Area Grazed by Horses: Areas Related to Circles

This lesson solves two common questions on areas related to circles. First we find how much of a triangular field three tethered horses can graze, then the area remaining when four quarter circles are cut from the corners of a rectangle.

Grazed area from the three sectors

Three horses are tethered with $7\text{ m}$ ropes at the corners of a triangular field. At each corner the horse sweeps out a sector of radius $r = 7\text{ m}$, with the corner angles $\theta_1$, $\theta_2$, $\theta_3$.

The total grazed area is the sum of the three sectors:

$A = \frac{\theta_1}{360^\circ}\pi r^2 + \frac{\theta_2}{360^\circ}\pi r^2 + \frac{\theta_3}{360^\circ}\pi r^2 = \frac{\pi r^2}{360^\circ}\left(\theta_1 + \theta_2 + \theta_3\right)$

The angles are not given, but the three angles of any triangle add to $180^\circ$, so we never need them individually:

$A = \frac{\pi r^2}{360^\circ}\times 180^\circ = \frac{\pi r^2}{2} = \frac{1}{2}\times\frac{22}{7}\times 7^2 = 77\text{ m}^2$

So the grazed area is $77\text{ m}^2$.

Total field area by Heron's formula

The field is a triangle with sides $a = 20\text{ m}$, $b = 34\text{ m}$, $c = 42\text{ m}$. First the semi-perimeter:

$s = \frac{a + b + c}{2} = \frac{20 + 34 + 42}{2} = \frac{96}{2} = 48$

Then $s - a = 28$, $s - b = 14$, $s - c = 6$, and Heron's formula gives:

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{48 \times 28 \times 14 \times 6}$

Factorising under the root, $48 = 6\times 8$, $28 = 4\times 7$, $14 = 2\times 7$, $6 = 6$, which pairs up as $6\times 6$, $8\times 8$, $7\times 7$:

$\text{Area} = \sqrt{6^2 \times 8^2 \times 7^2} = 6 \times 8 \times 7 = 336\text{ m}^2$

Ungrazed area

The part the horses cannot reach is the whole field minus the grazed sectors:

$336 - 77 = 259\text{ m}^2$

Quadrants cut from a rectangle

A rectangular piece $ABCD$ has length $l = 20\text{ m}$ and breadth $b = 15\text{ m}$, so its area is:

$l \times b = 20 \times 15 = 300\text{ m}^2$

From each of the four corners a quadrant of radius $r = 3.5\text{ m}$ is removed. Every corner of a rectangle is $90^\circ$, so the four quadrants together make one full circle:

$4 \times \frac{90^\circ}{360^\circ}\,\pi r^2 = \pi r^2 = \frac{22}{7}\times 3.5 \times 3.5 = 38.5\text{ m}^2$

Remaining area

The area left after the cuts is the rectangle minus the four quadrants:

$300 - 38.5 = 261.5\text{ m}^2$

Key takeaways

• The three corner sectors of any triangle add up to a half circle, so the grazed area is $\tfrac{1}{2}\pi r^2$ no matter the angles.
• Heron's formula, $\sqrt{s(s-a)(s-b)(s-c)}$, gives a triangle's area straight from its three sides.
• Four corner quadrants of equal radius combine into one full circle, which makes the cut-out area easy to subtract.