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Question

On a large ground, there is a straight tall vertical wall of length $28$ m. A goat is tied to a point on the ground which is at the middle of the wall, using a rope. If the length of the rope is $21$ m, what is the area of the region (in sq. m) around the wall that the goat can access?

The correct answer is
$847$

To determine the area the goat can access, we need to understand the setup. The goat is tied to a point on the ground which is at the middle of a vertical wall of length \(28\) m, and the rope length is \(21\) m.

Since the wall is vertical, the goat has access to the area on either side of the midpoint of the wall, as long as it does not exceed the reach of the rope.

  1. \(x_m\) and extend the length of the rope on the ground. Since the goat is tied at the midpoint of the wall, it can be \(0.5 \times 28 = 14\) m from either end. Thus, the goat can access the ground on both sides of the wall.
  2. When the goat is lined parallel with the wall, it can form a semi-circle on the ground with a radius of \(21\) m, which is the length of the rope. The wall acts as the straight-line boundary in this semicircle.
  3. The area of this semicircular accessible region is given by the formula for the area of a semicircle: \[ \text{Area} = \frac{1}{2} \pi r^2 \] where \( r = 21 \) m. Simplifying, we get: \[ \text{Area} = \frac{1}{2} \times \pi \times (21)^2 = \frac{1}{2} \times \pi \times 441 \] \[ \text{Area} = \frac{441 \pi}{2} \] It's approximately \[ \frac{441 \times 3.14}{2} = 693.18 \text{ sq. m} \]
  4. Since the goat can also reach out of the enclosed semicircle at either end of the wall by \(7\) m (because \(21 - 14 = 7\)), forming two quarter circles at each end, the area for one quarter circle is given by: \[ \frac{1}{4} \pi r^2 \] Using the remaining rope length \(r = 7\) m : \[ \text{Area of one quarter circle} = \frac{1}{4} \times \pi \times (7)^2 = \frac{49 \pi}{4} \] Simplified, the area becomes: \[ \text{Area of one quarter circle} = \frac{49 \times 3.14}{4} = 38.465 \text{ sq. m} \] Adding this like area for the two ends: \[ 2 \times 38.465 = 76.93 \, \text{sq. m} \]
  5. To find the total area accessible by the goat, sum the semicircle area and the two quarter-circle areas: \[ \text{Total area} = 693.18 + 76.93 = 847 \, \text{sq. m} \]

Thus, the total area accessible by the goat is \(847\) sq. m.

The correct option is 847.

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Important Questions from Mensuration

  1. A solid cube is painted yellow, blue and black such that opposite faces are of same colour. The cube is then cut into 36 cubes of two different sizes such that 32 cubes are small and the other four cubes are Big. None of the faces of the bigger cubes is painted blue. How many cubes have only one face painted?

  2. A and B are two heavy steel blocks. If B is placed on the top of A, the weight increases by 60%. How much weight will reduce with respect to the total weight of A and B, if B is removed from the top of A?

  3. A gardener increased the area of his rectangular garden by increasing its length by 40% and decreasing its width by 20%. The area of the new garden

  4. A village having a population of 4000 requires 150 liters of water per head per day. It has a tank measuring 20 m x 15 m x 6 m. The water of this tank will last for

  5. The centroid of an equilateral triangle ABC is G. If AB is 6 cms, the length of AG is

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