The perimeter of the rectangular garden
Perimeter include two lengths each of 16 m and two semi circular arcs of diameter 8 m.
Circumference of the semicircle = \(\frac { \pi d }{ 2 } \) units
= \(\frac { \pi \times 8 }{ 2 } \) = 4π
= 4 x 3.14
= 12.56 m
Therefore, circumference of two semicircles = 2 x 12.56
= 25.12 m
Total perimeter = length + length + circumference of two semicircles 
= 16 +16 + 25.12
= 32 + 25.12
= 57.12 m

Given that, the area of the circle = 9856 sq.m
  πr² = 9856
   \(\frac { 22 }{ 7 } \)x r² = 9856
  r² = 9856 x \(\frac { 7 }{ 22 } \)
  r² = 448 x 7 = 3136
  
  r² = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7
  r² = 8 x 8 x 7 x 7 = 8² x 7² = (8 x 7)²
  r = 8 x 7 = 56 m
Therefore the length of the rope should be 56 m.

The distance covered by the man is nothing but the circumference of the circle.
Given that the distance covered = 154 m
Therefore circumference of the circle = 154 m
That is, 2πr = 154
2 x \(\frac { 22 }{ 7 } \)x r = 154
r = 154 x \(\frac { 7 }{ 44 } \)
r = 3.5 x 7
= 24.5
Area of the park = πr² sq.units
= \(\frac { 22 }{ 7 } \)× 24.5 x 24.5
= 22 x 3.5 x 24.5
= 1886.5 m²
Cost of levelling the park per sq.m = Rs.25.
Cost of levelling the park of 1886.5 m² = 1886.5 x 25 = Rs.47,162.50

Given that the area of the circular region = 2464 cm ²
πr² = 2464
\(\frac { 22 }{ 7 } \)x r² = 2464
 r² = 2464 x  \(\frac { 7 }{ 22 } \)
  
 r² = 112 × 7 = 784
 r² = 2 x 2 x 2 x 2 x 7 x 7
 = 4 x 4 x 7 x 7
 = 4² x 7²
 r² = (4 x 7)² [r x r = (4 x 7) x (4 x 7)]
 r = 4 x 7
 = 28 cm
Diameter (d) = 2 x r = 2 x 28 = 56 cm.

Given the diameter (d) = 28 cm
 Radius (r) = \(\frac { 28 }{ 2 } \) = 14 cm
Area of a circle = πr² sq.units
 = \(\frac { 22 }{ 7 } \) x 14 x 14
So, the area of the circle = 616 cm².