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 = 9 m, b = 7 m
Area of the Room = l x b
= 9 x 7
= 63 m²
L = l + 2w = 8 + 2(3) = 8 + 6 = 14 m
B = b + 2w = 5 + 2(3) = 5 + 6 = 11m
Area of the room including verandah = L × B
= 14 x 11
= 154 m²
The area of the verandah = Area of the room including verandah − Area of the room
= 154 − 63
= 91m²
The cost of cementing the floor for 1sq.m = Rs.15
Therefore, the cost of cementing the floor of the verandah = 91 x 15 = Rs.1365.

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 inner circle (cricket ground), r = 76 m
A drainage is constructed around the cricket ground.
Therefore, the radius of the outer circle, R = 76 + 2 = 78 m
We have, area of the circular path = π(R² - r²) sq. units 
=\(\frac { 22 }{ 7 } \) x (78² - 76²)
= \(\frac { 22 }{ 7 } \) x (6084 - 5776)
= \(\frac { 22 }{ 7 } \)x 308
= 22 x 44 = 968 m²
Given, the cost of constructing the drainage per sq.m is Rs.180.
Therefore, the cost of constructing the drainage = 968 x 180 = Rs.1,74,240.

The radius of the inner circle r = 21m
The path is around the inner circle.
Therefore, the radius of the outer circle, R = 21 + 14 = 35 m
The area of the circular path = π(R² − r² ) sq.units
= \(\frac { 22 }{ 7 } \)x(35² - 21²)
= \(\frac { 22 }{ 7 } \) x (35 x 35) - (21 x 21)
= \(\frac { 22 }{ 7 } \) x (1225 - 441)
= \(\frac { 22 }{ 7 } \) x 784
= 22 x 112 = 2464 m²

The radius of the outer circle, R = 10 m
The radius of the inner circle, r = 3 m
The area of the circular path =Area of outer circle −Area of inner circle
= πR² − πr²
= π(R² − r² ) sq.units
= \(\frac { 22 }{ 7 } \) x (10² - 3²)
= \(\frac { 22 }{ 7 } \) x (10 x 10)-(3 x 3))
= \(\frac { 22 }{ 7 } \)x (100 - 9)
= \(\frac { 22 }{ 7 } \) x 91
= 286 m²

To find the perimeter of the quadrant disc, we need to find the circumference of quadrant shape.
Given that radius (r) = 14 cm.
We know that the circumference of circle = 2πr units.
So, the circumference of the quadrant arc = \(\frac { 1 }{ 4 } \) x 2πr
 =\(\frac { \pi r }{ 2 } \)
 =\(\frac { 22 }{ 7 } \times \frac { 14 }{ 2 } \)
 = 22 cm
Given, the radius of the circle = 14 cm
Thus, perimeter of required quadrant shaped disc = 14 +14 + 22
 = 50 cm.

In this shape, we have to calculate the circumference of semicircle on each side of the rectangle. The outer boundary of this figure is made up of semicircles of two different sizes. Diameters of each of the semicircles are 7 cm and 14 cm. We know that the circumference of the circle = πd units.
Circumference of the semicircular part = \(\frac { 1 }{ 2 } \)πd units
Hence, the circumference of the semicircle having diameter 7 cm is,
= \(\frac { 1 }{ 2 } \) x \(\frac { 22 }{ 7 } \) x 7 = 11 cm
Circumference of the pair of semicircular parts (II and IV) = 2 x 11 = 22 cm
Similarly, circumference of the semicircle having diameter 14 cm is,
= \(\frac { 1 }{ 2 } \) x \(\frac { 22 }{ 7 } \) x 14 = 22 cm
Circumference of the pair of semicircular parts (I and III) = 2 x 22 = 44 cm.
Perimeter of the given shape = 22 + 44 = 66 cm.

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.

The distance covered in one rotation
=  the circumference of the circle
= 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 77
= 2 x 22 x 11
= 484 cm
The distance covered in one rotation = 484 cm
The distance covered in 35 rotations = 484 x 35
= 16940 cm

Here the distance travelled by the tip of the seconds hand of a clock in 1 minute is the circumference of the circle and the length of the seconds hand is the radius of the circle. So, r = 56 mm
Circumference of the circle, C = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 56
= 2 x 22 x 8
= 352 mm
Therefore, distance travelled by the tip of the seconds hand of a clock in 1 minute is 352 mm.

Circumference of the circle, C = 2πr units
The circumference of the given circle = 132 m
\(\frac { C }{ 2\pi } =r\)
r = \(\frac { 132 }{ 2\times \frac { 22 }{ 7 } } \)
= \(\frac { 132 }{ 2 } \times \frac { 7 }{ 22 } \)
= 21 m
d = 2r
= 2 x 21
= 42 m

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

Given, d = 6 cm, d = 2r = 6 cm, r = 3 cm
Circumference of a circle = 2πr units
 = 2π × 3
 = 18.84
The circumference is 18.84 cm.