(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.

(i) The length of the iron frame required = length of the arc + 4r
= πr + 4r
\(=\left( \frac { 22 }{ 7 } \times 49 \right) +(\times 49)\)
= 154 + 196
= 350 cm (approximately)
(ii) Area of each of the mirror sectors
\(=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times { \pi r }^{ 2 }\)
\(=\frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times 49\times 49\)
= 1257.67 sq.cm (approximately)

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

Area of the combined shape = Area of the rectangle + Area of 2 triangles
Length of the rectangle I = 10cm
Breadth b = 8cm
Base of the triangle = 8cm
Height h = 6cm
∴ Area of the shape = (l x b) + (2 x \(\frac12\) x base x h) cm²
= (10 x 8) + (8 x 6) cm² = 80 + 48 cm² = 128 cm²
Area of the given shape = 128 cm²

Area of the card = Area of the rectangle + area of 2 semicircles
Length of the rectangle I = 30 cm
Breadth b = 21 cm
Radius of the semicircle = \(\frac{21}{2} \) cm
∴ Area of the card = (l x b) + \(\frac { 1 }{ 2 } \times 2\pi { r }^{ 2 }\)
= 30 x 21 + \(\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \) cm² = 630 + 346.5
= 976.5cm² (approximately)
∴ Area of the Invitation card = 976.5 cm²

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 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

Central angle of the quadrant = 90°
Radius of the circle = 2 feet
Area of the quadrant = \(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times { \pi r }^{ 2 }sq.units=\frac { { 90 }^{ o } }{ { 360 }^{ o } } \times \pi \times 2\times 2\times \)sq. feet
= \(\frac14\) x 3.14 x 4 = 3.14 sq. feet
Area of the quadrant = 3.14 sq. feet (approximately)

i) Radius of the sector = 21 cm
Area of the sector = 462 cm²
\(\frac { lr }{ 2 } =462\)
\(\frac { l\times 21 }{ 2 } =\) 462
l = \(\frac { 462\times 2 }{ 21 } \)
l = 22 x 2
Length of the arc I = 44 cm
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r=44cm\)
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\times \frac { 22 }{ 7 } \times 21\) = 44 cm
θ^o = \(\frac { 44\times 360\times 7 }{ 2\times 22\times 21 } \)
θ^o = 120^o
(ii) Radius of the sector = 35 m
Length of the arc I = 44 m
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r=44cm\)
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\times \frac { 22 }{ 7 } \times 35=44cm\)
θ^o = \(\frac { 44\times 360\times 7 }{ 2\times 22\times 35 } \)
θ^o = 72^o

(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

The given field has four triangles (I, III, IV & V) and a trapezium (II).
Area of the triangle (I) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 5\times 6=15{ m }^{ 2 }\)
Area of the trapezium (II) \(=\frac { 1 }{ 2 } h(a+b)=\frac { 1 }{ 2 } \times 13\times (6+4)=65{ m }^{ 2 }\)
Area of the triangle (III) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 8\times 4=\frac { 32 }{ 2 } =16{ m }^{ 2 }\)
Area of the triangle (IV) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
Area of the triangle (V) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
∴ The total area of the field = 15 + 65 + 16 + 65 + 65 = 226 m²

Mark the unshaded parts of the given figure as I, II, III and IV
Area of the I and III parts = Area of the square – Area of 2 semicircles
\(={ a }^{ 2 }-\left( 2\times \frac { 1 }{ 2 } { \pi r }^{ 2 } \right) \)
\(=(10\times 10)-\left( \frac { 22 }{ 7 } \times 5\times 5 \right) \) = 100 - 78.57 = 21.43 cm²
Similarly, the area of the II and IV parts = 21.43 cm²
∴ Area of the unshaded parts (I, II, III and IV)
= 21.43 x 2 = 42.86 cm² 
∴ Area of the shaded part = area of the square – area of the unshaded parts
= 100 – 42.86 = 57.14 cm² 

The mat given in the figure can be split into two rectangles as follows
∴ Area of the entire mat
= area of the I rectangle + area of the II rectangle
= (l₁ x b₁) + (l₂ x b₂)
= 5 x 2 + 9 x 2 = 10 + 18 = 28 sq.feet
Cost per sq. foot = Rs. 20
∴ The total cost of the entire mat = 28 x Rs. 20 = Rs. 560.

Figures I and II are trapeziums separately as well as combinedly.
The parallel sides of the combined trapezium (I and II) are 5 cm and 16 cm. Its height, h = 8 + 8 = 16cm
Length of the rectangle = 16 cm
Breadth of the rectangle = 8 cm
∴  Area of the combined invitation card
= area of the combined trapezium + area of the rectangle
\(=\left( \frac { 1 }{ 2 } h\times (a+b) \right) +(l\times b)\)
\(\left( \frac { 1 }{ 2 } \times 16\times (5+16) \right) +(16\times 8)\)
= 168 + 128 = 296cm²
Aliter:
Area of the invitation card = area of the outer rectangle – area of the right angled triangle
\(= l\times b - \frac { 1 }{ 2 } \times h \times b \)
= \(24 \times16 \frac{1}{ 2} \times11 \times16\)
= − 384 88 296 = cm²

Side of the square = 5 cm
Diameter of the semi-circle = 5 cm
∴ Radius = 2.5 cm
Side of the equilateral triangle = 5 cm
∴  Area of the keychain = area of the semi circle + area of the square + area of the equilateral triangle
\(=\frac { 1 }{ 2 } { \pi r }^{ 2 }+{ a }^{ 2 }+\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
\(=\left( \frac { 1 }{ 2 } \times 3.14\times 2.5\times 2.5 \right) +\left( 5\times 5 \right) +\left( \frac { \sqrt { 3 } }{ 4 } \times 5\times 6 \right) \)
= 9.81 + 25 + 10.83
= 45.64cm² (approx.)

(i) Area of the blue shaded part = Area of the quadrant of a circle
\(=\frac { 1 }{ 4 } \times { \pi r }^{ 2 }\)
\(=\frac { 1 }{ 4 } \times 3.14\times 2\times 2\)
= 3.14 cm² (approximately)
(ii) Area of the grey shaded part =  Area of the square –  Area of the blue shaded part
a² - \(\frac{1}{4}\pi r^2\)
= 6 x 6 - 3.14
= 36 - 3.14
= 32.86 cm² (approximately)