Mix Examples - Areas Related to Circles Questions in English

(A) Diameter of the wheel $= 80\, cm$.
Radius $(r)$ of the wheel $= 40\, cm$.
Circumference of the wheel $= 2\pi r = 2 \times \frac{22}{7} \times 40 = \frac{1760}{7}\, cm$.
Speed of the car $= 66\, km/h = \frac{66 \times 100000\, cm}{60\, min} = 110000\, cm/min$.
Distance travelled in $10\, minutes = 110000 \times 10 = 1100000\, cm$.
Let the number of revolutions be $n$.
$n \times \text{Circumference} = \text{Total distance}$.
$n \times \frac{1760}{7} = 1100000$.
$n = \frac{1100000 \times 7}{1760} = \frac{110000 \times 7}{176} = 4375$.
Thus,each wheel makes $4375$ complete revolutions.

(A) Let the radius of the circle be $r$ and the side of the square be $a$.
According to the given condition,the circumference of the circle equals the perimeter of the square:
$2 \pi r = 4a$
$\pi r = 2a$
$a = \frac{\pi r}{2}$
Now,the area of the circle is $A_1 = \pi r^2$.
The area of the square is $A_2 = a^2 = (\frac{\pi r}{2})^2 = \frac{\pi^2 r^2}{4}$.
Comparing $A_1$ and $A_2$:
$\frac{A_1}{A_2} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}} = \frac{4}{\pi}$.
Since $\pi \approx 3.14$,$\frac{4}{3.14} > 1$.
Therefore,$A_1 > A_2$,which means the area of the circle is greater than the area of the square.

(B) To find the area of the largest triangle inscribed in a semi-circle,we consider the base of the triangle as the diameter of the semi-circle.
Let the diameter be $AB = 2r$.
For the triangle to have the maximum area,its height must be the maximum possible,which is the radius of the semi-circle,$r$.
Let $C$ be the vertex on the circumference such that the altitude $CD$ is perpendicular to $AB$. Here,$CD = r$.
Thus,the area of the triangle $ABC = \frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times (2r) \times r = r^{2}$ sq. units.

(C) Let the radius of the circle be $r$ and the side of the square be $a$.
According to the given condition,
Perimeter of a circle = Perimeter of a square
$2 \pi r = 4a \Rightarrow a = \frac{\pi r}{2}$ ...............$(i)$
Now,the ratio of their areas is:
$\frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\frac{\pi r}{2})^2}$ [from Eq. $(i)$]
$= \frac{\pi r^2}{\frac{\pi^2 r^2}{4}} = \frac{4}{\pi} = \frac{4}{\frac{22}{7}} = \frac{4 \times 7}{22} = \frac{28}{22} = \frac{14}{11}$
Thus,the ratio of their areas is $14: 11$.

(B) Given that the radius of the circle is $r = OC = 8 \, cm$.
Therefore,the diameter of the circle is $AC = 2 \times OC = 2 \times 8 = 16 \, cm$.
This diameter is equal to the length of the diagonal of the square.
Let the side of the square be $x$.
In the right-angled $\triangle ABC$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
$(16)^2 = x^2 + x^2$.
$256 = 2x^2$.
$x^2 = 128$.
Therefore,the area of the square is $x^2 = 128 \, cm^2$.

(C) Let the diameter of the first circle be $d_1 = 36 \, cm$ and the diameter of the second circle be $d_2 = 20 \, cm$.
Circumference of the first circle $C_1 = \pi d_1 = 36 \pi \, cm$.
Circumference of the second circle $C_2 = \pi d_2 = 20 \pi \, cm$.
Let the diameter of the new circle be $D$ and its circumference be $C$.
According to the problem,$C = C_1 + C_2$.
$\pi D = 36 \pi + 20 \pi$.
$\pi D = 56 \pi$.
$D = 56 \, cm$.
The radius of the new circle is $r = \frac{D}{2} = \frac{56}{2} = 28 \, cm$.

(A) Yes,it is true.
Let the side of the square be $a = 5 \, cm$.
The diameter of the inner circle is equal to the side of the square,so $d_{inner} = a = 5 \, cm$. The radius is $r = \frac{5}{2} \, cm$.
The area of the inner circle is $A_{inner} = \pi r^2 = \pi \left(\frac{5}{2}\right)^2 = \frac{25\pi}{4} \, cm^2$.
The diameter of the outer circle is equal to the diagonal of the square,so $d_{outer} = a\sqrt{2} = 5\sqrt{2} \, cm$. The radius is $R = \frac{5\sqrt{2}}{2} \, cm$.
The area of the outer circle is $A_{outer} = \pi R^2 = \pi \left(\frac{5\sqrt{2}}{2}\right)^2 = \pi \left(\frac{25 \times 2}{4}\right) = \frac{50\pi}{4} = \frac{25\pi}{2} \, cm^2$.
Comparing the two areas: $\frac{A_{outer}}{A_{inner}} = \frac{25\pi / 2}{25\pi / 4} = \frac{4}{2} = 2$.
Thus,the area of the outer circle is indeed two times the area of the inner circle.

(N/A) False.
Let $ABCD$ be a square of side $a$.
The circle is inscribed in the square,so its diameter is equal to the side of the square.
$\therefore \text{Diameter of circle} = a \, cm$.
$\therefore \text{Radius of circle} (r) = \frac{a}{2} \, cm$.
Now,the area of the circle is given by the formula $\pi r^2$.
$\therefore \text{Area} = \pi \left( \frac{a}{2} \right)^2 = \pi \left( \frac{a^2}{4} \right) = \frac{\pi a^2}{4} \, cm^2$.
Since $\frac{\pi a^2}{4} \neq \pi a^2$,the given statement is false.

(A) True.
Given,radius of the circle,$r = a \, cm$.
Therefore,the diameter of the circle,$d = 2 \times \text{radius} = 2a \, cm$.
Since the square circumscribes the circle,the side of the square is equal to the diameter of the circle.
Therefore,the side of the square $= 2a \, cm$.
Now,the perimeter of a square $= 4 \times \text{side} = 4 \times 2a = 8a \, cm$.
Hence,it is true that the perimeter of the square is $8a \, cm$.

(N/A) The statement is False.
Given the diameter of the circle is $d$.
$1$. For the inner square inscribed in the circle,the diagonal of the square is equal to the diameter of the circle.
Diagonal $= d$.
Let the side of the inner square be $x$.
Using the property of a square,the diagonal is $\sqrt{2} \times \text{side}$.
So,$\sqrt{2}x = d \Rightarrow x = \frac{d}{\sqrt{2}}$.
Area of the inner square $= x^2 = (\frac{d}{\sqrt{2}})^2 = \frac{d^2}{2}$.
$2$. For the outer square circumscribing the circle,the side of the square is equal to the diameter of the circle.
Side $= d$.
Area of the outer square $= d^2$.
$3$. Comparing the areas:
Ratio $= \frac{\text{Area of outer square}}{\text{Area of inner square}} = \frac{d^2}{d^2/2} = 2$.
Thus,the area of the outer square is twice the area of the inner square,not four times.

(B) False.
The distance travelled by a circular wheel in one revolution is equal to its circumference.
The circumference of a circle with diameter $d$ is given by the formula $\pi d$.
Since $d = 2r$ (where $r$ is the radius),the circumference is $\pi(2r) = 2 \pi r$.
Therefore,the distance travelled in one revolution is $\pi d$,not $2 \pi d$.

(A) The statement is true.
The distance covered by a wheel in one complete revolution is equal to its circumference,which is given by the formula $2 \pi r$ metres.
To find the total number of revolutions made in covering a distance of $s$ metres,we divide the total distance by the distance covered in one revolution:
$\text{Number of revolutions} = \frac{\text{Total distance}}{\text{Distance in one revolution}} = \frac{s}{2 \pi r}$.

(A) Yes, it is necessary that their areas be equal.
Let the circumferences of two circles be $C_1$ and $C_2$, and their radii be $r_1$ and $r_2$ respectively.
Given that $C_1 = C_2$, we have $2\pi r_1 = 2\pi r_2$.
Dividing both sides by $2\pi$, we get $r_1 = r_2$.
The area of a circle is given by $A = \pi r^2$.
Since $r_1 = r_2$, it follows that $\pi r_1^2 = \pi r_2^2$, which means $A_1 = A_2$.
Therefore, if the circumferences of two circles are equal, their areas must also be equal.

(A) Yes,it is necessary that their circumferences are equal.
Let the radii of the two circles be $r_1$ and $r_2$ respectively.
The area of a circle is given by the formula $A = \pi r^2$.
Given that the areas are equal,we have $\pi r_1^2 = \pi r_2^2$.
Dividing both sides by $\pi$,we get $r_1^2 = r_2^2$,which implies $r_1 = r_2$ (since radius must be positive).
The circumference of a circle is given by $C = 2\pi r$.
Since $r_1 = r_2$,it follows that $2\pi r_1 = 2\pi r_2$.
Therefore,the circumferences of the two circles must be equal.

(A) Given,diameter of the first circle $d_{1} = 20 \, cm$,so its radius $r_{1} = \frac{20}{2} = 10 \, cm$.
Diameter of the second circle $d_{2} = 48 \, cm$,so its radius $r_{2} = \frac{48}{2} = 24 \, cm$.
Sum of the areas of the two circles $= \pi r_{1}^{2} + \pi r_{2}^{2} = \pi(10)^{2} + \pi(24)^{2} = \pi(100 + 576) = 676\pi \, cm^{2}$.
Let the radius of the new circle be $r$. Its area is $\pi r^{2}$.
According to the problem,$\pi r^{2} = 676\pi$.
$r^{2} = 676$.
$r = \sqrt{676} = 26 \, cm$.
Therefore,the diameter of the new circle $= 2r = 2 \times 26 = 52 \, cm$.

(C) Area of the circle $= \pi r^2$
$= 3.14 \times (7.5)^2 \, cm^2$
$= 3.14 \times 56.25 \, cm^2 = 176.625 \, cm^2$
Clearly, the side of the square $=$ diameter of the circle $= 2 \times 7.5 \, cm = 15 \, cm$.
So, the area of the square $= (\text{side})^2 = 15^2 \, cm^2 = 225 \, cm^2$.
Therefore, the area of the shaded region $=$ Area of the square $-$ Area of the circle
$= 225 \, cm^2 - 176.625 \, cm^2 = 48.375 \, cm^2$.

(N/A) Let the side of the square be $a$ and the radius of the circle be $r$.
Given that,the length of the diagonal of the square $= 8\, cm$.
Since the diagonal of a square $= a\sqrt{2}$,we have:
$a\sqrt{2} = 8$
$a = \frac{8}{\sqrt{2}} = 4\sqrt{2}\, cm$.
When a square is inscribed in a circle,the diagonal of the square is equal to the diameter of the circle.
Therefore,the diameter of the circle $= 8\, cm$.
Radius of the circle $r = \frac{\text{Diameter}}{2} = \frac{8}{2} = 4\, cm$.
Area of the circle $= \pi r^2 = \pi(4)^2 = 16\pi\, cm^2$.
Area of the square $= a^2 = (4\sqrt{2})^2 = 16 \times 2 = 32\, cm^2$.
The area of the shaded region is the difference between the area of the circle and the area of the square:
Area of shaded region $= \text{Area of circle} - \text{Area of square}$
$= (16\pi - 32)\, cm^2$.
Hence,the required area of the shaded region is $(16\pi - 32)\, cm^2$.

(D) Given,radius of the wheel,$r = 35 \, cm$.
The circumference of the wheel is given by $C = 2 \pi r$.
$C = 2 \times \frac{22}{7} \times 35 = 220 \, cm$.
The speed of the motorcycle is $66 \, km/h$. We need to convert this into $cm/min$:
$66 \, km/h = \frac{66 \times 1000 \, m}{60 \, min} = 1100 \, m/min$.
Since $1 \, m = 100 \, cm$,the speed is $1100 \times 100 = 110000 \, cm/min$.
The number of revolutions per minute is calculated by dividing the total distance covered in one minute by the circumference of the wheel:
$\text{Number of revolutions} = \frac{\text{Total distance per minute}}{\text{Circumference}} = \frac{110000}{220} = 500$.
Therefore,the wheel must make $500$ revolutions per minute.

(N/A) The flower bed consists of a central rectangular part and two semi-circular ends.
The length of the rectangular part is $38 \, cm$ and its breadth is $10 \, cm$.
Area of the rectangle $= \text{Length} \times \text{Breadth} = 38 \times 10 = 380 \, cm^2$.
The two semi-circular ends together form one complete circle with a diameter of $10 \, cm$.
Radius of the circle $(r) = \frac{10}{2} = 5 \, cm$.
Area of the two semi-circles $= \pi r^2 = \pi \times (5)^2 = 25 \pi \, cm^2$.
Total area of the flower bed $= \text{Area of rectangle} + \text{Area of two semi-circles} = (380 + 25 \pi) \, cm^2$.

(C) Given,$AC = 6 \, cm$ and $BC = 8 \, cm$.
We know that an angle in a semi-circle is a right angle. Therefore,$\angle C = 90^{\circ}$.
In the right-angled $\triangle ACB$,using the Pythagoras theorem:
$AB^2 = AC^2 + BC^2$
$AB^2 = 6^2 + 8^2 = 36 + 64 = 100$
$AB = \sqrt{100} = 10 \, cm$ (since length cannot be negative).
Area of $\triangle ACB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24 \, cm^2$.
The diameter of the circle is $AB = 10 \, cm$,so the radius $r = \frac{10}{2} = 5 \, cm$.
Area of the circle $= \pi r^2 = 3.14 \times (5)^2 = 3.14 \times 25 = 78.5 \, cm^2$.
Area of the shaded region = Area of the circle - Area of $\triangle ACB$
$= 78.5 - 24 = 54.5 \, cm^2$.

(N/A) In the given figure,join $DE$.
From the figure,the diameter of the semi-circle $DFE$ is $6 - 4 = 2 \, m$. Therefore,the radius $r = 2 / 2 = 1 \, m$.
Now,the area of the rectangle $ABCD$ is $BC \times AB = 8 \times 4 = 32 \, m^2$.
The area of the semi-circle $DFE$ is $\frac{\pi r^2}{2} = \frac{\pi}{2}(1)^2 = 0.5 \pi \, m^2$.
Therefore,the total area of the shaded region = Area of rectangle $ABCD$ + Area of semi-circle $DFE$.
Total area = $(32 + 0.5 \pi) \, m^2$.

(B) The sum of the areas of the three sectors at vertices $A, B,$ and $C$ is given by:
Area $= \frac{\angle A}{360^{\circ}} \times \pi r^2 + \frac{\angle B}{360^{\circ}} \times \pi r^2 + \frac{\angle C}{360^{\circ}} \times \pi r^2$
$= \frac{(\angle A + \angle B + \angle C)}{360^{\circ}} \times \pi r^2$
Since the sum of angles in a triangle is $180^{\circ}$,the area of the three sectors is:
$= \frac{180^{\circ}}{360^{\circ}} \times 3.14 \times (5)^2 = 0.5 \times 3.14 \times 25 = 39.25 \, cm^2$
Now,we calculate the area of $\triangle ABC$. Since $14^2 + 48^2 = 196 + 2304 = 2500 = 50^2$,the triangle is a right-angled triangle with base $48 \, cm$ and height $14 \, cm$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 48 \times 14 = 336 \, cm^2$
The area of the shaded region is the area of the triangle minus the area of the three sectors:
Area $= 336 - 39.25 = 296.75 \, cm^2$

(A) Given,diameter of the front wheel,$d_{1} = 80 \,cm$.
Diameter of the rear wheel,$d_{2} = 2 \,m = 200 \,cm$.
Circumference of the front wheel $C_{1} = \pi d_{1} = 80\pi \,cm$.
Circumference of the rear wheel $C_{2} = \pi d_{2} = 200\pi \,cm$.
Total distance covered by the front wheel in $1400$ revolutions is $D = 1400 \times C_{1} = 1400 \times 80\pi \,cm$.
Let the number of revolutions made by the rear wheel be $n$.
Then,$n \times C_{2} = D$.
$n \times 200\pi = 1400 \times 80\pi$.
$n = \frac{1400 \times 80\pi}{200\pi} = 7 \times 80 = 560$.
Thus,the rear wheel makes $560$ revolutions.

(A) Given,a triangular field with sides $a = 15 \, m$,$b = 16 \, m$,and $c = 17 \, m$.
Each animal is tied at a corner with a rope of length $r = 7 \, m$. Each animal grazes a sector of the field at its respective corner.
The sum of the areas of the three sectors is given by:
Sum of areas $= \frac{\angle A}{360^{\circ}} \pi r^2 + \frac{\angle B}{360^{\circ}} \pi r^2 + \frac{\angle C}{360^{\circ}} \pi r^2 = \frac{(\angle A + \angle B + \angle C)}{360^{\circ}} \pi r^2$
Since the sum of angles in a triangle is $180^{\circ}$,the sum of areas $= \frac{180^{\circ}}{360^{\circ}} \times \pi \times (7)^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = 77 \, m^2$.
Now,calculate the area of the triangular field using Heron's formula:
Semi-perimeter $s = \frac{15 + 16 + 17}{2} = \frac{48}{2} = 24 \, m$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{24(24-15)(24-16)(24-17)} = \sqrt{24 \times 9 \times 8 \times 7} = \sqrt{12096} = 24\sqrt{21} \, m^2$.
The area which cannot be grazed is the difference between the total area and the grazed area:
Area not grazed $= (24\sqrt{21} - 77) \, m^2$.

(D) Given that,a circular pond is surrounded by a path of width $2 \, m$.
Diameter of the circular pond $= 17.5 \, m$.
Radius of the circular pond $(r_i) = \frac{17.5}{2} = 8.75 \, m$.
Width of the path $= 2 \, m$.
Radius of the outer circle including the path $(r_e) = 8.75 + 2 = 10.75 \, m$.
Area of the circular path $= \pi r_e^2 - \pi r_i^2 = \pi(r_e^2 - r_i^2) = \pi(r_e + r_i)(r_e - r_i)$.
Area $= \frac{22}{7} \times (10.75 + 8.75) \times (10.75 - 8.75) = \frac{22}{7} \times 19.5 \times 2$.
Area $= \frac{22 \times 39}{7} \approx 122.57 \, m^2$.
Using $\pi \approx 3.14$,Area $= 3.14 \times 19.5 \times 2 = 122.46 \, m^2$.
Cost of constructing the path $= \text{Area} \times \text{Rate} = 122.46 \times 25 = Rs. \, 3061.50$.

(B) Given that,three circles are drawn such that each of them touches the other two.
Let the centers of the three circles be $A, B,$ and $C$. Since each circle has a radius $r = 3.5\, cm$,the distance between the centers of any two touching circles is $2r = 2 \times 3.5 = 7\, cm$.
Thus,$AB = BC = CA = 7\, cm$. This forms an equilateral triangle $\triangle ABC$ with side length $a = 7\, cm$.
The area of the equilateral triangle $\triangle ABC$ is given by:
$\text{Area} = \frac{\sqrt{3}}{4} \times a^{2} = \frac{\sqrt{3}}{4} \times (7)^{2} = \frac{49\sqrt{3}}{4} \approx \frac{49 \times 1.732}{4} \approx 21.2176\, cm^{2}$.
The area enclosed between the three circles is the area of the triangle minus the area of the three sectors inside the triangle. Each angle of the equilateral triangle is $60^{\circ}$.
Total area of the three sectors $= 3 \times \left( \frac{60^{\circ}}{360^{\circ}} \times \pi r^{2} \right) = 3 \times \frac{1}{6} \times \pi \times (3.5)^{2} = \frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5 = 11 \times 0.5 \times 3.5 = 19.25\, cm^{2}$.
Therefore,the area enclosed between the circles $= \text{Area of } \triangle ABC - \text{Total area of the three sectors}$
$= 21.2176 - 19.25 = 1.9676\, cm^{2}$.
Rounding to three decimal places,the required area is $1.967\, cm^{2}$.

(A) Given, area of square sheet $= 784 \, cm^{2}$.
Since, area of square $= (\text{side})^{2}$, we have $(\text{side})^{2} = 784$.
Therefore, side $= \sqrt{784} = 28 \, cm$.
Since four congruent circular plates are placed such that each touches the other two and the sides of the square, the diameter of each circular plate is half the side of the square.
Diameter of each circular plate $= 28 / 2 = 14 \, cm$.
Radius of each circular plate $(r) = 14 / 2 = 7 \, cm$.
Area of one circular plate $= \pi r^{2} = \frac{22}{7} \times (7)^{2} = 22 \times 7 = 154 \, cm^{2}$.
Area of four circular plates $= 4 \times 154 = 616 \, cm^{2}$.
Area of the square sheet not covered by the circular plates $= \text{Area of square} - \text{Area of four circular plates}$.
$= 784 - 616 = 168 \, cm^{2}$.

(B) Given,the floor of a room is covered with circular tiles.
Length of the floor $(l) = 5 \, m$.
Breadth of the floor $(b) = 4 \, m$.
Area of the floor $= l \times b = 5 \times 4 = 20 \, m^2$.
Diameter of each circular tile $= 50 \, cm = 0.5 \, m$.
Radius of each circular tile $(r) = \frac{0.5}{2} = 0.25 \, m = \frac{1}{4} \, m$.
Area of one circular tile $= \pi r^2 = 3.14 \times (0.25)^2 = 3.14 \times 0.0625 = 0.19625 \, m^2$.
From the figure,the number of tiles along the length $= \frac{5 \, m}{0.5 \, m} = 10$ tiles.
The number of tiles along the breadth $= \frac{4 \, m}{0.5 \, m} = 8$ tiles.
Total number of tiles $= 10 \times 8 = 80$ tiles.
Total area covered by $80$ tiles $= 80 \times 0.19625 = 15.7 \, m^2$.
Area of the floor that remains uncovered $=$ Total area of floor $-$ Total area of tiles.
Area uncovered $= 20 - 15.7 = 4.3 \, m^2$.

(C) Let the radius of the circle be $r$.
Given that,Area of the circle $= 1256 \, cm^{2}$.
$\pi r^{2} = 1256$
$r^{2} = \frac{1256}{3.14} = 400$
$r = \sqrt{400} = 20 \, cm$.
Since all the vertices of a rhombus lie on a circle,the rhombus must be a square because its diagonals are diameters of the circle and they bisect each other at $90^{\circ}$.
Thus,the diagonals $d_{1}$ and $d_{2}$ are equal to the diameter of the circle.
$d_{1} = d_{2} = 2 \times r = 2 \times 20 = 40 \, cm$.
Area of the rhombus $= \frac{1}{2} \times d_{1} \times d_{2}$
$= \frac{1}{2} \times 40 \times 40$
$= 20 \times 40 = 800 \, cm^{2}$.
Hence,the required area of the rhombus is $800 \, cm^{2}$.

(A) The formula for the circumference of a circle is $C = \pi d$,where $C$ is the circumference and $d$ is the diameter.
Given that $C = 251.2 \, cm$ and $\pi = 3.14$.
Substituting the values into the formula: $251.2 = 3.14 \times d$.
To find the diameter $d$,divide the circumference by $\pi$: $d = \frac{251.2}{3.14}$.
$d = 80 \, cm$.
Therefore,the diameter of the circle is $80 \, cm$.

(D) The distance covered in $1$ rotation is equal to the circumference of the wheel.
Circumference $= 2 \pi r = 2 \times \frac{22}{7} \times 21 = 132 \, cm$.
Distance covered in $1$ minute $= 800 \times 132 \, cm = 105600 \, cm$.
Distance covered in $1$ hour ($60$ minutes) $= 105600 \times 60 \, cm = 6336000 \, cm$.
To convert centimeters to kilometers,we divide by $100$ (to get meters) and then by $1000$ (to get kilometers).
Speed $= \frac{6336000}{100 \times 1000} \, km/h = 63.36 \, km/h$.
Thus,the speed of the car is $63.36 \, km/h$.

(A) Given: Radius $(r) = 12 \, cm$ and $\pi = 3.14$.
Circumference of a circle $= 2 \pi r = 2 \times 3.14 \times 12 = 75.36 \, cm$.
Area of a circle $= \pi r^2 = 3.14 \times (12)^2 = 3.14 \times 144 = 452.16 \, cm^2$.
Thus,the circumference is $75.36 \, cm$ and the area is $452.16 \, cm^2$.

(N/A) $1$. Cost of fencing: The boundary of the circular ground is its circumference,given by $C = 2 \pi r$.
Using $r = 63 \, m$ and $\pi = 22/7$,we get $C = 2 \times (22/7) \times 63 = 2 \times 22 \times 9 = 396 \, m$.
The cost of fencing is $396 \, m \times ₹ 50/m = ₹ 19,800$.
$2$. Cost of levelling: The area of the circular ground is $A = \pi r^2$.
$A = (22/7) \times 63 \times 63 = 22 \times 9 \times 63 = 12,474 \, m^2$.
The cost of levelling is $12,474 \, m^2 \times ₹ 40/m^2 = ₹ 4,98,960$.

(D) The area of the circular sheet is given by $A_{circle} = \pi r^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386\, cm^2$.
$A$ regular hexagon with side length $a = 21\, cm$ consists of $6$ equilateral triangles with side $21\, cm$.
The area of the regular hexagon is $A_{hexagon} = 6 \times \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} \times (21)^2$.
Substituting $\sqrt{3} = 1.73$ and $a = 21$:
$A_{hexagon} = \frac{3 \times 1.73}{2} \times 441 = 1.5 \times 1.73 \times 441 = 2.595 \times 441 = 1144.395\, cm^2$.
The area of the remaining sheet is $A_{remaining} = A_{circle} - A_{hexagon} = 1386 - 1144.395 = 241.605\, cm^2$.

(A) $1$. The diagonal of the rectangle $ABCD$ is the diameter of the circle.
$2$. Using the Pythagorean theorem in $\triangle ABC$,the diagonal $AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, cm$.
$3$. The diameter of the circle is $10 \, cm$,so the radius $r = 5 \, cm$.
$4$. The area of the circle is $\pi r^2 = 3.14 \times 5^2 = 3.14 \times 25 = 78.5 \, cm^2$.
$5$. The area of the rectangle $ABCD$ is $AB \times BC = 8 \times 6 = 48 \, cm^2$.
$6$. The area of the shaded region is the area of the circle minus the area of the rectangle: $78.5 - 48 = 30.5 \, cm^2$.

(C) $1$. Area of rectangle $ABCD = \text{length} \times \text{breadth} = 20 \times 14 = 280 \, cm^2$.
$2$. Area of the semicircle with diameter $BC = 14 \, cm$ (radius $r = 7 \, cm$): $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77 \, cm^2$.
$3$. Area of the sector with centre $A$ and radius $AD = 14 \, cm$ (angle $\theta = 90^{\circ}$): $\frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \, cm^2$.
$4$. Area of the remaining sheet = Area of rectangle - (Area of semicircle + Area of sector) = $280 - (77 + 154) = 280 - 231 = 49 \, cm^2$.

(C) Given: $AB = BC = 14 \text{ cm}$ and $\angle B = 90^{\circ}$.
First,find the length of the hypotenuse $AC$ using the Pythagorean theorem:
$AC^2 = AB^2 + BC^2 = 14^2 + 14^2 = 196 + 196 = 392$.
$AC = \sqrt{392} = 14\sqrt{2} \text{ cm}$.
The radius of the semicircle with diameter $AC$ is $r = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \text{ cm}$.
Area of the semicircle with diameter $AC = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times 98 = 11 \times 14 = 154 \text{ cm}^2$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2$.
Area of the segment of the circle (bounded by chord $AC$ and arc $APC$) = Area of sector $BAPC$ - Area of $\Delta ABC$.
Area of sector $BAPC = \frac{90}{360} \times \pi \times (14)^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = 154 \text{ cm}^2$.
Area of segment $APC = 154 - 98 = 56 \text{ cm}^2$.
The shaded region is the area of the semicircle minus the area of the segment $APC$.
Area of shaded region = $154 - 56 = 98 \text{ cm}^2$.

(B) Given: Radius of the large circle $R = 7\, cm$.
Diameter of the small circle is $OD = R = 7\, cm$,so its radius $r = 3.5\, cm$.
Area of the small shaded circle = $\pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5\, cm^2$.
Area of $\triangle ABC$ = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OC = \frac{1}{2} \times 14 \times 7 = 49\, cm^2$.
Area of the semi-circle $ACB$ = $\frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77\, cm^2$.
Area of the shaded segment (excluding the small circle) = Area of semi-circle $ACB$ - Area of $\triangle ABC = 77 - 49 = 28\, cm^2$.
Total shaded area = Area of small circle + Area of shaded segment = $38.5 + 28 = 66.5\, cm^2$.

(D) The side length of the square $ABCD$ is $s = 35 \, cm$.
The area of the square $ABCD = s^2 = 35^2 = 1225 \, cm^2$.
There are two semicircles drawn on sides $\overline{AB}$ and $\overline{CD}$ with diameter $d = 35 \, cm$,so the radius $r = \frac{35}{2} = 17.5 \, cm$.
The area of two semicircles = $2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \frac{22}{7} \times 17.5 \times 17.5 = 22 \times 2.5 \times 17.5 = 962.5 \, cm^2$.
The area of the shaded region = (Area of square) - (Area of two semicircles) = $1225 - 962.5 = 262.5 \, cm^2$.

(A) The radius of each circle is $r = 21\, cm$.
Diameter of each circle is $d = 2r = 2 \times 21 = 42\, cm$.
Since there are $3$ circles along each side of the square,the side length of the square is $S = 3 \times d = 3 \times 42 = 126\, cm$.
Area of the square $ABCD = S^2 = 126^2 = 15876\, cm^2$.
Area of one circle = $\pi r^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386\, cm^2$.
Area of $9$ circles = $9 \times 1386 = 12474\, cm^2$.
Area of the region without design = Area of square - Area of $9$ circles.
Area = $15876 - 12474 = 3402\, cm^2$.

(B) The figure consists of one large semicircle with diameter $AB$ and two smaller semicircles with diameters $OA$ and $OB$.
Given $OA = 70\, cm$ and $OB = 70\, cm$,the diameter of the large semicircle is $AB = OA + OB = 70 + 70 = 140\, cm$.
The radius of the large semicircle is $R = 140 / 2 = 70\, cm$.
The radius of each smaller semicircle is $r = 70 / 2 = 35\, cm$.
The area of the large semicircle is $\frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 70 \times 70 = 11 \times 10 \times 70 = 7700\, cm^2$.
The area of the two smaller semicircles is $2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3850\, cm^2$.
The total area of the figure is the sum of the area of the large semicircle and the two smaller semicircles: $7700 + 3850 = 11550\, cm^2$.

(A) Let the initial radius of the circle be $r$. The initial area is $A_1 = \pi r^2$.
If the radius is increased by $10 \%$,the new radius $r'$ becomes $r + 0.10r = 1.1r$.
The new area $A_2$ is $\pi (1.1r)^2 = \pi (1.21r^2) = 1.21 \pi r^2$.
The increase in area is $A_2 - A_1 = 1.21 \pi r^2 - \pi r^2 = 0.21 \pi r^2$.
The percentage increase in area is $\frac{0.21 \pi r^2}{\pi r^2} \times 100 \% = 21 \%$.

(B) Let the side of the square be $a$ and the diagonal be $d$.
The formula for the diagonal of a square is $d = a\sqrt{2}$.
Given $d = 10$,we have $a\sqrt{2} = 10$.
Therefore,$a = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.
The area of a square is given by $Area = a^2$.
$Area = (5\sqrt{2})^2 = 25 \times 2 = 50$.
Alternatively,the area of a square can be calculated using the diagonal as $Area = \frac{1}{2} \times d^2$.
$Area = \frac{1}{2} \times (10)^2 = \frac{1}{2} \times 100 = 50$.

(C) Let the side of the square be $a$.
The length of the diagonal of a square is given by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 8 \sqrt{2}$.
Equating the two,we get $a \sqrt{2} = 8 \sqrt{2}$.
Dividing both sides by $\sqrt{2}$,we find the side length $a = 8$.
The area of a square is calculated as $\text{Area} = a^2$.
Substituting the value of $a$,we get $\text{Area} = 8^2 = 64$.
Alternatively,the area of a square can be calculated using the diagonal $d$ as $\text{Area} = \frac{1}{2} \times d^2$.
Substituting $d = 8 \sqrt{2}$,we get $\text{Area} = \frac{1}{2} \times (8 \sqrt{2})^2 = \frac{1}{2} \times (64 \times 2) = \frac{1}{2} \times 128 = 64$.

(D) The area of a square can be calculated using the length of its diagonal $d$ with the formula: $\text{Area} = \frac{d^2}{2}$.
Given the length of the diagonal $d = 6$.
Substituting the value into the formula:
$\text{Area} = \frac{6^2}{2} = \frac{36}{2} = 18$.

(C) The length of a minor arc of a circle with radius $r$ and central angle $\theta$ (in degrees) is given by the formula $l_{minor} = \frac{\theta}{360} \times 2 \pi r = \frac{\pi r \theta}{180}$.
Since the total circumference of the circle is $2 \pi r$,the length of the major arc is the total circumference minus the length of the minor arc.
Therefore,the length of the major arc $l_{major} = 2 \pi r - \frac{\pi r \theta}{180}$.

(D) For the circular ground,the diameter $d = 105 \, m$.
The length of the fence is equal to the circumference of the circular ground.
Circumference $= \pi d = \frac{22}{7} \times 105$.
$= 22 \times 15 = 330 \, m$.
Therefore,the length of the fence is $330 \, m$.

(D) The perimeter of a semicircular garden is the sum of its semi-circumference and its diameter.
Perimeter $= \pi r + 2r$
Given radius $r = 35 \, m$.
Perimeter $= \frac{22}{7} \times 35 + 2 \times 35$
Perimeter $= 22 \times 5 + 70$
Perimeter $= 110 + 70$
Perimeter $= 180 \, m$.
Therefore,one has to walk $180 \, m$ to make one complete round of the garden.

(C) The formula for the circumference of a circle is $C = 2\pi r$.
Given the radius $r = 8.4\,cm$.
Substituting the values,we get $C = 2 \times \frac{22}{7} \times 8.4$.
$C = 2 \times 22 \times 1.2$.
$C = 44 \times 1.2 = 52.8\,cm$.

(C) Distance covered in one rotation $= 440 \, cm$.
Distance covered in $250$ rotations $= 440 \times 250 \, cm$.
Since the wheel makes $250$ rotations per minute,the distance covered in $1$ minute is $440 \times 250 \, cm$.
Distance covered in $1$ hour ($60$ minutes) $= 440 \times 250 \times 60 \, cm$.
To convert this distance into kilometers,we divide by $100$ (to convert $cm$ to $m$) and then by $1000$ (to convert $m$ to $km$):
Distance in $km = \frac{440 \times 250 \times 60}{100 \times 1000} \, km$.
$= \frac{6,600,000}{100,000} \, km = 66 \, km$.
Therefore,the speed of the truck is $66 \, km/h$.