The given field has four triangles and a trapezium.
Area of the triangle AGE
\(=\frac{1}{2} \mathrm{bh}=\frac{1}{2} 8 \times 14=56 \mathrm{~m}^{2} \)
Area of the triangle EGD
\(=\frac{1}{2} b h=\frac{1}{2} 8 \times 15=60 \mathrm{~m}^{2} \)
Area of the triangle HDC
\(=\frac{1}{2} b h=\frac{1}{2} 8 \times 5=20 \mathrm{~m}^{2} \)
Area of the trapezium FHCB
\(=\frac{1}{2} \times 18 \times 11 \)
= 99 m²
Area of the triangle AFB
\(=\frac{1}{2} \mathrm{bh}=\frac{1}{2} 6 \times 3 \)
= 9 m²
Total  Area of the field
= 56 + 60 + 20 + 9 + 99
= 244 m²

Area of the unshaded region = Area of the square - Area of the circle
= a² - πr²
= \({ a }^{ 2 }-\pi { \left( \frac { a }{ 2 } \right) }^{ 2 }={ a }^{ 2 }-\frac { 22 }{ 7 } \times \frac { { a }^{ 2 } }{ 4 } \)
= \(\frac { { 14a }^{ 2 }-{ 1 }1a^{ 2 } }{ 14 } =\frac { 3 }{ 14 } { a }^{ 2 }\)sq. units
Area of the area = \(\frac { 3 }{ 14 } { a }^{ 2 }\)sq. units unshaded
The unshaded portion of the figures are the same.

\(\frac{22}{7}\) and 3.14 are rational numbers π has non-terminating and non-repeating decimal expansion. So it is not a rational number. It is an irrational number. 

Net for the cuboid is:

One of the possible nets for a cuboid of length = 5 cm, breadth = 4 cm, height = 3 cm is given above
Area of the cuboid = 20 cm² + 15 cm² + 20 cm² + 15cm² + 12cm² + 12cm²
= 94cm²
Using formula,
Surface area of a cuboid = 2 (lb + bh + lh) unit²
= 2(5 x 4 + 4 x 3 + 5 x 3) cm² = 2 (20 + 12 + 15) cm²
= 94cm²

(i) F, T, S
(ii) T, S,F
(iii) S, F, T

Triangular Prism

The given figure is the combination of the square and two triangle
b = 7cm, b = 70cm ,h = 10cm
The area of the shaded part
= Area of the square + Area of the two triangle
\(=\mathrm{b}^{2}+2 \times \frac{1}{2} \mathrm{bh} \)
\(=(70 \times 70)+2 \times \frac{1}{2} \times 70 \times 10 \)
= 4900 + 700
= 5600 cm²

From the figure
diameter = 10 cm
radius = 5 cm
Area of the shaded part
= Area of the square - Area of the four circular quadrant
\( =a^{2}-\left(4 \times \frac{\theta}{360} \times \pi r^{2}\right) \)
\(=(10 \times 10)-\left(4 \times \frac{90}{360} \times 3.14 \times 5 \times 5\right) \)
= 100 - 78.5 = 21.5 cm² (approximately)

From the given data the circle is divided into 8 equal sectors.
The central angle is 45^o and radius = 56 cm
Area of each of the sectors
\(=\frac{\theta}{360} \times \pi \mathrm{r}^{2} \)
\(=\frac{45}{360} \times \frac{22}{7} \times 56 \times 56 \)
= 1232 cm²

Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64cm
l+ 2r = 64cm
44 + 2r = 64
2r = 64-44
2r = 20
r = \(\frac{20}{2}\) = 10cm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 44\times 10 }{ 2 } \) cm² = 22 x 10 cm² = 220 cm²
Area of the sector = 220 cm²

Length of the arc of the sector l = 50 mm
Radius r = 14mm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 14 }{ 2 } \) mm² = 50 x 7 mm² = 350 mm²
Area of the sector = 350mm²

(i) Area of the sector A = \(\frac { lr }{ 2 } \) sq. units
I = 48m
r = 10m
= \(\frac { 48\times 10 }{ 2 } \) m²
= 24 x 10 m²
= 240m²
Area of the sector = 240m²
(ii) Length ofthe arc l = 50 cm
Radius r = 13.5 cm
Area of the sector A = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 13.5 }{ 2 } \) cm² = 25 x 13.5 cm²
= 337.5 cm²
Area of the sector = 337.5 cm²

Area of I + Area of III = Area of square - Area of two semi circles having centres as Q and S.

= \({ a }^{ 2 }-2\frac { 1 }{ 2 } \pi { r }^{ 2 }\)
= \({ a }^{ 2 }-\pi { r }^{ 2 }={ a }^{ 2 }-\pi { \left( \frac { a }{ 2 } \right) }^{ 2 }\)
= \({ a }^{ 2 }-\frac { 22 }{ 7 } \times \frac { { a }^{ 2 } }{ 4 } =\frac { { 14a }^{ 2 }-{ 1 }1a^{ 2 } }{ 14 } \)
= \(\frac { { 14a }^{ 2 }-{ 1 }1a^{ 2 } }{ 14 } \)
Area of I + Area of III = \(\frac { 3{ a }^{ 2 } }{ 14 } \)sq. units ...(1)
Area of II + Area of IV = Area of square -  Area of 2 semicircles having centres at P and R
= \({ a }^{ 2 }-2\frac { 1 }{ 2 } \pi { r }^{ 2 }={ a }^{ 2 }-\pi { \left( \frac { a }{ 2 } \right) }^{ 2 }\)
Area of II + Area of IV = \(\frac { 3{ a }^{ 2 } }{ 14 } \) sq. units ....(2)
∴  Area of unshaded region = Area of I + Area of II + Area of III + Area of IV
= \(\frac { 3{ a }^{ 2 } }{ 14 } +\frac { 3{ a }^{ 2 } }{ 14 } \) sq. units = \(\frac { { 6a }^{ 2 } }{ 14 } \) sq. units
= \(\frac { 3{ a }^{ 2 } }{ 7 } \) sq. units
∴ Area of un shaded region = \(\frac { 3{ a }^{ 2 } }{ 7 } \) sq. units
Area of the shaded portion = Area of the square - Area of the unshaded region
= \({ a }^{ 2 }-\frac { 3{ a }^{ 2 } }{ 7 } =\frac { 3{ 7a }^{ 2 }-3{ a }^{ 2 } }{ 7 } =\frac { 4{ a }^{ 2 } }{ 7 } \) sq. units
Area of the shaded part = \(\frac { 4{ a }^{ 2 } }{ 7 } \) sq. units