Circumference of the circle = 4\(\pi\) cm \(\Rightarrow\) r = 2 cm.
Increase circumference = 8\(\pi\) cm. \(\Rightarrow\) r = 4 cm.
Area of the 1^st circle = \(\pi\) \(\times\) (2)²= 4\(\pi\) cm²
Area of the new circle = \(\pi\) (4)²= 16\(\pi\) = 4 \(\times\) 4\(\pi\)
\(\therefore\)  Area of the new circle = 4 times the area of first circle.

Perimeter of the circle = Perimeter of square 
Let side of square be x cm.
\(\begin{aligned} & 2 \pi \mathrm{r}=4 \mathrm{x} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad 2 \times \frac{22}{7} \times 77=4 x \\ \end{aligned}\)
\(\begin{aligned} & \therefore \quad x=\frac{2 \times 22 \times 11}{4}=121 \end{aligned}\)
Side of the square = 121 cm.

A line intersecting the circle at two distinct points is called a secant.

\(\because\) Area of circle = \(\pi { r }^{ 2 }\)
\(\therefore\) Area of the shaded region = \(\pi \)(2)² -  \(\pi \)(1)²
4\(\pi \) - \(\pi \) = 3\(\pi \) sq units.

56 cm

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.

n \(\times\) Circumference of circular track = Length of race
\(\Rightarrow n\times (2\pi r)=1540\)
where, n = number of rounds taken on a circular track
\(\Rightarrow n\times 2\times \frac { 22 }{ 7 } \times 3.5=1540\quad \quad [\because r=3.5cm]\\ \Rightarrow n=\frac { 1540 }{ 2\times \frac { 22 }{ 7 } \times 3.5 } =\frac { 1540 }{ 22\times 2\times 3.5 } \times 7=70\)
Hence, the number of rounds is 70.

\(\frac { Circumference\ of\ one\ circle({ C }_{ 1 }) }{ Circumference\ of\ other\ circle({ C }_{ 2 }) } =\frac { 3 }{ 1 } \\ \Rightarrow \frac { { 2\pi r }_{ 1 } }{ { 2\pi r }_{ 2 } } =\frac { 3 }{ 1 } \Rightarrow \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 3 }{ 1 } \)
Now, ratio of their areas \(=\frac { { \pi r }_{ 1 }^{ 2 } }{ { \pi r }_{ 2 }^{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 2 }=\left( \frac { 3 }{ 1 } \right) ^{ 2 }=\frac { 9 }{ 1 } \)

30.77 cm²

9.8 cm

Area of square field = 484 m² 
∴ Side x Side = 22 x 22 
⇒ Side = 22 m 
∴ Perimeter of square field = 4 X 22 = 88 m 
Since circumference of circular field = Perimeter of square field
\(\Rightarrow \pi d=88\Rightarrow \frac { 22 }{ 7 } d=88\)
\(\Rightarrow d=88\times \frac { 7 }{ 22 } =28m\)
Hence, the length of the diameter of the circular field is 28 m.

Distance covered by train in 1 revolution 
= Circumference of the wheel 
= 4\(2\over7\) m = \(30\over7\)3
ஃ Distance covered in 4 seconds 
= 7 x \(30\over7\)=30 m
ஃ Distance covered in 1 second 
= \(30\over4\) = 7.5 m
ஃ Speed = 7.5 m/s.

d = 90 cm, circumference = \(\pi\)d = 90\(\pi\) cm = Distance covered in 1 revolution 
Number of revolutions per minute = 315 
ஃ Number of revolution per hour = 315 x 60 cm 
Distance covered in 1 hour = 90\(\pi\) x 315 x 60 cm 
= 90 x \({22\over7}\times{315\times60\over100000}km={9\times22\times45\times6\over1000}km\)
∴ Speed of the bus = 53.46 Km/h

Area of portion that horse can graze area of the shaded portion.
Shaded portion is a sector of radius 21 m = length of the rope 
Angle of this sector = angle of the corners of the rectangle = 90°
Area the shaded portion that horse can graze 
= \(\frac { \theta }{ 360° } \times { \pi r }^{ 2 }=\frac { 90° }{ 360° } \times \frac { 22 }{ 7 } \times { (21) }^{ 2 }{ m }^{ 2 }\)
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 21\times 21 \ { m }^{ 2 } \ = \ \frac { 11 }{ 2 } \times 3\times 21 \ { m }^{ 2 }=346.5 \ \ { m }^{ 2 }\)

Let radius of outer circle = \(R_1\) and radius of Inner circle = \(R_2\)
A.T.Q.  \(\pi R_1^2-\pi R^2_2=286\)
\(\Rightarrow\) \(\pi (R^2_1-R^2_2)=286 \Rightarrow {22\over 7}\times (R_1-R_2)(R_1+R_2)=286\)
\(\Rightarrow\) \({22\over 7}\times 7\times (R_1+R_2)=286 \Rightarrow R_1+R_2=13 cm\)

Distance covered in 5000 revolutions = 11 km.
Distance covered in 1 revolution
Distance covered in 1 revolution \(={11000\over 5000}m={11\over 5}m\)
Distance covered in 1 revolution = circumference of the wheel
\(\Rightarrow\)\(2\ \pi r ={11\over 5} \Rightarrow 2\times {22\over 7}\times r = {11\over 5}\)
\(\Rightarrow\)  \(r \Rightarrow {11\over 5}\times 7\times {1\over 2\times 22}={7\over 20}m\)
\(\therefore\) Diameter \(=2\times r=2\times {7\over 20}={7\over 10}\times 100 cm=70cm\)

\({\theta \over 720^o}\times 2\pi R^2\)

Let r be the radius of the circular track.
Now, area of the circular track = 5544 m²  [given]
⇒ \(\pi\)r² = 5544
⇒ r² = \(5544\over \pi\) ⇒  r² = \(5544 \times7 \over22\)
⇒ r² = 1764 ⇒ r = \(\sqrt{1764}\) = 42 m
Circumference of the circular track = 2\(\pi\)r
= 2 x \(22\over7\)x42
= 264 m
ஃ Distance covered in 12\(1\over2\) rounds = 264 x \(25\over2\)
= 3300 m
= 3.3 km
Time taken to cover 3.3 km = 1 hour [given] Hence, time taken to cover 3300 m (12\(1\over2\) rounds) around the circular track is 1 hour.

Let AO = OB = r
Then perimeter of semicircle APB
= \(2\pi r \over 2\) = \(\pi\)r
Perimeter of semicircle AQO
= \(\frac { 2\pi \frac { r }{ 2 } }{ 2 } =\frac { \pi r }{ 2 } \)
(∵ radius of semicircle AQO = \(r\over2\))
∴ Perimeter of shaded region
\(=\ \pi r+\frac { \pi r }{ 2 } +r=\frac { 2\pi r+\pi r+2r }{ 2 } \)
But perimeter of shaded region
ஃ \(\frac { 2\pi r+\pi r+2r }{ 2 } =40\)
⇒ r(2\(\pi\) + \(\pi\) + 2) = 80
⇒  r(3\(\pi\) + 2) = 80
⇒ \(r\left( 3\times \frac { 22 }{ 7 } +2 \right) =80\)
⇒ \(r\left( \frac { 66 }{ 7 } +2 \right) =80\)
⇒  \(r\left( \frac { 80 }{ 7 } \right) =80\)
ஃ  \(r=\frac { 80\times 7 }{ 80 } =7cm\)
Now, Area of APB = \(\frac { \pi { r }^{ 2 } }{ 2 } =\frac { 22\times 7\times 7 }{ 7\times 2 } \)
= 77 cm2
Area of AQO = \(\frac { \pi { r }^{ 2 } }{ 2 } \)
 \(=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times \frac { 1 }{ 2 } \)
= \(77\over4\) cm²
ஃ Area of shaded region
= 77 + \(\frac { 77 }{ 4 } =\frac { 308+77 }{ 4 } \)
= \(\frac { 385 }{ 4 } \) = 96.25 cm²

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

In right-angled ΔABC, right angled at A
AB² + BC²=BC²
(14)² + (14)² = BC²
392 = BC²
BC=14√2cm

Area of shaded region = Area of ΔABC, + Area of semicircle on BC - area of quadrant ABDC
Area of ΔABC=\({1\over 2}\times AC\times AB={1\over 2}\times14\times14=98cm^2\)
Area of semicircle on BC\(={\pi r^2\over 2}\)
\(={22\over 7}\times7\sqrt2\times7\sqrt2\times{1\over 2}=154cm^2\)
Area of quadrant ABDC=\({\pi r^2\over 4}={22\over 7}\times14\times14\times{1\over 4}=154cm^2\)
∴ Area of shaded region=(98+154-154)cm²=98cm²

Length of semicircle ADC = \(\pi\times{4.2\over 2}cm=2.1\pi cm\)
Length of semicircle AEB\(=\pi\times{2.8\over2}cm=1.4\pi cm\)
Length of semicircle BFC=\(\pi\times{1.4\over2}cm=0.7\pi cm\)
∴ Perimeter of shaded region = 2.1π+1.4π+0.7π=4.2π cm=13.2cm

In right-angled triangle OAP,
\({OA\over OP}=cos\angle AOP\)

\(⇒\ \ {5\over 10}= cosㄥAOP\)
⇒ ㄥAOP=60⁰
Similarly, ㄥAOP=60⁰
ㄥAOB=60⁰+60⁰=120⁰
Length of arc (AB)=\({\theta\over 360^0}\times2\pi r\)
\(={120^0\over 360^0}\times2\times\pi\times5\)
\(={10\pi\over 3}cm\)
Length of the belt that is in the contact with the rim of pulley
= circumference of circle - length of arc AB
\(=2\pi\times5-{10\pi\over 3}={20\pi\over 3}cm\)
Area of sector
OAQB=\(={\theta\over 360^0}\times\pi r^2\)
\(={120^0\over 360^0}\times\pi5\times\times5cm^2={25\over 3}\pi cm^2\)
Area of quadrilateral OAPB
= 2 x area of triangle OAP
\(2\times{1\over 2}\times OA\times AP\)
\(=5\times5\sqrt3cm=25\sqrt3cm^2\)
Area of shaded region
\(=\left(25\sqrt3-{25\pi\over 3}\right)cm^2\)