Radius, r = 7.5 cm and n = 6
Area of each of the sectors, A = \(\frac{1}{n}\times \pi r^2\) sq. units
\(\frac{1}{6}\pi \times 7.5\times 7.5\)
= 9.375 x \(\frac{22}{7}\)

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

length of the arc ,\(l=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times 2\pi r\)
\(=\frac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \pi \times 21\)
\(=\frac { 1 }{ 3 } \times 2\times \pi \times 21\)
\(l=14\pi cm(or)\)
\(=14\times \frac { 22 }{ 7 } \)
= 44 cm (approximately)

Radius of the sector = 27 cm
Area of the sector = 324 cm²
\(\frac { lr }{ 2 } \) = 324
\(\frac { l\times 27 }{ 2 } =342\)
l = \(\frac { 234\times 2 }{ 27 } \)
l = 24cm
Perimeter of the sector P = (l + 2r) units
= 24 + 2(27) cm = (24 + 54) cm = 78 cm

Area of the irregular field = Area of ΔAHF + Area of trapezium FHIE + Area of  + triangle EID + Area of ΔJDC + Area of rectangle BGJC + Area of ΔAGB
Area of the triangle = (\(\frac12\)x base x height) sq. units
Area of ΔAHF = \(\frac12\) (80 + 60) x 50 m² = 25 x 140 m²
= 3500 m²   ....(1)
Area of trapezium FHIE = \(\frac12\) h (a + b) sq. units =( \(\frac12\) x 35 x (50 + 40)) m²
= \(\frac12\)(35 x 90) = 1575 m² .....(2)
Area of ΔEID = \(\frac12\) x (65 + 25) x 40 m² = 90 x 20 m² = 1800 m² ....(3)
Area of ΔJDC = \(\frac12\) x 25 x 50 m² = 625 m² ....(4)
Area of rectangle BGJC = (60 + 35 + 65) x 50
= 160 x 50 = 8000 m² ....(5)
Area of the triangle AGB = \(\frac12\) 80 x 50 = 2000 m²
∴ Area of the field = (1) + (2) + (3) + (4) + (5) + (6)
= 3500 + 1575 + 1800 + 625 + 8000 + 2000 m²
= 17,500 m²
∴ Area of the field = 17,500 m²

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

(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

Area of the field where none of the cow can graze = Area of the rectangle - [Area of 4 quadrant circles] - Area of a circle
Area of the rectangle = I x b units²
= 15 x 8 m² = 120 m²
Area of 4 quadrant circles = 4 x \(\frac12 \) πr² units
Radius of the circle = 3m
Area of 4 quadrant circles = 4 x \(\frac14\)x 3.14 x 3 x 3 = 28.26m²
Area of the circle at the middle = πr² units
= 3.14 x 3 x 3 m² = 28.26m²
∴ Area where none of the cows can graze
= [120 - 28.26 - 28.26] m² = 120 - 56.52 m²
= 63.48m²

Given that \(\theta \) = 90^o and radius r = 6 feet
The distance moved by the wheel \(=\frac{\theta}{360} \times 2 \pi \mathrm{r} \)
\(=\frac{90}{360} \times 2 \times 3.14 \times 6 \)
= 9.42 feet

Triangular Prism

From this figure
Perimeter = 2l + 2 x 6 + 2r
\(=2 \times \frac{\theta}{360} \times 2 \pi \mathrm{r}+12+2 \times 3.5 \)
\(=2 \times \frac{90}{360} \times 2 \times \frac{22}{7} \times 3.5+12+7 \)
= 11+ 12 + 7
= 30 cm
The area of shaded part
= Area of two circular quadrants + Area of the rectangle
\(=\left(2 \times \frac{\theta}{360} \times \pi r^{2}\right)+(l \times b) \)
\(=\left(2 \times \frac{90}{360} \times \frac{22}{7} \times 3.5 \times 3.5\right)+(6 \times 3.5) \)
= 19. 25 + 21
= 40.25 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

Radius of the sector r = 4.2 cm
Area of the sector = 9.24 cm²
\(\frac { lr }{ 2 } \) = 9.24
\(\frac { l\times 4.2 }{ 2 } =9.24\)
l x 2.1 = 9.24
l = \(\frac { 9.24 }{ 2.1 } \)
l = 4.4cm
Perimeter of the sector = 1+ 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 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