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?
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.
- \(x_m\) 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.
- 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.
- 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 \[ \frac{441 \times 3.14}{2} = 693.18 \text{ sq. m} \]
- 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: \[ 2 \times 38.465 = 76.93 \, \text{sq. m} \]
- 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.
Important Questions from Mensuration
The areas of three adjacent faces of a cuboidal tank are 3 m 2, 12 m 2 and 16 m 2. the capacity of the tank, in litres, is:
Volume of a cuboid is 4800 cm 3. If the height of this cuboid is 20 cm, then what will be the area of the base of cuboid ?
Two similar cubes have heights of 8 cm and 12 cm, respectively. If the capacity of the smaller cube is 80 cm 3, what is the capacity of the bigger cube (in cm 3)?
Three circles of radius 7 cm are kept touching each other. The string is tightly tied around these three circles. What is the length of the string?
Three circles of radius 6 cm are kept touching each other. The string is tightly tied around these three circles. What is the length of the string?
