From the dimensions of the ground we have,
L = 30 m; B = 19 m; l = 27 m; b = 16 m
Area of the kho kho ground = L × B
= 30 x 19
= 570 m²
Area of the play field = l x b
= 27 x 16
= 432 m²
Area of the lobby = Area of Kho-Kho ground − Area of the play field
= 570 − 432
= 148 m²

Here L = 23 cm B = 11 cm
Area of the chart = L x B
= 23 x 11
= 253 cm²
l = L − 2w = 23 − 2(3) = 23 − 6 = 17 cm
b = B − 2w = 11− 2(3) = 11− 6 = 5 cm
Area of the picture 17 x 5 = 85 cm²
Therefore, the area of the margin = 253 − 85
  = 168 cm².

Here, L = 10 m B = 8 m
Area of the floor = L x B
= 10 x 8
= 80 m²
Area of the carpet = l x b
= 7 x 5
= 35 m²
Therefore, the total area of the floor not covered by the carpet = 80 − 35
= 45 m²

The radius of the poultry farm is = 420 m
The length of the barbed wire for fencing the poultry farm is equal to the circumference of the circle.
We know that the circumference of the circle = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 420
= 2 x 22 x 60
The length of the barbed wire to fence the poultry farm = 2640 m
The cost of fencing the poultry farm at the rate of Rs.12 per metre = 2640 x 12
= Rs.31,680
Given that he has Rs.30,000 with him.
The excess amount required = Rs.31,680 − Rs.30,000 = Rs.1,680.

Radius of circular disc (r) = 14 cm
Circumference of the disc = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 14
= 88 cm

Given, Radius of lawn (r) = 28 m
Width of path = 7 m
Outer radius (R) = 28 + 7 = 35 m
Area of the path \(=\pi\left(R^{2}-r^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}\left(35^{2}-28^{2}\right) \)
\(=\frac{22}{7}(35 \times 35-28 \times 28) \)
\(=\frac{22}{7} \times(1225-784) \)
\(=\frac{22}{7} \times 441=1386 \mathrm{~m}^{2} \)

Diameter (d) = 105 cm
\(\mathrm{r}=\frac{105}{2} \mathrm{~cm}\)
Area of the dining table = \(\pi r^{2}\) sq. units
\(=\frac{22}{7} \times \frac{105}{2} \times \frac{105}{2}\)
\(=\frac{11 \times 15 \times 105}{2}\)
Area of the dining table = 8662.5 cm²

r = 63 m
Circumference of garden (length of fence) - 2r units
\(=2 \times \frac{22}{7} \times 63\)
= 396 m
The cost of fencing per 1 m = Rs 150
The cost of fencing 396 m = Rs 150 x 396
= Rs 59,400

d = 28 mm
Circumference C = πd \(=\frac{22}{7} \times 28=88 \mathrm{~mm}\)
 

d = 56 cm
Circumference C = πd \(=\frac{22}{7} \times 56=176 \mathrm{~m}\)

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