A horse is tied to a peg at one corner of a square shaped grass field of side 15 m15\text{ m} by means of a 5 m5\text{ m} long rope (see the figure). Find (i) the area of the part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m10\text{ m} long instead of 5 m5\text{ m}. [take π=3.14\pi = 3.14] **Diagram Details:** The figure shows a square grass field with a horse tethered to a peg at the bottom-left corner. A circular arc representing the boundary of the grazing area with a radius of 5 m5\text{ m} forms a quadrant of a circle inside the square corner. Another dashed circular arc with a larger radius of 10 m10\text{ m} is also shown originating from the same corner.

Answer: (i) 19.625 m219.625\text{ m}^2, (ii) 58.875 m258.875\text{ m}^2

Step-by-step solution

Step 1: Identify Sector Parameters for Initial Grazing Area

Since the horse is tethered at the corner of a square field, the angle of the sector is the vertex angle of the square, which is θ=90∘\theta = 90^\circ. The rope length forms the radius of the circular sector, so r1=5 mr_1 = 5\text{ m}.

Step 2: Calculate the Initial Grazing Area

Using the formula for the area of a sector θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2, we substitute θ=90∘\theta = 90^\circ, r1=5 mr_1 = 5\text{ m}, and π=3.14\pi = 3.14. Simplifying 90360=14\frac{90}{360} = \frac{1}{4} gives an area of 19.625 m219.625\text{ m}^2.

Step 3: Calculate the Grazing Area with a 10 m Rope

When the rope length increases to r2=10 mr_2 = 10\text{ m}, the horse can graze over a larger quadrant of radius 10 m10\text{ m}. Substituting into the formula gives 14×3.14×100=78.5 m2\frac{1}{4} \times 3.14 \times 100 = 78.5\text{ m}^2.

Step 4: Calculate the Increase in Grazing Area

Subtract the initial grazing area from the new grazing area: 78.5−19.625=58.875 m278.5 - 19.625 = 58.875\text{ m}^2.

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