Outer rectangle
L= 30 cm,B = 20 cm
Inner rectangle
l = 30 - 2 x 4 = 22 cm
.b = 20 - 2 x 4 = 12 cm
Area of removed portion = 22 x 12 = 264 cm²
Area of remaining cardboard = Area of outer rectangle - Area of inner rectangle
= LB - lb
= 30 x 20 - 22 x 12
= 600 - 264 = 336 cm²

R =.52 m, r = 45 m
Area of the grass land that the goat cannot graze \(=\pi\left(R^{2}-r^{2}\right) \)
\(=\pi\left(52^{2}-45^{2}\right) \)
\(=\pi(2704-2025) \)
\(=\frac{22}{7}(679) \)
= 22 x 97 = 2134 m².

Area of the path = Area of outer rectangle - Area of inner rectangle
= LB - lb
= 124 x 92 - 120 x 90
= 11408 - 10800 = 608 m²

Distance covered by a wheel in one rotation = circumference of a wheel circles
Distance covered by a
wheel in 20 rotations = 3520 cm
20 x 2r = 3520
\(20 \times 2 \times \frac{22}{7} \times r =3520 \)
\(r =3520 \times \frac{1}{20} \times \frac{1}{2} \times \frac{7}{22} \)
Radius of the wheel r = 28 cm

Area of the path = Area of outer rectangle - Area of inner rectangle
= LB - lb
= 11 x 8 - 7 x 4
= 88 - 28 = 60 m²

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

Given diameter of flower garden = 49 m
d = 49 m
\(2 r=49 \Rightarrow r=\frac{49}{2}\)
Area of the circular flower garden = \(\pi r^{2}\) sq. units
\(=\frac{22}{7} \times \frac{49}{2} \times \frac{49}{2} \)
\(=\frac{3773}{2} m^{2} \)
The cost of levelling the flower garden per sq.m = Rs. 150
The cost of levelling the flower garden of \(\frac{3773}{2} sq.m \)
\(=Rs. 150 \times \frac{3773}{2}=Rs. 282975 \)

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

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

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