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.

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.

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, Radius of circular carpet (r) = 106 crn
Radius of circular hall (R) = 120 cm
Area of the hall uncovered by the carpet
\(=\pi\left(R^{2}-r^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}\left(120^{2}-106^{2}\right) \)
\(=\frac{22}{7}(120 \times 120-106 \times 106) \)
\(=\frac{22}{7} \times(14400-11236) \)
\(=\frac{22}{7} \times 3164=9944 \mathrm{~cm}^{2} \)

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.

91mm
Radius r = 91mm
Circumference C = 2πr
 
Circumference = 572 mm

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

The total area of the garden
Total area of the garden = Area of rectangle + Area of 2 semicircle
= Area of rectangle + Area of the circle
Here, the area of the rectangle = l x b sq. units
= 16 x 8
= 128 m² ...(1)
Area of the circle = πr² sq.units
= 3.14 x 4 x 4
= 3.14 x 16
= 50.24 m² ...(2)
From (1) and (2), the total area of garden = 128 + 50.24
= 178.24 m²

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.