Tag: area between two concentric circles

Questions Related to area between two concentric circles

What is the area of the semicircle if its circumference is $12.5\ cm$?

  1. $6.28$ $cm^2$

  2. $4.28$ $cm^2$

  3. $2.28$ $cm^2$

  4. $1.28$ $cm^2$


Correct Option: A
Explanation:

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

Calculate the area of a circular ring whose outer and inner radii are $12$ and $10\ cm.$

  1. $118.16$ $cm^2$

  2. $128.16$ $cm^2$

  3. $138.16$ $cm^2$

  4. $148.16$ $cm^2$


Correct Option: C
Explanation:

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

The area of the semicircle is $200\ m^2$. What is the radius of the semicircle?

  1. $10\ m$

  2. $11\ m$

  3. $12\ m$

  4. $13\ m$


Correct Option: B
Explanation:

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

Jackson measured the button on his shirt. Then he calculated that it has a semicircle of $25.12\ mm$. What is the button's radius? (Use $\pi = 3.14$).

  1. $1\ mm$

  2. $2\ mm$

  3. $3\ mm$

  4. $4\ mm$


Correct Option: D
Explanation:

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

Using $\pi = 3.14$ (given)
$16 = r^2$
$r = 4\ mm$

A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost graveling the path at Rs 4 per square meter.

  1. $Rs 2002$

  2. $Rs 2003$

  3. $Rs 2004$

  4. $Rs 2000$


Correct Option: A
Explanation:

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

             Radius ($r _2$) $=24.5m$

Required area $=\pi (r _2^{2}-r _1^{2})$

Required area $=\pi (24.5^{2}-21^{2})=500.5$

Cost $=500.5\times 4$

         $=2002Rs$

The front wheels of a wagon are $2$ m in circumference and the back wheels are $3$ m feet in circumference. When the front wheels have made $10$ more revolutions than the back wheels, how many metres has the wagon travelled?

  1. $40$ feet

  2. $50$ feet 

  3. $60$ feet

  4. $80$ feet


Correct Option: C

The area in ( ${cm^2}$) of the largest triangle that can be inscribed in a semicircle of radius r cm is 

  1. ${\cfrac{1}{3}\pi r^2}$

  2. ${2r^2}$

  3. ${r^2}$

  4. ${\cfrac{1}{2}\pi r^2}$


Correct Option: C
Explanation:

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

base = diameter of the circle
$\therefore$ Area of a triangle $=\dfrac{1}{2}\times base\times height$
                                  $=\dfrac{1}{2}\times 2r \times r=r^2$
Hence, option C is correct.

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.

  1. $13\ swings$.

  2. $11\ swings$.

  3. $15\ swings$.

  4. None of these


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

  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


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

  1. $1450\ m^2$

  2. $1576\ m^2$

  3. $1694\ m^2$

  4. $3368\ m^2$


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