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.

Area that the Bull can graze = 2464 m²
\(\pi r^{2} =2464 \)
\(\frac{22}{7} \times r^{2} =2464 \)
\(r^{2} =2464 \times \frac{7}{22} \)
= 112 x 7 
r² = 4 x 4 x 7 x 7
Length of the rope r = 4 x 7 = 28 m

Given length of rope = radius = 4.9 m
Area that the sheep can graze = \(\pi r^{2}\)
\(=\frac{22}{7} \times 4.9 \times 4.9\)
= 75.46 m²

Given that, circumference of a circular park = 352 m
= 352
\(2 \times \frac{22}{7} \times r =352 \)
\(r =352 \times \frac{1}{2} \times \frac{7}{22}=56 \mathrm{~m} \)
Area of circular park = \(\pi r^{2}\) sq.units
\(=\frac{22}{7} \times 56 \times 56\)
= 22 x 8 x 56 = 9856 m²

Area of watered space = 1386
\(\pi r^{2}\)  = 1386
\(\frac{22}{7} \times r^{2} =1386 \)
\(r^{2} =1386 \times \frac{7}{22}\)
r² = 63 x 7 = 3 x 3 x 7 x 7
r² = 3² x 7²
Radius r = 3 x 7 = 21 cm
Diameter d = 2r = 2 x 21
d = 42 cm

radius r = 2.135 m
Area of the shotput circle = \(\pi r^{2}\) sq. units
\(=\frac{22}{7} \times 2.135 \times 2.135=14.32 \mathrm{~m}^{2}\)

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

Length of wire = 1320 cm
Circumference of a circular frame - 2r units
\(=2 \times \frac{22}{7} \times 7=44 \mathrm{~cm}\)
Number of circular
\(\text { frames }= \frac{\text { Length of wire }}{\text { Circumference of a circular frame }} \)
\(=\frac{1320 \mathrm{~cm}}{44 \mathrm{~cm}} \)
= 30 frames
30 circular frames can be made.

d = 350 m
One revolution = Perimeter of ground
\(=\pi d=\frac{22}{7} \times 350=1100 \mathrm{~m}\)
Distance covered by the Athlete = 4 revoltitions
= 4 x 1100 m = 4400 m

d = 1.4 m
Distance covered in 1 rotation = Circumference of wheel
\(=\pi d=\frac{22}{7} \times 1.4\)
Distance covered 1 rotation = 4.4 m
Distance covered by 150 rotations = 4 x 150 = 660 m

d = 4,.2 m
Circumference of well (C) = d
\(=\frac{22}{7} \times 4.2\)
Circumference of circular well = 13.2 m

49 cm
Radius r = 49 cm
Circumference C = 2πr

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

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

d = 70 cm
Circumference C = πd \(=\frac{22}{7} \times 70=220 \mathrm{~cm}\)
 

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.

Radius (r) = 21 cm
Area of a circle = πr² sq. units
 = 3.14 x 21 x 21
 = 1384.74
Area of the circle = 1384.74 cm²