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)

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

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

(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

= r²

Let the length of each side of the square be a cm and radius of the circle be r cm
By given, area of square = area of circle
⇒ a² = π r²
⇒ a = r 
∴ Required ratio = \(\frac{2\pi r}{4a}=\frac{2\pi r}{4r\sqrt \pi}=\frac{\sqrt \pi }{2} \)
\(\pi:\sqrt2\)

Area of  a circleA = π × r²
The radius is decreased by 10%, hence new radius r′ = 0.9r
New Area A′ = π × (r′)²
A′ = π × (0.9r)² = π × 0.81 r² = 0.81A
Hence, the area is decreased by 19%

Area of the mat = Area of I trapezium + Area of II trapezium
= [ \(\frac12\)x h₁ x (a₁ + b₁)] + [ \(\frac12\)x h₂ x (a₂ + b₂)]

= [ \(\frac12\)x 2 x (7 + 5)] + \(\frac12\) x 2 x (9 + 7) sq. feet
= 12 + 16 = 28 sq.feet
∴ Cost per sq.feet = Rs.20
Cost for 28 sq. feet = Rs.20 x 28 = Rs.560
∴ Total cost for the entire mat = Rs.560
Both the answers are the same.

For a regular polygon all sides and all the angles must be equal. But in a rhombus all the sides are equal, But all the angles are not equal.
∴ It is not a regular polygon.

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

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
\(=\frac{1}{n} 2 \pi \mathrm{r}=\frac{1}{8} \times 2 \pi \times 120=30 \pi \mathrm{m}\)