Since, the sectors are cut from an equilateral triangle, the central angle of each of them is 60°.
∴ Area of the shape formed, \(A=3\times \left( \frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times { \pi r }^{ 2 } \right) \)
\(=3\times \frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \pi \times 5\times 5\)
\(=3\times \frac { 1 }{ 6 } \times \pi \times 5\times 5\)
= 12.5 π sq.cm

Perimeter of the palm leaf fan = 43 cm
That is, l + 2r = 43
l + 2 x (10, 5) = 43
l = 43 - 21
∴ the length of the arc l = 22 cm.
Length of the arc \(l=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times 2\pi { r }\) units

\(=\frac { { 360 }^{ 0 } }{ { 3 } } \times { 120 }^{ 0 }\)
Also, area of the palm leaf fan
A = \(\frac{lr}{2}\) sq.units
\(=\frac { 22\times 10.5 }{ 2 } \)
A = 115.5cm² (approximately)

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)

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.)

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)

(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

Triangular Prism

The given figure is a combination of a semicircle and a triangle.
diameter = 6 cm
radius = 3 cm
base = 6 cm
height = 9 cm
The shaded area of the figure = Area of the semicircle + Area of the triangle
\(=\frac{1}{2} \pi \mathrm{r}^{2}+\frac{1}{2} \mathrm{bh} \)
\(=\frac{1}{2} \times 3.14 \times 3 \times 3+\frac{1}{2} \times 6 \times 9 \)
= 14.13 + 27 = 41.13 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

Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64cm
l+ 2r = 64cm
44 + 2r = 64
2r = 64-44
2r = 20
r = \(\frac{20}{2}\) = 10cm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 44\times 10 }{ 2 } \) cm² = 22 x 10 cm² = 220 cm²
Area of the sector = 220 cm²

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