The door mat has two semi-circles attached to a rectangle.
Length of the rectangle = 90 – 40 = 50 cm
Its breadth = 40 cm
Diameter of the semi-circle = 40 cm
Its radius = 20 cm
∴ Area of the door mat = area of the rectangle + 2 × area of the semicircle
\(=(l\times b)+\left( 2+\frac { 1 }{ 2 } { \pi r }^{ 2 } \right) \)
= (50 x 40) + (3.14 x 20 x 20)
= 2000 + 1256 = 3256 sq.cm (approximately)

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.)

The door is made of semicircle and rectangular shapes.
Length of the rectangle = 9 –3.5 = 5.5 feet
Its breadth = 3.5 feet
Diameter of the semicircle = 3.5 feet
∴ Radius =\(\frac{3.5}{2}=1.75\)feet
∴ Area of the door = Area of the rectangle + Area of the semicircle
\(=(l\times b)+\left( \frac { 1 }{ 2 } { \pi r }^{ 2 } \right) \)
\(=(5.5\times 3.5)+\left( \frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 1.75\times 1.75 \right) \)
= 19.25 + 4.81 = 24.06 sq.feet (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)

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)