(A = \frac{\theta}{360}\times\pi r^2\) with \(\theta\) in degrees.
  • Full circle \(\pi r^2\), semicircle \(\frac12\pi r^2\), quarter circle \(\frac14\pi r^2\), three-quarter circle \(\frac34\pi r^2\).
  • An animal tied outside a wall corner of angle \(\alpha\) can sweep the reflex angle \(360^\circ - \alpha\). At a shed corner \(\alpha = 90^\circ\), so the sweep is \(270^\circ\).

Rope round a corner

  • Draw the shed and mark the hook first. Then add each arc: its centre, its radius and its angle.
  • When the rope reaches a corner it wraps round. The rope left is the rope length minus the length of wall it used.
  • Each wrap gives a new quarter circle, centred on that corner, with the rope left as its radius.
  • If the rope left is 0 or negative, there is no new sector. If it is longer than the next wall, the rope wraps again.
  • Check the arcs do not overlap. At a corner of a \(a\) by \(b\) shed, a rope \(L\) with \(L < a + b\) gives no overlap.

The hook's position x

  • Call the distance of the hook from the corner x. The two rope lengths left are written with x.
  • Write the radius of each sector in x, then add the areas. Simplify to one expression A(x).
  • State the domain: every radius is between 0 and the length of the wall it sweeps along. Outside it the formula is wrong.
  • Use the calculator to solve A(x) = a given area, and to find the maximum and minimum on the domain. A U-shaped graph has its maximum at an end of the domain.

On the calculator

Solving A(x) = a given area

  • Type the area function as f1(x).
  • Use Solve with the given area, or graph y = A(x) and y = the area and find the intersections.
  • Check each answer is inside the domain. A quadratic can give two positions.

Screenshot and video to come

Maximum and minimum on an interval

  • Graph A(x) with the window set to the domain.
  • Find the minimum at the lowest point of the graph.
  • Compare the values at the two ends of the domain for the maximum.

Screenshot and video to come

Explore: the grazing goat

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.

Grass the goat can graze

On the long wall, drag the hook X along the wall, or click it and use the arrow keys. Angles are in degrees.

Move the rope slider to change the grass the goat can reach.

  1. Slide the rope from 6 m to 14 m with the post at a corner. At which lengths does a new quarter circle appear? Why those two?
  2. Why is the first sector three quarters of a circle and not a whole circle?
  3. When the rope wraps round a corner, why is the new radius the rope minus the wall, and not the whole rope?
  4. Tie the rope on the long wall with a rope of 9 m. Where should the hook go for the most grass? For the least? Compare with the area function in the lesson.

Explore: arc length and sector area

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.



Fraction of the circle
Arc length, s
Sector area, A
Perimeter of the sector

Drag the point to change the angle.

  1. Double the radius and keep the angle. What happens to the arc length? What happens to the area? Why are they different?
  2. Set the angle to 180°. What fraction of the circle is the sector? Check the area against half of πr².
  3. Tick the segment box. Why is the segment area the sector area minus the triangle area?
  4. Find an angle and a radius that give a sector with an area of exactly 10π cm². Is there more than one pair?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A rope at the corner of a shed

A goat is tied by a 15 m rope to the corner X of a shed. The shed is a rectangle 12 m by 8 m. The goat can graze all the open ground it can reach. Find the area of grass the goat can graze.

12 m8 m15 mX
Working
  1. Three-quarter circle

    \(\frac{270}{360} \times \pi \times 15^2 = \frac34 \times \pi \times 15^2 = 168.75\pi\)

    The shed blocks a right angle at X, so the goat sweeps 360° − 90° = 270° with the full 15 m.

  2. Round the 12 m side

    \(\text{rope left} = 15 - 12 = 3\text{ m},\quad \frac14 \pi \times 3^2 = 2.25\pi\)

    The rope runs along the 12 m wall and wraps at the far corner. A quarter circle of radius 3.

  3. Round the 8 m side

    \(\text{rope left} = 15 - 8 = 7\text{ m},\quad \frac14 \pi \times 7^2 = 12.25\pi\)

    Same idea along the other wall. Check: 3 + 7 < 12 + 8, so the two quarter circles do not overlap.

  4. Add

    \(A = 168.75\pi + 2.25\pi + 12.25\pi = 183.25\pi = 576\text{ m}^2\)

    Three sectors, one total.

Video walkthrough coming
A rope hooked on the side of a shed

A shed is a rectangle 12 m by 9 m. A goat is tied by a 10 m rope to a hook X on a 12 m wall, 3 m from the left-hand corner. Find the area of grass the goat can graze.

12 m9 m10 mX
Working
  1. Semicircle

    \(\frac12 \times \pi \times 10^2 = 50\pi\)

    The wall is straight at X, so the goat sweeps a half turn with the full rope.

  2. Left corner

    \(\text{rope left} = 10 - 3 = 7\text{ m},\quad \frac14\pi\times 7^2 = 12.25\pi\)

    X is 3 m from the left corner. 7 m of rope are left to swing along the 9 m side wall.

  3. Right corner

    \(\text{rope left} = 10 - (12-3) = 1\text{ m},\quad \frac14\pi\times 1^2 = 0.25\pi\)

    X is 12 − 3 = 9 m from the right corner.

  4. Add

    \(A = 50\pi + 12.25\pi + 0.25\pi = 62.5\pi = 196.3\text{ m}^2\)

    Both left-over lengths are shorter than the 9 m side wall, so the rope does not wrap again.

Video walkthrough coming