Diameter, d = 7 cm, therefore r = \(\frac{7}{2}\) cm
Area of the shaded region = Area of the square – 4 x Area of the circular quadrant
\(={ a }^{ 2 }-4\times \frac { 1 }{ 4 } { \pi r }^{ 2 }\)
\(=(7\times 7)-\left( 4\times \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \right) \)
= 49 – 38.5 = 10.5 sq.cm. (approx.)

Radius, r = 35 cm and n = 5
Length of each of the arcs, l = \(\frac{1}{n}\times 2\pi r\) units
\(=\frac{1}{5}\times2\times \pi\times35\)
l = 14 π cm

Perimeter of the sector,P = l + 2r units
= 44 + 2 x 21
= 44 + 42
P = 86 cm (approximately)

(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)

(i) Central angle 45⁰, r = 16 cm
Length of the are l = \(\frac{θ^o}{360^o}\) x 2πr units
l = \(\frac{45^o}{360^o}\) x 2 x 3.14 x 16 cm
l = \(\frac18\) x 2 x 3.14 x 16 cm
l = 12.56 cm
Area of the sector = \(\frac{θ^o}{360^o}\) x  πr² sq.units
A = \(\frac{45^o}{360^o}\) x 3.14 x 16 x 16
A = 100.48 cm²
Perimeter of the sector P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm
(ii) Central angle 120⁰, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3 cm
Length of the are l = \(\frac{θ^o}{360^o}\) x  2πr units
l = \(\frac{120^o}{360^o}\) x 2 x 3.14 x 6.3 cm
l = 13.188 cm
l = 13.19 cm.
Area of the sector A = \(\frac{θ^o}{360^o}\) x 2πr units
A = \(\frac{120^o}{360^o}\) x 3.14 x 6.3 x 1.2 cm²
A = 3.14 x 6.3 x 2.1 cm²
A = 41.54 cm²
Perimeter of the sector P = l+ 2r cm
P = 13.19 + 2 (6.3) cm
= 13.19 + 12.6 cm
P = 25.79 cm

Length of the arc of the sector l = 50 mm
Radius r = 14mm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 14 }{ 2 } \) mm² = 50 x 7 mm² = 350 mm²
Area of the sector = 350mm²

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