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

By Euler's formula F + V - E = 2 for a polyhedron
Here F = 12, V = 17, E = 22
F + V - E = 12 + 17 - 22
= 29 - 22
= 7≠ 2
∴ The polyhedron cannot have 12 faces 22 edges and 17 vertices.

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

Cylinder

Area of the glass painting = Area of the square + Area of the triangle
Side of the square a = 30 cm
Base of the triangle b = 30 cm
Height of the triangle h = 8 cm
∴ Area of the painting (a)² + (\(\frac { 1 }{ 2 } \) bh) sq. units
= (30 x 30) +  (\(\frac { 1 }{ 2 } \) 30 x 8) cm²  = 900 + 120 cm²
= 1020 cm²
Area of the glass painting = 1020 cm²

Area of the irregular field = Area of ΔAHF + Area of trapezium FHIE + Area of  + triangle EID + Area of ΔJDC + Area of rectangle BGJC + Area of ΔAGB
Area of the triangle = (\(\frac12\)x base x height) sq. units
Area of ΔAHF = \(\frac12\) (80 + 60) x 50 m² = 25 x 140 m²
= 3500 m²   ....(1)
Area of trapezium FHIE = \(\frac12\) h (a + b) sq. units =( \(\frac12\) x 35 x (50 + 40)) m²
= \(\frac12\)(35 x 90) = 1575 m² .....(2)
Area of ΔEID = \(\frac12\) x (65 + 25) x 40 m² = 90 x 20 m² = 1800 m² ....(3)
Area of ΔJDC = \(\frac12\) x 25 x 50 m² = 625 m² ....(4)
Area of rectangle BGJC = (60 + 35 + 65) x 50
= 160 x 50 = 8000 m² ....(5)
Area of the triangle AGB = \(\frac12\) 80 x 50 = 2000 m²
∴ Area of the field = (1) + (2) + (3) + (4) + (5) + (6)
= 3500 + 1575 + 1800 + 625 + 8000 + 2000 m²
= 17,500 m²
∴ Area of the field = 17,500 m²

(i) Euler's formula is given by F + V - E = 2
(i) V = 6, E = 14
By Euler's formula = F + 6 -14 = 2
F = 2 + 14 - 6
F = 10
(ii) F = 8, E = 10
By Euler's formula 8 + V -10 = 2
V = 2- 8 + 10
V = 4
(iii) F = 20, y.= 10
By Euler's formula = 20 + 10 - E = 2
30-E = 2
E = 30- 2
E = 28
Tabulating the required unknowns

Guna fixed a single door of 3 feet wide.
Radius of this single door = 3 feet.
Nathan fixed a double door each of 1 1/2 feet wide.
Radius of each of this double door
\(=\frac{3}{2} \text { feet. } \)
The area required for the single door
\(=\frac{\theta}{360} \times \pi r^{2} \)
\(=\frac{120}{360} \times 3.14 \times 3 \times 3 \)
= 9.42 m²    ............(i)
The area required for the double door
\(=2 \times \frac{\theta}{360} \times \pi r^{2} \)
\(=2 \times \frac{120}{360} \times 3.14 \times \frac{3}{2} \times \frac{3}{2} \)
= 4.71 m²       ........................(ii)
From (i) and (ii), the double door requires minimum area.