Area of a sector of a circle - class-X

Area of a sector with angle $p$ $=\dfrac{\theta}{360}\times\pi R^2$

$=\dfrac{\theta}{360\times2}\times\pi R^2\times2$

$=\dfrac{\theta}{720}\times2\pi R^2$

Hence, Option $D$ is correct

Since the Central angle is $90^{\circ}$, it means it is a Quad-circle.

So the Area of this sector is $\dfrac{1}{4}$ th of the Circle's Area $= \dfrac{1}{4}* \pi r^2$

Area $= \pi r^{2}=64\pi cm^{2}$  

The area of circle is $\pi r^2$

The area of the field in which the horse can graze is one fourth of the circle of radius $ 20  cm $
Area of a circle $ = \pi { r }^{ 2 } $
Hence, area of the field in which the horse can graze $ = \cfrac {1}{4} \times
\cfrac{22}{7} \times 20 \times 20 = 314  $ sq m

$A, =, \displaystyle \frac {1}{2}, lr, =, \displaystyle \frac {1}{2}, \times, 18.5, \times, 16, =, 148, cm^2$

$\displaystyle \frac{x}{360}=\frac{1}{18}\Rightarrow x=20^{\circ}$

Now $\frac{c}{circle}=\frac{120}{360}=\frac{1}{3}$

To change the formula in terms of $A$ and $S$, Isolate $r^2$, we get the formula as
$\dfrac{A\times 360}{S\pi}=r^2$
Now taking square root on both the sides to get the value of $r$ in terms of $A$ and $S$.
$\therefore r=\sqrt{\dfrac{360A}{S\pi}}$

Area of sector $=$ $\dfrac { \theta  }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }=\dfrac { { 42 }^{ 0 } }{ { 360 }^{ 0 } } \times \dfrac { 22 }{ 7 } \times 6\times 6=13.2{ m }^{ 2 }$

Arc Length : Perimeter = Area of Sector : Area of Circle

$21: 2\pi r = \; Area \; of \;  Sector : \pi r^2$

$21:24\pi = \; Area \; of \;  Sector :144\pi$

Area of Sector $= \dfrac{144 \pi *21}{24 \pi} = 126cm^2$

Arc length of the Circle  : Area of the Sector = Perimeter of the Circle :Area of the Circle

Let the radius of the circle be 'r'.

Hence, $20 : 60$=$ 2\pi r : \pi r^2$

$ 1:3 = 2: r$

$ r= 6 $ (Product of Means = Product of Extremes)

Therefore, $Diameter = 2r = 12cm$

Perimeter of the Sector =37cm

We know that minute hand covers $180^{o}$ in half an hour, which is a semicircle, hence area is

Area of a sector with angle $p = \dfrac{p}{360} \times \pi \times R^2$ ,which matches with option D.

Length PQR $= \displaystyle  \frac{1}{2} \times 2 \pi r = \frac{1}{2} \times 2 \times 3.14 \times 9.8 = 30.772 cm$
Length PSR $ = \displaystyle \frac{\theta}{360} \times 2 \pi = \frac{48.2}{360}\times 2 \pi r= \frac{48.2 }{360} \times 2 \times 3.14 \times24$
$= 20.1797 cm$
Perimeter of crescent $= 30.772 + 20. 1797 = 51$
Area of sector POR $= \displaystyle \frac{\theta}{360} \times \pi \times r^2 = \frac{48.2}{360} \times 3.14 \times 14^2$
$= 242.16 cm^2 = 242 cm^2$
In triangle POR height, $ h = 24 cos24.1 = 21.91 cm$
Area $= \displaystyle \frac{1}{2} bh = \frac{1}{2} \times 19.6 \times 21.91= 214.70 cm^2$
$\therefore $ Area of shape PSR remaining $= 242.16 - 214.70 = 27.46 cm^2$
Area of semicircle $= \displaystyle \frac{1}{2} \times \pi \times 9.8^2 = 150.859$
$\therefore$ Area of crescent $=150.859 - 27.46 = 123.4 cm^2$

For $180^\circ$, we have $\cfrac{1}{2}$ of the total area.

Area of sector = (1/2) * r^2 * x. Arc length = r * x. Given: (1/2) * r^2 * x = (r * x)^2. This simplifies to (1/2) * r^2 * x = r^2 * x^2. Dividing by r^2 * x (assuming x is not 0), we get 1/2 = x.

This is a geometry problem involving a circle inscribed in a right triangle. The sector area outside the triangle is calculated based on the geometry of the circle touching the sides. The result is pi - 2.

The circle x^2 + y^2 = 16 has a radius of 4. The line y = x makes an angle of 45 degrees (pi/4 radians) with the positive x-axis. The area of the sector in the first quadrant bounded by this line is (pi/4) / (2pi) of the total area, or 1/8 of the circle's area, which is (1/8) * pi * 4^2 = 2pi square units.

The perimeter of a sector is given by P = 2r + l = 4r, which simplifies to the arc length l = 2r. The area of a sector is given by Area = (1/2) * r * l. Substituting l = 2r gives Area = (1/2) * r * (2r) = r^2 square units.

Area of a sector of a circle of radius '$r$' and angle $ = \dfrac { \theta  }{ 360 } \pi {r}^{2}$
Hence, area of the sector of the circle of  radius $ 7 $ cm and angle $ = \dfrac { 30 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times 7 = 12.83 \ \text{cm}^{2} $

Area of a sector of a circle of radius 'r' and angle  $ \theta = \dfrac { \theta  }{ 360 } \pi {r}^{2}$

Hence,area of the sector of the circle of  radius $ 7 $ cm and angle $ { 210 }^{

0 } = \dfrac { 210 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times

7\quad = 89.83  {cm}^{2} $

Given:
Area of circle = $314 $$cm^2$
Area of minor sector = $31.4 $$cm^2$
Area of major sector = Area of a circle - Area of minor sector
= $314 - 31.4 cm^2$
= $282.6$ $cm^2$

Let the angle of centre made by the sector be $\theta$
Therefore,
Area of the sector=$\pi r^2\dfrac{\theta }{360}$
                        $=>3.85=\dfrac{\pi(3.5)^2\theta}{360}$

Angle made the centre by each $5$ minutes =$\dfrac{360}{12}$
                                                              =$30^o$

Area of sector $\displaystyle =\frac{42}{360}\times \pi r^{2}$
$\displaystyle =\frac{42}{360}\times \frac{22}{7}\times6\times6=13.2m^{2}$

$\displaystyle A=\frac{1}{2}l r=\frac{1}{2}\times 18.5\times 16=148cm^{2}$

We have,
Angle described by the minute hand in one minute $=6^o$

The total angle of $ 12 $ hours in a clock is 
$ { 360 }^{ 0 } $.
$ => 24 $ half hours $ = { 360 }^{ 0 } $.
This means for one half an hour, angle $ = \frac {{ 360 }^{ 0 }}{24} = { 15 }^{ 0 } $

Area of a sector of a circle of radius 'r' and angle 
$ \theta = \frac { \theta  }{ 360 } \pi {r}^{2}$
Hence, area of the sector of the circle of  radius $ 7 $ cm and angle $

{ 15 }^{ 0 } = \frac { 15 }{ 360 } \times \frac { 22 }{ 7 } \times 7 \times

7\quad = 6.42  {cm}^{2} $

$\displaystyle \theta =60^{\circ}$, r = 6 cm
Area of minor segment = $\displaystyle \dfrac{36}{2}\left [ \dfrac{60\times 3.14}{180}-\dfrac{\sqrt{3}}{2} \right ]$
                                     = 3.27 $\displaystyle cm^{2}$

$\Rightarrow$  Perimeter of a sector $=45,cm$

The area of a sector is proportional to the angle it subtends at the center. It is (angle/360) * (area of the circle).

Given: $\theta = 25^o = \dfrac{\pi}{4}$
Sector area $=$ $\dfrac{\theta}{2}r^2$
$=$ $\dfrac{\frac{\pi}{4}}{2}\times 2^2$ $=$ $\dfrac{\pi}{2}$

Area of sector $=$ $\dfrac { Arc.length }{ 2\pi r } \times \pi { r }^{ 2 }$

Area = $\dfrac{n}{360}\pi r^2$
= $\dfrac{100}{360}\pi 5^2$
= $6.944\pi$
= 21.80

$A _{sector}= \dfrac{n}{360}\pi r^2$
$120\pi= \dfrac{160}{360}\pi r^2$
$270=r^2$
$r = 16.43$

Area of a sector is given by $\cfrac{\theta}{360} \times \pi \times r^2$ where $\theta$ is the angle made by the sector, $r$ is the radius of the circle.

Time $= 20$ min

The total area of the seven sectors is (1/8) * pi * r^2. With r=1, total area = pi/8. The areas form a geometric progression: a, 2a, 4a, 8a, 16a, 32a, 64a. Sum = a(2^7 - 1)/(2 - 1) = 127a. 127a = pi/8, so a = pi / (127 * 8) = pi / 1016.

Radius of sector $=28 cm$

Area of sector = $\cfrac{120}{360} \times \pi \times (\cfrac{21}{2})^2$

Area = (theta/360) * pi * r^2. Area = (80/360) * 3.14159 * (16.5)^2 = (2/9) * 3.14159 * 272.25 = 0.2222 * 855.6 = 189.97. This is approximately 190 km^2.

Area of a sector $=\dfrac{\theta}{360^0} \times \pi r^2 =\dfrac{30}{360} \times \pi (7)^2 = \dfrac{49 \pi}{12}$

Area of sector formed from angle $\theta$ = $\frac{\theta}{260} \pi r^2$, where r is the radius of the circle
Now, if angle is 120, 150 then the ratio of area of sector will be:
= $\frac{\frac{120}{360} \pi r^2}{\frac{150}{360} \pi r^2}$
= $\frac{120}{150}$ = 4:5

Area of a sector of a circle of radius 'r' and angle  $ \theta = \frac {

\theta  }{ 360 } \pi {r}^{2}$
Given, $ \frac { 60 }{ 360 } \times \frac {22}{7} \times {r}^{2} = \frac {66}{7}  {cm}^{2} $
$ {r}^{2} = 18 $
Now, area of sector with angle {300}^{o} $ = \frac

{ 300 }{ 360 } \times \frac {22}{7} \times {r}^{2} = \frac

{ 300 }{ 360 } \times \frac {22}{7} \times 18 = \frac {330}{7}  {cm}^{2} $

Required area $\displaystyle =2\times \frac{115^{\circ}}{360^{\circ}}\pi \left ( 25 \right )^{2}$
$\displaystyle =2\times \frac{115}{360}\times \frac{22}{7}\times 625$