Tag: semicircle and ring

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

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

  1. $\displaystyle \frac{\pi s^{2}}{9}$
  2. $\displaystyle \frac{3s^{2}}{\pi }$
  3. $\displaystyle \frac{3s}{\pi }$
  4. $\displaystyle \frac{3\sqrt{3}s}{\pi }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

  1. $\dfrac {1}{\Pi }$
  2. $\dfrac {100}{\Pi }$
  3. $\dfrac {500}{\Pi }$
  4. $\dfrac {1000}{\Pi }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$n=\frac {1000}{\Pi }$

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

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.

  1. $27\ ft^{2}$
  2. $36\ ft^{2}$
  3. $56.55\ ft^{2}$
  4. $84.82\ ft^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $= \cfrac{270}{360} \times \pi \times 6^2$
$= \cfrac{3}{4} \times \pi \times 36$
$= 84.82 \ \ ft^2$