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

Radius of a circular quadrant, r = 3.5 cm and side of a square, a = 3.5 cm.
The given figure is formed by the joining of 4 quadrants of a circle with each side of a square. The boundary of the given figure consists of 4 arcs and 4 radii.
(i) Perimeter of the given combined shape
= 4 x length of the arcs of the quadrant of a circle + 4 x radius
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\pi r \right) +4r\)
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\times 3.5 \right) +(1\times 3.5)\)
= 22 + 14 = 36 cm (approximately)
(ii) Area of the given combined shape
= area of the square + 4 x area of the quadrants of the circle
\({ a }^{ 2 }=\left( 4\times \frac { 1 }{ 4 } \times \pi { r }^{ 2 } \right) \)
\(=(3.5\times 3.5)+\left( \frac { 22 }{ 7 } \times 3.5\times 3.5 \right) \)
A = 12.25 + 38.5 = 50.75 cm² (approximately)