Horse Grazing in a Square Field: Area of Sector Problem | Class 10 Maths

Quick answer: A horse tied at the corner of a square field grazes a quadrant (a $90^\circ$ sector), so its grazing area is $\frac{1}{4}\pi r^2$. With a $5\text{ m}$ rope the area is $19.625\text{ m}^2$. Lengthening the rope to $10\text{ m}$ increases the area by $58.875\text{ m}^2$ (using $\pi = 3.14$).

Many geometry word problems hide a simple shape inside a real-life situation. When a horse, cow or goat is tied by a rope to a peg, it can graze in every direction up to the full length of the rope, so the edge of its grazing region is an arc of a circle.

If the peg is at the corner of a square field, the two fences meeting at that corner block the animal on two sides. Its grazing region becomes a $90^\circ$ sector of a circle, called a quadrant.

In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles), the horse grazing in a square field problem is a well-known NCERT question and a common board exam question.

In this guide, we explain why the grazing region is a quadrant, solve the NCERT problem step by step, and cover the common variations.

What You Will Learn

• How to turn a tethered-animal problem into an area of a sector problem
• Why the corner of a square field makes the sector angle $\theta = 90^\circ$
• NCERT Part (i): grazing area with a $5\text{ m}$ rope ($19.625\text{ m}^2$)
• NCERT Part (ii): increase in grazing area with a $10\text{ m}$ rope ($58.875\text{ m}^2$)
• A shortcut for the increase: $\Delta A = \frac{1}{4}\pi\left(r_2^2 - r_1^2\right)$
• Variations: triangular, rectangular and other fields, including three animals at the corners of a triangle
• Common mistakes to avoid

1. Why the Grazing Region Is a Quadrant

Consider a square field $ABCD$ of side $15\text{ m}$. A horse is tied to a peg at corner $A$ with a rope of length $r$.

  D ┌─────────────────────────┐ C
    │                         │
    │      SQUARE FIELD       │
    │      side = 15 m        │
    │                         │
    │ * *                     │
    │     *   grazing         │
    │ r    *  quadrant        │
    │       *  (θ = 90°)      │
  A ●───────*─────────────────┘ B
    peg     r = 5 m

Key Observations

1. The boundary is an arc. The horse can move anywhere within the rope's length, so the edge of its grazing region is an arc of a circle of radius $r$ centred at peg $A$.

2. The sector angle is $90^\circ$. Every angle of a square is a right angle, so the two fences at $A$ limit the horse to an angle of $\theta = 90^\circ$.

3. The region is a quarter circle.

$\text{Fraction of circle} = \frac{\theta}{360^\circ} = \frac{90^\circ}{360^\circ} = \frac{1}{4}$

$\text{Grazing area} = \frac{1}{4}\pi r^2$

4. Check the rope against the side. Both rope lengths ($5\text{ m}$ and $10\text{ m}$) are less than the side of the field ($15\text{ m}$). So the arc never reaches the other fences, and the grazing region is a complete quadrant. The $15\text{ m}$ side is only given to confirm this.

2. NCERT Problem: Horse Tied in a Square Field

Question

A horse is tied to a peg at one corner of a square-shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ long rope. Find:

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were $10\text{ m}$ long instead of $5\text{ m}$. (Use $\pi = 3.14$)

Part (i): Grazing Area with a 5 m Rope

Sector angle $\theta = 90^\circ$ (corner of a square) and radius $r_1 = 5\text{ m}$.

$\text{Area}_1 = \frac{\theta}{360^\circ} \times \pi r_1^2 = \frac{90^\circ}{360^\circ} \times \pi r_1^2 = \frac{1}{4}\pi r_1^2$

Substituting $\pi = 3.14$ and $r_1 = 5$:

$\text{Area}_1 = \frac{1}{4} \times 3.14 \times 5^2 = \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = 19.625\text{ m}^2$

Answer (i): The horse can graze an area of $19.625\text{ m}^2$.

Part (ii): Increase in Grazing Area with a 10 m Rope

With the longer rope, $r_2 = 10\text{ m}$:

$\text{Area}_2 = \frac{1}{4}\pi r_2^2 = \frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = 78.5\text{ m}^2$

The increase in grazing area is:

$\text{Increase} = \text{Area}_2 - \text{Area}_1 = 78.5 - 19.625 = 58.875\text{ m}^2$

Answer (ii): The grazing area increases by $58.875\text{ m}^2$.

Shortcut: Find the Increase Directly

Instead of finding both areas and subtracting, take $\frac{1}{4}\pi$ common:

$\Delta A = \frac{1}{4}\pi\left(r_2^2 - r_1^2\right) = \frac{1}{4} \times 3.14 \times \left(10^2 - 5^2\right) = \frac{1}{4} \times 3.14 \times 75 = \frac{235.5}{4} = 58.875\text{ m}^2$

This matches the step-by-step answer and is a quick way to check your work.

3. Variations: Animals Tied in Other Fields

The method is always the same: find the angle at the corner where the animal is tied, then use $\text{Area} = \frac{\theta}{360^\circ}\pi r^2$ (as long as the rope is shorter than the sides meeting at that corner).

Three Animals at the Three Corners of a Triangle

Suppose three cows are tied, each with a rope of length $r$, at the three corners of a triangular field, and the ropes are short enough that the grazing regions do not overlap.

The angles of any triangle add up to $180^\circ$:

$\theta_1 + \theta_2 + \theta_3 = 180^\circ$

So the total area grazed by the three cows is:

$\text{Total area} = \left(\frac{\theta_1}{360^\circ} + \frac{\theta_2}{360^\circ} + \frac{\theta_3}{360^\circ}\right)\pi r^2 = \frac{180^\circ}{360^\circ}\pi r^2 = \frac{1}{2}\pi r^2$

Key result: The three corner sectors together always make exactly one semicircle, $\frac{1}{2}\pi r^2$, whatever the shape of the triangle. By the same logic, four animals at the corners of any quadrilateral graze a total of one full circle, $\pi r^2$, because the angles of a quadrilateral add up to $360^\circ$.

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: Part (ii) asks for the increase in grazing area, not the new area. Writing $78.5\text{ m}^2$ as the final answer instead of $58.875\text{ m}^2$ will cost you marks.

Common Mistake: Do not subtract the rope lengths first. $\frac{1}{4}\pi(10 - 5)^2 = 19.625\text{ m}^2$ is wrong. Subtract the squares of the radii instead: $r_2^2 - r_1^2 = 100 - 25 = 75$, not $(10 - 5)^2 = 25$.

Common Mistake: Always check which value of $\pi$ the question asks for. Using $\frac{22}{7}$ when the question says $\pi = 3.14$ gives a slightly different answer.

Frequently Asked Questions

What shape does a horse graze when tied at the corner of a square field?

It grazes a quadrant, which is a sector of a circle with angle $90^\circ$, because the corner of a square is a right angle. Its area is $\frac{1}{4}\pi r^2$, where $r$ is the length of the rope.

What is the grazing area of a horse tied with a 5 m rope at the corner of a square field?

Using $\pi = 3.14$, the area is $\frac{1}{4} \times 3.14 \times 5^2 = 19.625\text{ m}^2$.

What is the increase in grazing area if the rope is 10 m instead of 5 m?

The new area is $78.5\text{ m}^2$, so the increase is $78.5 - 19.625 = 58.875\text{ m}^2$.

What is the area of the field the horse cannot graze with a 5 m rope?

The field's area is $15^2 = 225\text{ m}^2$, so the ungrazed area is $225 - 19.625 = 205.375\text{ m}^2$.

Why is the 15 m side of the field given in the question?

It confirms that both ropes ($5\text{ m}$ and $10\text{ m}$) are shorter than the side, so the grazing region is a complete quadrant that does not cross the other fences.