Thew intercepted arc for an angle inscribed in a semi-circle is ${180^0}.$ Therefore the measure of the angle must be half of ${180^0}$  or ${90^0}.$

Perimeter of Protractor =$ \pi r +2r$

Area of outer circle $=\pi{(19)}^2$
                                 $=\cfrac{22}{7}\times 361$
Area of inner circle $=\pi{(16)}^2$
                                 $=\cfrac{22}{7}\times 256$
$\therefore$ area of the ring $=\dfrac{22}{7}\times 361-\dfrac{22}{7}\times 256$
                              $=\cfrac{22}{7}(361-256)$
                              $=\cfrac{22}{7}(105)$
                              $=330\ cm^2$

Let radius of the inner circle be $r$ and radius of the outer circle be $R=21cm$

Given: $r _1=5+r _2$
$\displaystyle \frac {2\pi r _1}{2\pi r _2}=\frac {23}{22}$

$\displaystyle \Rightarrow \frac {r _1}{r _2}=\frac {23}{22}$

$\displaystyle \Rightarrow \frac {5+r _2}{r _2}=\frac {23}{22}$

$\displaystyle \Rightarrow 110+22r _2=23r _2$

$\displaystyle \Rightarrow r _2=110\ m$
$\therefore $ Radius of inner circle$=110\ m$

$New\ radius=r+1\ cm\\ Ratio=2\pi (r+1):2(r+1)=\pi :1$'

Given the diameter of wheel is 98 cm

 Area of the circular ring = $\frac { 22 }{ 7 } \times \left( { R }^{ 2 } - { r }^{ 2 } \right)$ = $ \frac { 22 }{ 7 } \times \left( { 14 }^{ 2 } - { 10.5 }^{ 2 } \right)$ = $269.5 \ { cm }^{ 2 }\approx 269\ { cm }^{ 2 } $

$Area=\pi (100^{2} - 8^{2}) = \pi 36 = 113.143\ cm^{2}$.

The area of a ring shaped region$=\pi(20)^2-\pi(15)^2$
                                                    $=(400-225)\times \dfrac{22}{7}$
                                                    $=(175)\times \dfrac{22}{7}$
                                                   $ =550$ sq. cm

Given $\displaystyle 2\pi r _{2}-2\pi r _{1}=132$
$\displaystyle \Rightarrow r _{2}-r _{1}=\frac{132}{2\pi }=\frac{132\times 7}{44}=21\ m$

Inner circumference $= \displaystyle 2\pi r=2\times \frac{22}{7}\times r=\frac{44r}{7}$
Given $\displaystyle \frac{44r}{7}=440\Rightarrow r=\frac{440\times 7}{44}=70\ m$
$\displaystyle \therefore $ Radius of outer circle $= 70\ m + 14\ m = 84\ m$ 

Let '$a$' be the length of each side of the square
Then $\displaystyle a^{2}=81\Rightarrow a=9$ cm
Length of wire $=$ Perimeter of square
=$ 4a= 36$ cm
$\displaystyle \Rightarrow $ Circuference of semicircle $= 36$ cm
$\displaystyle \Rightarrow \pi r+2r=36$

$R = 15\ cm, r = 13\ cm$
Area of the circulating ring 
$\displaystyle = \pi \left(R^2-r^2\right)$

Let O be the centre of the circle. Then, $OA = OB = OD= r$
Now, $OC = r - 2$
and $CD = 6$
Thus, in $\triangle ODC$
$OC^2 + CD^2 = OD^2$
$(r - 2)^2 + 6^2 = r^2$
$r^2 + 4 - 4r + 36 = r^2$
$4r = 40$
$r = 10$ $cm$
Area of semicircle $= \dfrac{\pi r^2}{2} = \dfrac{\pi (10)^2}{2} = 50 \pi$

Let the side of the square be $a$ cm

Circumference of inner side = $220 m$

Breadth of the ring is equal to the difference between the radius of the outer circle and the radius of the inner circle
Given the area of outer circle=616$\displaystyle cm^{2}$
$\displaystyle \Rightarrow  \pi r _{2}^{2}=616 cm^{2}$
$\displaystyle \Rightarrow r _{1}^{2}=\frac{616\times 7}{22}=196$
$\displaystyle \therefore r _{1}=14 cm $
and the area of the inner circle $\displaystyle =154 cm^{2}$
$\displaystyle \Rightarrow \pi r _{2}^{2}= 154$
$\displaystyle \Rightarrow r _{2}^{2}=\frac{154\times 7}{22}=49$
$\displaystyle \therefore r _{2}=7 cm.$
$\displaystyle \therefore $ The required answer $\displaystyle =r _{1}-r _{2}=14-7=7 cm.$

Consider the difference between circumference and radius.

Taking$\pi $ as $3.14$ will give an approximate answer. we can do it by taking $22/7$ as wel

Let, the radius be$ r cm$

Circumference will be,

$2r=2\times 3.14r=6.28r$

ATQ,  $6.28r-1.r=37$

$5.28r=37$

 $r=\dfrac{37}{5.28}$

$r=$approx. $7 cm$ 

Area$=\pi {{r}^{2}}=3.14\times 7\times 7=154c{{m}^{2}}$

Hence, this is the answer.

The perimeter of a semi circle is $= \pi r+2r$

Diameter $= 2 \times$ radius
radius $= \dfrac{4}{2}=2$ $cm$
Circumference of the semicircle $= \pi r+ 2r$
= $3.14 \times  2 +4= 10.28$ $cm$

Area of a semicircle $=\dfrac{1}{2}\pi r^2$
$=\dfrac{1}{2}\pi (20)^2$
$=628\ cm^2$

Area of a semicircle $=\dfrac{1}{2}\pi r^2$
$ = \dfrac{1}{2}\pi (2)^2$
$ = 2\pi  $

radius $= 2\ m$
Area of a semicircle $= \dfrac{1}{2}\pi r^2$
$ = \dfrac{1}{2}\pi (2)^2$
$ = 2\pi  $

Circumference of a semicircle $=\pi r$
$15.15= \pi r$
$r = 5.05$
Area of the semicircle $=\dfrac{1}{2}\pi r$
$=\dfrac{1}{2}\pi \times 5.05$
$= 7.575$ $cm^2$

Circumference of a semicircle $=\pi r$
$12.5= \pi r$
$r = 3.98 \approx 4$
Area of the semicircle $=\dfrac{1}{2}\pi r$
$=\dfrac{1}{2}\pi (4)$
$= 6.28$ $cm^2$

Area of a circular ring $=\pi(R^2-r^2)$
$=3.14(12^2-10^2)$
$= 3.14 \times 44$
$= 138.16$ $cm^2$

Area of a semicircle $=\dfrac{1}{2}\pi r^2$
$200 = \dfrac{1}{2}\pi r^2$
$400 = \pi r^2 $
$r^2=127.38\ m^2$
$r \approx 11$ $m$

Area of a semicircle $=\dfrac{1}{2}\pi r^2$
$ 25.12= \dfrac{1}{2}\pi r^2$
$ 50.24= \pi r^2  $

Given : Radius ($r _1$) $=21m$

The largest triangle inscribed in a semicircle has a height = radius of the circle

Radius of the merry go round, r = 7m

The area of a triangle is equal to the base times the height.
In a semi circle, the diameter is the base of the semi-circle.
This is equal to $2\times r$ (r = the radius)
If the triangle is an isosceles triangle with an angle of $45^\circ$ at each end, then the height of the triangle is also a radius of the circle.
A = $\frac{1}{2} \times b \times h$ formula for the area of a triangle becomes
A = $\frac{1}{2}\times 2 \times r \times r$ because:
The base of the triangle is equal to $2\times r$
The height of the triangle is equal to r
A = $\frac{1}{2} \times 2 \times r \times r$ becomes:
A = $r^2$

Radius of bigger circle(with the path) = $35 + 7 = 42\ m.$
Thus area of the path $=$ Area of bigger circle $-$ Area of smaller circle
$\therefore$ Required area $= \pi (42)^2 - \pi (35)^2 = \dfrac{22}{7} \times (42 + 35)(42 - 35) = 22 \times 77 = 1694\ m^2$

Area of equilateral triangle $= s$ sq.cm
$\Rightarrow  \dfrac{\sqrt3}{4} a^2 = s$, [where $a$, the side of equilateral triangle]
$\Rightarrow a= \sqrt{\dfrac{4s}{\sqrt3}}$
Now perimeter of equilateral triangle $ 3\times a =3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$ cm 
Circumference of circle $=$ perimeter of equilateral triangle
$\Rightarrow 2\pi r= 3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$, [where $r$ the radius of circle]
Solve the above expression for $r$, we get 
$r= \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}}$
Area of circle $=\pi r^2 = \pi \times \left ( \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}} \right )^2$
After simplification, we get
Area of circle $=\dfrac{3s\sqrt3}{\pi}$ sq.cm

Let the number of revolution of the wheel is n.
Then,
n $\times$ circumference of wheel = Distance travelled by bicycle
$n \times  2\Pi  \times \frac {1}{2}=1$ kilometer 
$n \times  \Pi $=1000 meter
$n=\frac {1000}{\Pi }$

A dog is chained on a $6$ ft leash to the corner of a rectangular building.
Since the building is rectangular, the dog is left with an angle of $360 - 90 = 270^o$ for it to roam around.
Also, the length of the leash will act as the radius of this sector.
$\therefore$ Area of the sector $= \cfrac{270}{360} \times \pi \times 6^2$
$= \cfrac{3}{4} \times \pi \times 36$
$= 84.82 \ \ ft^2$