Tag: circle measures

Questions Related to circle measures

Radius of a marry-go-round is $7\ m$. At its edge, at equal distances swings are suspended. If length of an arc between two successive swings is $4$ metre, then find the number of swings that marry-go-round has.

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $13\ swings$.

  2. $11\ swings$.

  3. $15\ swings$.

  4. None of these

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Correct Option: B
Explanation:

Radius of the merry go round, r = 7m

Perimeter of the merry go round = 2 x pi x r = 2 x 22 / 7 x 7 = 44 m
Arc distance between two swings = 4 m
Hence nos. of swings, n on a circle, should satisfy the following equation:
nx arc distance between two swings = Perimeter of the merry go round
Hence, n x 4 = 44
Hence, n = 44/4 = 11
Hence there are 11 swings on the merry go round.
Correct answer is option (B)

Area of the largest triangle that can be inscribed in a semi-circle of radius $r$ units is

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $r^2$ sq. units

  2. $\dfrac{1}{2} r^2$ sq. units

  3. $2r^2$ sq. units

  4. $\sqrt{2} r^2$ sq. units

Show answer
Correct Option: A
Explanation:

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 = $r^2$

If a circular grass lawn of $35\ m$ in radius has a path $7\ m$ wide running around it on the outside, then the area of the path is

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $1450\ m^2$

  2. $1576\ m^2$

  3. $1694\ m^2$

  4. $3368\ m^2$

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Correct Option: C
Explanation:

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 $= \pi (42)^2 - \pi (35)^2 = \dfrac{22}{7} \times (42 + 35)(42 - 35) = 22 \times 77 = 1694\ m^2$

A wire in the shape of an equilateral triangle encloses an area $s$ sq. cm  If the same wire is bent to form circle, the area of the circle will be

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $\displaystyle \frac{\pi s^{2}}{9}$

  2. $\displaystyle \frac{3s^{2}}{\pi }$

  3. $\displaystyle \frac{3s}{\pi }$

  4. $\displaystyle \frac{3\sqrt{3}s}{\pi }$

Show answer
Correct Option: D
Explanation:

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

A bicycle wheel has diameter 1m. If the bicycle travels one kilometer, then the number of revolutions the wheel make is.

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $\dfrac {1}{\Pi }$

  2. $\dfrac {100}{\Pi }$

  3. $\dfrac {500}{\Pi }$

  4. $\dfrac {1000}{\Pi }$

Show answer
Correct Option: D
Explanation:

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, fastened to the corner of a rectangular building. Calculate, about how much area does the dog have to move in.

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $27\ ft^{2}$

  2. $36\ ft^{2}$

  3. $56.55\ ft^{2}$

  4. $84.82\ ft^{2}$

Show answer
Correct Option: D
Explanation:

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 $= \cfrac{270}{360} \times \pi \times 6^2$
$= \cfrac{3}{4} \times \pi \times 36$
$= 84.82 \ \ ft^2$

Tick the correct answer in the following:
Area of a sector of angle $\theta$ (in degrees) of a circle with radius R is

maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

  1. $\dfrac {\theta}{180}\times 2\pi R$

  2. $\dfrac {\theta}{180}\times \pi R^{2}$

  3. $\dfrac {\theta}{3600}\times 2\pi R$

  4. $\dfrac {\theta}{720}\times 2\pi R^{2}$

Show answer
Correct Option: D
Explanation:

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

The perimeter of a sector of a circle is $56$ cms and the area of the circle is $64\pi$ sq. cms  Find the area of sector.

maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

  1. $360cm^2$

  2. $160cm^2$

  3. $260cm^2$

  4. None of these

Show answer
Correct Option: B
Explanation:

Area $= \pi r^{2}=64\pi cm^{2}$  


$\Rightarrow r=8cm$ 


perimeter $=2r+r\theta $ 

perimeter of sector $=r(\theta +2)=56cm$ 

$\Rightarrow \theta =5rad$ 

Area of sector $=\dfrac{r^{2}\theta }{2}=\dfrac{64}{2}\times 5cm^{2}$

                        $=160cm^{2}$

In a circle with radius $5.7\ cm$, the perimeter of a sector is $27.2\ cm$. Find the area of this sector.

maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

  1. $97.52cm^2$

  2. $57.52cm^2$

  3. $77.52cm^2$

  4. $87.52cm^2$

Show answer
Correct Option: C
Explanation:
$R=5.7 cm$
Perimeter = $R\theta =27.2 cm$
$\therefore R\theta = 27.2 cm$
$\theta = \left(\dfrac{27.2}{5.7}\right)^{c}$
$\therefore $ Area of sector $=\dfrac{1}{2}R^{2}\theta $
$=\dfrac{1}{2}\times (5.7)^{2}\times \dfrac{27.2}{5.7}$
$=\dfrac{5.7}{2}\times 27.2 cm^{2}$
$ = 5.7 \times 13.6 = 77.52 cm^{2}$

The angle of sector with area equal to one fifth of total area of whole circle 

maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

  1. 72

  2. 80

  3. 60

  4. 45

Show answer
Correct Option: A
Explanation:

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

The area of sector is $\dfrac 15\pi r^2$
The area of sector is given as $\dfrac{x}{360}\times \pi r^=\pi r^2\\dfrac x{360}=\dfrac 15\x=72$