Circumference of a circle = 44 cm
Sum of two sides of a square is 22 cm.
Radius of the circle = \(\frac { 44 }{ 2\times \frac { 22 }{ 7 } } =7\quad cm\)
Area of the circle = \(\pi { r }^{ 2 }=\frac { 22 }{ 7 } \times 7\times 7\)
= 154cm²

9.625 cm²

Because Sum of circumference of a circle and radius = 51 cm
\(\Rightarrow \ 2\pi r+r=51\\ \Rightarrow \ r(2\pi +1)=51\\ \Rightarrow \ r\left( 2\times \frac { 22 }{ 7 } +1 \right) =51\\ \Rightarrow\ r\times \frac { 51 }{ 7 } =51\Rightarrow r=7cm\)
Hence, the radius of the circle is 7 cm.

Length of the minute hand = Radius of a circle (r) = 14 cm
Area swept in 1h = Area of sector = \(\frac { { \pi r }^{ 2 }\theta }{ { 360 }^{ o } } \)
\(=\frac { { \pi r }^{ 2 }\times { 360 }^{ o } }{ { 360 }^{ o } } ={ \pi r }^{ 2 }=\frac { 22 }{ 7 } \times 14\times 14\\ =22\times 28=616{ cm }^{ 2 }\)
Hence, the area swept out by the minute hand in 1h is 616 cm²

Let radius of the circle = R 
∴ Circumference = 2πR and area = πR²
New circumference
\(=2\pi R+\frac { 50 }{ 100 } 2\pi R=3\pi R\)
∴ New Radius \(=\frac { 3 }{ 2 } R\)
New area \(=\pi \times \left( \frac { 3 }{ 2 } R \right) ^{ 2 }=\frac { 9 }{ 4 } \pi { R }^{ 2 }\)
Increase in area\(=\frac { 9 }{ 4 } \pi { R }^{ 2 }-\pi { R }^{ 2 }=\frac { 5 }{ 4 } \pi { R }^{ 2 }\)
%increase in area\(=\frac { \frac { 5 }{ 4 } \pi { R }^{ 2 } }{ \pi { R }^{ 2 } } \times 100\)=125%

Radius of wheel=70cm
Distance covered in 1 revolution=2πr
\(=2\times{22\over 7}\times70=440cm=4.4m\)
\(Speed=66km/h={66000\over 60}m/minute\)
=1100m/minute
∴ No. of revolutions\(={1100\over 4.4}=250\)

Let radius = r units
A.T.O
\(2\pi r=3\pi { r }^{ 2 }\)
\(\Rightarrow r=\frac { 2 }{ 3 } \) units

Area of the sector of angle P and radius R
\(=\frac { P\times { \pi R }^{ 2 } }{ { 360 }^{ ° } } sq\quad units\)

Inner radius \(= {220\times 7\over 22\times 2}=35 cm\) and Width of the track = 15 m
Outer radius = 35 + 15 = 50 m
Area of the track \(=\ \pi(50^2-35^2)={22\over 7}(50-35)(50+35)={22\over 7}\times 1585\ m^2=4007.14\ sq.m\)

Radius of circle = 28 cm
Circumference of the circle = \(2\pi\ r=2\times {22\over 7}\times 28 = 176 \ cm\)
\(\therefore\)  Length of circle = 176 cm
\(\Rightarrow\) Perimeter of square = 176 cm
Side of square = \({176\over 4}=44\ cm\)

Let length of the pendulum he I cm.
Length of the arc = 8.8 cm
⇒ \(\frac { \theta \pi i }{ 180 } =8.8cm\Rightarrow \frac { 30°\times \pi \times i }{ 180° } =8.8\)
⇒ \(\pi i=8.8\times 6\Rightarrow \frac { 22 }{ 7 } \times i=52.8cm\)
⇒ \(i=\frac { 52.8\times 7 }{ 22 } cm\quad =\quad 16.8\quad cm\)

Area of sector OAPB \(={x\over 360^o}\times \pi r^2\)
According to the question,
Ares of sector \(OAPB = {5\over 18}\)of the area of the circle.
\(\Rightarrow\)  \({x\over 360^o} \times \pi r^2={15\over 18}\pi r^2\Rightarrow {x\over 360^o}={5\over 18} \Rightarrow x=100^o\).

Given, length of wiper blade = 25 cm = r (say)
and angle made by this blade, \(\theta\) = 115°
\(\therefore\) Area cleaned by one blade = Area of sector formed by blade
\(\begin{aligned} & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{115^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(25)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{23 \times 22}{72 \times 7} \times 625=\frac{23 \times 11 \times 625}{36 \times 7}=\frac{158125}{252} \mathrm{~cm}^2 \end{aligned}\)
\(\therefore\) Total area cleaned by both blades
= 2 \(\times\) Area cleaned by one blade
\(=\frac{2 \times 158125}{252}=\frac{158125}{126} \mathrm{~cm}^2\)

Given, radius of a circle, AO = 10 cm and \(\angle\)AOC = 90°
Area of \(\triangle A O C=\frac{1}{2} \times O A \times O C=\frac{1}{2} \times 10 \times 10=50 \mathrm{~cm}^2\)
\(\begin{aligned} \text { Area of sector } O A E C O & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{314}{4}=78.5 \mathrm{~cm}^2 \end{aligned}\)

(i) Area of minor segment AECDA
= Area of sector OAECO - Area of \(\Delta\)AOC
= 78.5 - 50 = 28.5 cm²
(ii) Area of major sector OAFGCO
= Area of circle - Area of sector OAECO
= 3.14 \(\times\)(10)² - 78.5
= 314 - 78.5 = 235.5 cm²

Rs. 82500

\(\frac{6600}{7} cm\)

Given, arc ABC subtends an angle 60° at the centre.
\(\therefore\) \(\theta\)= 60° and radius, r = 21 cm
(i) Length of arc \(\begin{aligned} A B C & =\frac{\theta}{360^{\circ}} \times 2 \pi r \end{aligned}\)
\(\begin{aligned} =\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 \\ \end{aligned}\)
\(\begin{aligned} =\frac{44}{6} \times 3=22 \mathrm{~cm} \end{aligned}\)

(i) Area of the sector formed by the arc
\(\begin{aligned} =\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(21)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{22}{6 \times 7} \times 21 \times 21=\frac{9702}{42}=231 \mathrm{~cm}^2 \end{aligned}\)
(iii) Given, \(\angle\)AOC = 60° and OA = OC (radii of circle)
Let \(\angle\)OAC = \(\angle\)OCA= x
In \(\Delta\)OAC, \(\angle\)OAC + \(\angle\)AOC + \(\angle\)OCA = 180°
\(\Rightarrow\) 60° + x + x = 180° \(\Rightarrow\) 60° + 2x = 180
\(\Rightarrow\) 2x = 120 \(\Rightarrow\) x = 60°
\(\therefore\) \(\angle\)A = \(\angle\)O = \(\angle\)C = 60°
\(\Rightarrow\) \(\Delta\)OAC is an equilateral triangle.
So, area of \(\Delta\)OAC = \(=\frac{\sqrt{3}}{4} \times(21)^2=\frac{441 \sqrt{3}}{4} \mathrm{~cm}^2\)
\(\left[\because \text { area of equilateral triangle }=\frac{\sqrt{3}}{4}(\text { side })^2\right]\)
hence, area of the segment
= Area of sector formed by the arc - Area of \(\Delta\) OAC
\(=\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^2\)

Area of the circle inscribed = 154 cm2
Let its radius be
⇒ r² = \(\frac { 7\times 154 }{ 22 } \) ⇒ r = 7 cm
Let side of the equilateral triangle be x unit
⇒ Its area = \(\frac { \sqrt { 3 } }{ 4 } \)x²,
Semiperimeter s = \(x+x+x \over2\) = \(3x \over2\)
Also radius of the inscribed circle = \(\frac { area\ of\ triangle }{ semiperimeter } \)
⇒ \(r=\frac { \frac { \sqrt { 3 } }{ 4 } { x }^{ 2 } }{ \frac { 3x }{ 2 } } \)
⇒ \(7=\frac { \sqrt { 3 } { x }^{ 2 }\times 2 }{ 4\times 3x } =\frac { \sqrt { 3 } x }{ 2\times 3 } =\frac { 1 }{ 2\sqrt { 3 } } x\)
⇒ x = 2 x 7\(\sqrt3\) cm = 14\(\sqrt3\) cm
Perimeter of the triangle
= 3 x side = 3 x 14\(\sqrt3\)  cm
= 42 x 1.732 cm = 72.7 cm

Area of the segment 
= area of the shaded portion 
= area of the sector - area of the triangle 
= \(\frac { 90° }{ 360° } \times \frac { 22 }{ 7 } \times { (10) }^{ 2 }-\frac { 1 }{ 2 } \times 10\times 10\)
= \(\left( \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 100-50 \right) \) cm²
= (78.57 - 50) cm²
= 28.57 cm²

Perimeter of 2 inner semicircle = (400 - 2 x 90) m = (400 - 180) m = 220 m
Radius of each semicircle = \(\frac { 220 }{ 2\pi } \) = \(\frac { 220\times 7 }{ 2\times 22 } \) = 35 m
Perimeter of 2 outer semicircle = (488 - 180) m = 308 m
Radius of each outer semicircle = \(\frac { 308 }{ 2\pi } =\frac { 308\times 7 }{ 2\times 22 } =\quad 49m\)
width of the track = outer radius - inner radius = 49 - 35 = 14 m
Area of rectangular tracks = 2 x area of rectangle
= 2 x i x b = 2 x 90 x 14       (i = 90, b = 14 m)
= 28 x 90 = 2520 m²
Area of two semicircle rings = area of one circle ring
\(=\quad \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) =\frac { 22 }{ 7 } \left( { 49 }^{ 2 }-{ 35 }^{ 2 } \right) { m }^{ 2 }\)
= \(22\over7\) x (49 - 35) (49 - 35) m²
= \(22\over7\) x 14 x 84 m² = 44 x 84 = 3696 m²
Total area of track = (2520 + 3696) m² = 6216 m²
Cost of developing the track at the rate of Rs.12.50/ m² = Rs. 6216 x 12.50 = Rs.77,700

In OBD
cos 60°=\(OD \over OB\)and sin 60°=\(BD\over OB\)
⇒ \(1\over2\)= \(OD\over6\) and \(\frac { \sqrt { 3 } }{ 2 } \) = \(BD\over6\)
⇒ \(6\over2\)=OD and \(\frac { \sqrt { 3 } }{ 2 } \)x6 = BD
⇒ OD = 3 and BD = 3\(\sqrt3\)
BC = 2BD = 2 x 3\(\sqrt3\)=6\(\sqrt3\)
Area of the shaded region = area of circle - area of \(\triangle\)ABC
= \(2\pi { (6) }^{ 2 }-\frac { \sqrt { 3 } }{ 2 } { \left( 6\sqrt { 3 } \right) }^{ 2 }\)
= 3.14 x 6 x - \(\frac { \sqrt { 3 } }{ 4 } \) x 6\(\sqrt3\) x 6 \(\sqrt3\)
=  113.04 - 27 x 1.732 = 113.04 - 46.76 = 66.28 cm²

Area of square ABCD = 14 × 14 cm² = 196 cm²
Diameter of each circle \(=\frac{14}{2} \mathrm{~cm}=7 \mathrm{~cm}\)
So, radius of each circle \(=\frac{7}{2} \mathrm{~cm}\)
So, area of one circle \(=\pi r^{2}=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2}\)
\(=\frac{154}{4} \mathrm{~cm}=\frac{77}{2} \mathrm{~cm}^{2}\)
Therefore, area of the four circles \(=4 \times \frac{77}{2} \mathrm{~cm}^{2}=154 \mathrm{~cm}^{2}\)
Hence, area of the shaded region = (196 – 154) cm² = 42 cm²