Given arc length of the sector = 5π cm
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r=\) 5π cm   .....(1)
Also area of the same sector = 20π cm²
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times \pi { r }^{ 2 }=20\pi { cm }^{ 2 }\).....(2)
Dividing equation (2) by equation (1), we have
\(\frac { \frac { { \theta }^{ o } }{ { 360 }^{ o } } \times \pi \times r\times r }{ \frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\times \pi \times r } =\frac { 20\pi }{ 2\pi } \)
\(\frac { r }{ 2 } =4\)
r = 4 x 2
r = 8
Radius of the circle = 8 cm

Central angle θ = 30°
Radius r = 4 cm
Area of the sector = \(\frac { { \theta }^{ o } }{ 360^{ o } } \times \pi { r }^{ 2 }\)sq. units
= \(\frac { { 30 }^{ o } }{ { 360 }^{ o } } \) x 3.14 x 4 x 4 cm²  =  \(\frac{3.14\times4}{3}\)cm²
= 4.153 cm²
θ = 30° belongs to minor sector
∴ Angle of the major sector = 360° - 30° = 330°
∴ Area of the major sector = \(\frac { { 330 }^{ o } }{ { 360 }^{ o } } \times 3.14\times 4\times 4{ cm }^{ 2 }\)
= \(\frac { 11 }{ 12 } \times 3.14\times 4\times 4=\frac { 3.14\times 4\times 11 }{ 3 } \)
= 46.05 cm²

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 

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

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.)

Given that r = 12cm
θ = 30°
Length of the arc l = \(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r\)
= \(\frac { { 30 }^{ o } }{ { 360 }^{ o } } \) x 2 x 3.14 x 12
= \(\frac12\) x 2 x 3.14 x 12
l = 6.28 cm
Perimeter of.a sector = l + 2r
= 6.28 + 2 (12) cm
= 6.28 + 24 cm = 30.28 cm
Perimeter of the sector = 30.28 cm

Radius of the circle = 14cm
Area of the circle = 63 cm²
Area of the circle = \(\frac { 1 }{ 2 } \)lr
\(\frac { 1 }{ 2 } \) l x r = 63
\(\frac { 1 }{ 2 } \) l x 14 = 63
l x 7 = 63
l = \(\frac{63}{7}\)
l = 9 cm
Length of the arc = 9 cm 

Length of the arc = \(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r\)
Central angle θ = 36°
Radius of the sector r = 10.5 cm
∴ Length of the arc = \(\frac { { 30 }^{ o } }{ { 360 }^{ o } } \times 2\times \frac { 22 }{ 7 } \times10.5\)
= \(\frac { 1 }{ 10 } \times 2\times 22\times 1.5=\frac { 66.0 }{ 10 } =6.6cm\)
∴ Length of the arc = 6.6 cm

Length of arc = \(\frac { { \theta }^{ o } }{ 360^{ o } } \times 2\pi r\) units
Given central angle θ = 60⁰
Radius of the sector r = 42 cm
l = \(\frac { { 60 }^{ o } }{ { 360 }^{ o } } \times 2\times \frac { 22 }{ 7 } \times 42\) cm = 44 cm
∴ Length of the arc = 44 cm