(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

From the figure, radius = 7 cm
diameter = 14 cm
Area of the shaded part = Area of the semicircle - Area of the triangle
\(=\frac{1}{2} \pi r^{2}-\frac{1}{2} b h \)
\(=\frac{1}{2} 3.14 \times 7 \times 7-\frac{1}{2} \times 14 \times 7 \)
= 76.93 - 49 = 27.93 cm²

From this figure, perimeter
= 10 m + 7 m + 10 m + L
\(\begin{equation} =27 \mathrm{~m}+\frac{\theta}{360} \times 2 \pi \mathrm{r} \end{equation}\)
\(=27+\frac{180}{360} \) \(\times 2 \times \frac{22}{7} \times \frac{7}{2}\)
= 27 + 11 = 38 m
Area of the shaded part - Area of the rectangle - Area of the semicircle
=\(​​(l\times b)-\frac { 1 }{ 2 } \times \pi { r }^{ 2 }\)
= \((10\times 7)-\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \)
= 50.75 m²

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)

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

Given that r = 70 cm ,n = 5
Since the circle is divided into 5 equal sectors, the area of each of the sector
\(=\frac{1}{n} \pi \mathrm{r}^{2}=\frac{1}{5} \times \pi \times 70 \times 70=980 \pi \mathrm{cm}^{2}\)

Given that r = 120m ; n = 8
Since the circle is divided into 8 equal sectors, the length of the arc of each of the sector
\(=\frac{1}{n} 2 \pi \mathrm{r}=\frac{1}{8} \times 2 \pi \times 120=30 \pi \mathrm{m}\)

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