Mix Examples - Areas Related to Circles Questions in English

(C) Let the calf be tied at the corner $A$ of the square lawn.
The increase in area is the difference between the areas of two sectors with a central angle of $90^{\circ}$ each and radii $R = 11.5 \,m$ $(6 \,m + 5.5 \,m)$ and $r = 6 \,m$.
Required increase in area $= \text{Area of sector with radius } R - \text{Area of sector with radius } r$
$= \frac{90^{\circ}}{360^{\circ}} \times \pi \times R^2 - \frac{90^{\circ}}{360^{\circ}} \times \pi \times r^2$
$= \frac{1}{4} \times \pi \times (R^2 - r^2)$
$= \frac{1}{4} \times \frac{22}{7} \times (11.5^2 - 6^2)$
$= \frac{1}{4} \times \frac{22}{7} \times (11.5 + 6) \times (11.5 - 6)$
$= \frac{1}{4} \times \frac{22}{7} \times 17.5 \times 5.5$
$= \frac{1}{4} \times \frac{22}{7} \times \frac{35}{2} \times \frac{11}{2}$
$= \frac{11 \times 5 \times 11}{8} = \frac{605}{8} = 75.625 \,m^2$.

(D) Given,area of a circular playground $= 22176 \, m^{2}$.
Using the formula for the area of a circle,$\pi r^{2} = 22176$.
Substituting $\pi = \frac{22}{7}$,we get $\frac{22}{7} r^{2} = 22176$.
$r^{2} = \frac{22176 \times 7}{22} = 1008 \times 7 = 7056$.
$r = \sqrt{7056} = 84 \, m$.
The circumference of the circular ground is $2 \pi r = 2 \times \frac{22}{7} \times 84 = 44 \times 12 = 528 \, m$.
The cost of fencing is calculated as $\text{Circumference} \times \text{Rate}$.
Cost $= 528 \times 50 = 26400 \, Rs.$

(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 $= (24\sqrt{21} - 77) \, m^2$.

(N/A) Given that,radius of a circle $(r) = 12 \, cm$ and central angle of sector $(\theta) = 60^{\circ}$.
Area of sector $= \frac{\pi r^2 \theta}{360^{\circ}} = \frac{3.14 \times 12 \times 12 \times 60^{\circ}}{360^{\circ}} = 3.14 \times 2 \times 12 = 75.36 \, cm^2$.
Since the triangle formed by the two radii and the chord is an isosceles triangle with a vertex angle of $60^{\circ}$,it is an equilateral triangle.
Area of equilateral triangle $= \frac{\sqrt{3}}{4} \times (\text{side})^2 = \frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} \, cm^2$.
Area of the segment $=$ Area of sector $-$ Area of triangle $= (75.36 - 36\sqrt{3}) \, cm^2$.

(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 $= \text{Area} \times \text{Rate} = 122.46 \times 25 = Rs. \, 3061.50$.

(A) Given: $AB = 18 \, cm$,$DC = 32 \, cm$,height $(h) = 14 \, cm$,and radius of each arc $(r) = 7 \, cm$.
Since $AB \parallel DC$,the sum of consecutive interior angles is $180^{\circ}$. Thus,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.
Area of a sector with angle $\theta$ is given by $\frac{\theta}{360^{\circ}} \times \pi r^2$.
Area of sectors at $A$ and $D = \frac{\angle A + \angle D}{360^{\circ}} \times \pi r^2 = \frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (7)^2 = \frac{1}{2} \times 22 \times 7 = 77 \, cm^2$.
Similarly,the area of sectors at $B$ and $C = \frac{\angle B + \angle C}{360^{\circ}} \times \pi r^2 = \frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (7)^2 = 77 \, cm^2$.
Area of trapezium $ABCD = \frac{1}{2} \times (AB + DC) \times h = \frac{1}{2} \times (18 + 32) \times 14 = \frac{1}{2} \times 50 \times 14 = 350 \, cm^2$.
Area of the shaded region = Area of trapezium $-$ (Sum of areas of the four sectors) = $350 - (77 + 77) = 350 - 154 = 196 \, cm^2$.

(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 $1.967\, cm^{2}$.

(C) The area of a sector of a circle can be calculated using the formula:
Area $= \frac{1}{2} \times l \times r$
where $l$ is the arc length and $r$ is the radius of the circle.
Given:
Radius $(r) = 5\, cm$
Arc length $(l) = 3.5\, cm$
Substituting the values into the formula:
Area $= \frac{1}{2} \times 3.5 \times 5$
Area $= \frac{17.5}{2} = 8.75\, cm^2$
Hence,the required area of the sector is $8.75\, cm^2$.

(D) Given that four circular cardboard pieces are placed on a paper such that each piece touches the other two pieces.
Let the centers of the four circles be $A, B, C,$ and $D$. Since the radius of each circle is $7 \, cm$, the distance between the centers of any two touching circles is $7 + 7 = 14 \, cm$.
Thus, $AB = BC = CD = DA = 14 \, cm$. This forms a square $ABCD$ with side length $14 \, cm$.
The area of the square $ABCD = (side)^2 = (14)^2 = 196 \, cm^2$.
Inside the square, there are four sectors, each with a central angle of $90^{\circ}$ and radius $r = 7 \, cm$.
The area of one sector $= \frac{90^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (7)^2 = \frac{1}{4} \times 22 \times 7 = 38.5 \, cm^2$.
The total area of the four sectors $= 4 \times 38.5 = 154 \, cm^2$.
The area of the portion enclosed between these pieces = (Area of square $ABCD$) - (Total area of four sectors)
$= 196 - 154 = 42 \, cm^2$.
Therefore, the required area is $42 \, cm^2$.

(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 $800 \, cm^{2}$.

(D) Let the diameters of the concentric circles be $k, 2k$,and $3k$.
Therefore,the radii of the concentric circles are $\frac{k}{2}, k$,and $\frac{3k}{2}$.
Area of the inner circle,$A_1 = \pi \left(\frac{k}{2}\right)^2 = \frac{\pi k^2}{4}$.
Area of the middle region,$A_2 = \pi(k)^2 - \frac{\pi k^2}{4} = \frac{3\pi k^2}{4}$.
Area of the outer region,$A_3 = \pi \left(\frac{3k}{2}\right)^2 - \pi(k)^2 = \frac{9\pi k^2}{4} - \pi k^2 = \frac{5\pi k^2}{4}$.
The required ratio of the areas is $A_1 : A_2 : A_3 = \frac{\pi k^2}{4} : \frac{3\pi k^2}{4} : \frac{5\pi k^2}{4} = 1 : 3 : 5$.

(A) We know that,in $60\, \text{min}$,the minute hand revolves $360^{\circ}$.
In $1\, \text{min}$,the minute hand revolves $\frac{360^{\circ}}{60} = 6^{\circ}$.
Therefore,in the time interval from $6:05\, a.m.$ to $6:40\, a.m.$,the time elapsed is $35\, \text{min}$.
The angle swept by the minute hand in $35\, \text{min}$ is $\theta = 6^{\circ} \times 35 = 210^{\circ}$.
Given that the length of the minute hand $(r) = 5\, \text{cm}$.
The area swept by the minute hand is the area of a sector with radius $r = 5\, \text{cm}$ and angle $\theta = 210^{\circ}$.
Area $= \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{210^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (5)^2$.
Area $= \frac{7}{12} \times \frac{22}{7} \times 25 = \frac{22 \times 25}{12} = \frac{11 \times 25}{6} = \frac{275}{6}$.
Converting to a mixed fraction,$\frac{275}{6} = 45 \frac{5}{6}\, \text{cm}^2$.
Hence,the required area swept by the minute hand is $45 \frac{5}{6}\, \text{cm}^2$.

(B) Let the radius of the circle be $r$ and the central angle be $\theta = 200^{\circ}$.
Area of the sector is given by the formula: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Substituting the given values: $770 = \frac{200}{360} \times \frac{22}{7} \times r^{2}$.
$770 = \frac{5}{9} \times \frac{22}{7} \times r^{2}$.
$r^{2} = \frac{770 \times 9 \times 7}{5 \times 22} = \frac{770 \times 63}{110} = 7 \times 63 = 441$.
$r = \sqrt{441} = 21 \, cm$.
Now,the length of the arc $l$ is given by: $l = \frac{\theta}{360^{\circ}} \times 2\pi r$.
$l = \frac{200}{360} \times 2 \times \frac{22}{7} \times 21$.
$l = \frac{5}{9} \times 2 \times 22 \times 3 = \frac{5 \times 2 \times 22}{3} = \frac{220}{3} \, cm$.
$l = 73 \frac{1}{3} \, cm$.

(N/A) Let the radii of the two sectors be $r_1 = 7 \, cm$ and $r_2 = 21 \, cm$,and their central angles be $\theta_1 = 120^{\circ}$ and $\theta_2 = 40^{\circ}$.
Area of a sector is given by $A = \frac{\theta}{360^{\circ}} \times \pi r^2$.
For the first sector:
$A_1 = \frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (7)^2 = \frac{1}{3} \times \frac{22}{7} \times 49 = \frac{22 \times 7}{3} = \frac{154}{3} \approx 51.33 \, cm^2$.
For the second sector:
$A_2 = \frac{40^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (21)^2 = \frac{1}{9} \times \frac{22}{7} \times 441 = \frac{22 \times 63}{9} = 22 \times 7 = 154 \, cm^2$.
Length of an arc is given by $l = \frac{\theta}{360^{\circ}} \times 2\pi r$.
For the first sector:
$l_1 = \frac{120^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 7 = \frac{1}{3} \times 44 = \frac{44}{3} \approx 14.67 \, cm$.
For the second sector:
$l_2 = \frac{40^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{9} \times 2 \times 22 \times 3 = \frac{132}{9} = \frac{44}{3} \approx 14.67 \, cm$.
Observation: The arc lengths of the two sectors are equal,but their areas are not equal.

(N/A) The side of the outer square is $14 \, cm$.
The distance from each side of the square to the inner figure is $3 \, cm$.
The inner figure consists of a central square and four semi-circles attached to its sides.
The side length of the inner square is $14 - (3 + 3) = 8 \, cm$.
Since there are four semi-circles attached to the sides of this $8 \, cm$ square,the diameter of each semi-circle is $8 / 2 = 4 \, cm$,so the radius $r = 2 \, cm$.
Area of the outer square $= 14^2 = 196 \, cm^2$.
Area of the inner square $= 8^2 = 64 \, cm^2$.
Area of the four semi-circles $= 4 \times (\frac{1}{2} \pi r^2) = 2 \pi r^2 = 2 \times \pi \times (2)^2 = 8 \pi \, cm^2$.
Area of the unshaded inner region $= \text{Area of inner square} + \text{Area of four semi-circles} = 64 + 8 \pi \, cm^2$.
Area of the shaded region $= \text{Area of outer square} - \text{Area of unshaded inner region} = 196 - (64 + 8 \pi) = 132 - 8 \pi \, cm^2$.

(A) Let the number of revolutions made by the circular wheel be $n$ and the radius be $r$.
Given,area of the circular wheel $= 1.54 \, m^2$.
Using the formula for the area of a circle,$\pi r^2 = 1.54$.
$\frac{22}{7} \times r^2 = 1.54 \Rightarrow r^2 = \frac{1.54 \times 7}{22} = 0.07 \times 7 = 0.49$.
Thus,$r = \sqrt{0.49} = 0.7 \, m$.
The distance covered in one revolution is equal to the circumference of the wheel.
Circumference $= 2 \pi r = 2 \times \frac{22}{7} \times 0.7 = 2 \times 22 \times 0.1 = 4.4 \, m$.
The total distance traveled is $176 \, m$.
Number of revolutions $n = \frac{\text{Total distance}}{\text{Circumference}} = \frac{176}{4.4} = \frac{1760}{44} = 40$.
Therefore,the wheel makes $40$ revolutions.

(D) Let the radius of the circle be $r$.
$OA = OB = r \, cm$.
Given that the length of the chord $AB = 5 \, cm$ and the central angle $\theta = 90^{\circ}$.
In $\triangle AOB$,by Pythagoras theorem,$AB^2 = OA^2 + OB^2 \implies 5^2 = r^2 + r^2 \implies 2r^2 = 25 \implies r^2 = 12.5$.
Area of $\triangle AOB = \frac{1}{2} \times OA \times OB \times \sin(90^{\circ}) = \frac{1}{2} \times r^2 \times 1 = \frac{12.5}{2} = 6.25 \, cm^2$.
Area of sector $OAB = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{90^{\circ}}{360^{\circ}} \times \pi \times 12.5 = \frac{12.5 \pi}{4} = 3.125 \pi \, cm^2$.
Area of minor segment = Area of sector - Area of $\triangle AOB = (3.125 \pi - 6.25) \, cm^2$.
Area of circle = $\pi r^2 = 12.5 \pi \, cm^2$.
Area of major segment = Area of circle - Area of minor segment = $12.5 \pi - (3.125 \pi - 6.25) = (9.375 \pi + 6.25) \, cm^2$.
Difference of areas = $(9.375 \pi + 6.25) - (3.125 \pi - 6.25) = 6.25 \pi + 12.5 = (6.25 \pi + 12.5) \, cm^2$ or $(\frac{25 \pi}{4} + \frac{25}{2}) \, cm^2$.

(C) Given that,the radius of the circle $(r) = 21 \, cm$ and the central angle of the minor sector $(\theta) = 120^{\circ}$.
First,calculate the total area of the circle:
Area $= \pi r^2 = \frac{22}{7} \times (21)^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386 \, cm^2$.
Next,calculate the area of the minor sector:
Area of minor sector $= \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{120^{\circ}}{360^{\circ}} \times 1386 = \frac{1}{3} \times 1386 = 462 \, cm^2$.
Then,calculate the area of the major sector:
Area of major sector $= \text{Total area} - \text{Area of minor sector} = 1386 - 462 = 924 \, cm^2$.
Finally,find the difference between the areas of the major sector and the minor sector:
Difference $= 924 - 462 = 462 \, cm^2$.

(N/A) Given,the radius of the circular ground is $r = 77\, m$.
Circumference of a circle is given by the formula $C = 2\pi r$.
Substituting the values,we get $C = 2 \times \frac{22}{7} \times 77 = 2 \times 22 \times 11 = 484\, m$.
Area of a circle is given by the formula $A = \pi r^2$.
Substituting the values,we get $A = \frac{22}{7} \times 77 \times 77 = 22 \times 11 \times 77 = 18634\, m^2$.
Thus,the circumference of the circular ground is $484\, m$ and the area is $18634\, m^2$.

(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 $80 \, cm$.

(B) The area of a circle is given by the formula $A = \pi r^{2}$.
Given,$A = 5544 \, cm^{2}$.
Substituting the value of $\pi = \frac{22}{7}$:
$5544 = \frac{22}{7} \times r^{2}$
$r^{2} = 5544 \times \frac{7}{22}$
$r^{2} = 252 \times 7$
$r^{2} = 1764$
Taking the square root on both sides:
$r = \sqrt{1764} = 42$
Thus,the radius of the circle is $42 \, cm$.

(C) Let the radius of the circular ground be $r_{1}$ and the radius of the circular ground including the road be $r_{2}$.
Given,the circumference of the ground $= 2 \pi r_{1} = 220\, m$.
Given,the circumference of the ground with the road $= 2 \pi r_{2} = 264\, m$.
The width of the road is given by $w = r_{2} - r_{1}$.
Subtracting the two circumferences:
$2 \pi r_{2} - 2 \pi r_{1} = 264 - 220$
$2 \pi (r_{2} - r_{1}) = 44$
$2 \times \frac{22}{7} \times w = 44$
$w = \frac{44 \times 7}{2 \times 22}$
$w = \frac{44 \times 7}{44}$
$w = 7\, m$.
Thus,the width of the road is $7\, m$.

(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) The radius of the outer circle is $r_{1} = 23 \, cm$ and the radius of the inner circle is $r_{2} = 16 \, cm$.
The area of the circular ring is given by the difference between the area of the outer circle and the area of the inner circle.
$\text{Area} = \pi r_{1}^{2} - \pi r_{2}^{2}$
$\text{Area} = \pi (r_{1}^{2} - r_{2}^{2})$
Using the identity $a^{2} - b^{2} = (a + b)(a - b)$:
$\text{Area} = \pi (r_{1} + r_{2})(r_{1} - r_{2})$
Substituting the values with $\pi = \frac{22}{7}$:
$\text{Area} = \frac{22}{7} \times (23 + 16) \times (23 - 16)$
$\text{Area} = \frac{22}{7} \times 39 \times 7$
$\text{Area} = 22 \times 39 = 858 \, cm^{2}$

(A) Given diameter $d = 42 \, cm$.
Radius $r = d / 2 = 42 / 2 = 21 \, cm$.
Circumference of the circle $= 2 \pi r = 2 \times (22 / 7) \times 21 = 2 \times 22 \times 3 = 132 \, cm$.
Area of the circle $= \pi r^2 = (22 / 7) \times 21 \times 21 = 22 \times 3 \times 21 = 1386 \, cm^2$.
Thus,the circumference is $132 \, cm$ and the area is $1386 \, cm^2$.

(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 $452.16 \, cm^2$.

(D) The formula for the circumference of a circle is $C = 2 \pi r$,where $C$ is the circumference and $r$ is the radius.
Given $C = 176 \, cm$ and taking $\pi = \frac{22}{7}$:
$176 = 2 \times \frac{22}{7} \times r$
$176 = \frac{44}{7} \times r$
$r = \frac{176 \times 7}{44}$
$r = 4 \times 7$
$r = 28 \, cm$
Therefore,the radius of the circle is $28 \, cm$.

(A) The area of a circle is given by the formula $A = \pi r^2$,where $r$ is the radius.
Given $A = 346.5 \, cm^2$ and taking $\pi = \frac{22}{7}$:
$346.5 = \frac{22}{7} \times r^2$
$r^2 = \frac{346.5 \times 7}{22}$
$r^2 = \frac{2425.5}{22}$
$r^2 = 110.25$
$r = \sqrt{110.25}$
$r = 10.5 \, cm$.

(B) Given,the circumference of the circular ground $C = 352\, m$.
We know that the formula for circumference is $C = 2\pi r$,where $r$ is the radius.
$2 \times \frac{22}{7} \times r = 352$
$r = \frac{352 \times 7}{2 \times 22}$
$r = \frac{352 \times 7}{44} = 8 \times 7 = 56\, m$.
Now,the area of the circular ground $A = \pi r^2$.
$A = \frac{22}{7} \times 56 \times 56$
$A = 22 \times 8 \times 56$
$A = 176 \times 56 = 9856\, m^2$.

(C) Given,the area of the circle $A = 75.46\, cm^{2}$.
We know that the area of a circle is given by the formula $A = \pi r^{2}$.
Taking $\pi = 3.14$,we have $75.46 = 3.14 \times r^{2}$.
$r^{2} = \frac{75.46}{3.14} = 24.0318... \approx 24.01$.
Taking the square root,$r = \sqrt{24.01} = 4.9\, cm$.
The circumference of the circle is given by $C = 2\pi r$.
$C = 2 \times 3.14 \times 4.9 = 6.28 \times 4.9 = 30.772\, cm$.
Rounding to one decimal place,the circumference is $30.8\, cm$.

(D) Let the radius of the circular ground be $r = 35 \, m$.
The width of the road is $w = 3.5 \, m$.
The radius of the outer circle (ground + road) is $R = r + w = 35 + 3.5 = 38.5 \, m$.
The area of the road is the difference between the area of the outer circle and the area of the inner circular ground.
Area of road $= \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi(R - r)(R + r)$.
Substituting the values: $\text{Area} = \frac{22}{7} \times (38.5 - 35) \times (38.5 + 35)$.
$\text{Area} = \frac{22}{7} \times 3.5 \times 73.5$.
$\text{Area} = 22 \times 0.5 \times 73.5 = 11 \times 73.5 = 808.5 \, m^2$.

(A) The radius of the outer circle $(R)$ is $56 \, m$.
The width of the road is $7 \, m$.
The radius of the inner circle $(r)$ is $R - \text{width} = 56 \, m - 7 \, m = 49 \, m$.
The area of the road is the difference between the area of the outer circle and the area of the inner circle.
Area of road $= \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi(R - r)(R + r)$.
Substituting the values: $\text{Area} = \frac{22}{7} \times (56 - 49) \times (56 + 49)$.
$\text{Area} = \frac{22}{7} \times 7 \times 105$.
$\text{Area} = 22 \times 105 = 2310 \, m^2$.

(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 $12,474 \, m^2 \times ₹ 40/m^2 = ₹ 4,98,960$.

(C) $1$. The circumference of the circular ground is given by: $\text{Circumference} = \frac{\text{Total Cost}}{\text{Rate per meter}} = \frac{26,400}{60} = 440 \ m$.
$2$. Let the radius of the ground be $r$. The formula for circumference is $2\pi r = 440$.
$3$. Using $\pi = \frac{22}{7}$, we get $2 \times \frac{22}{7} \times r = 440$, which simplifies to $r = \frac{440 \times 7}{44} = 70 \ m$.
$4$. The area of the circular ground is $A = \pi r^2 = \frac{22}{7} \times 70 \times 70 = 22 \times 10 \times 70 = 15,400 \ m^2$.
$5$. The cost of levelling the ground at $₹ 50/m^2$ is $15,400 \times 50 = ₹ 7,70,000$.

Given: Radius of the circle $r = 6 \, cm$ and the central angle $\theta = 60^{\circ}$.
Area of the minor sector $= \frac{\pi r^2 \theta}{360^{\circ}}$
$= \frac{22}{7} \times 6 \times 6 \times \frac{60}{360}$
$= \frac{22}{7} \times 36 \times \frac{1}{6}$
$= \frac{22 \times 6}{7} = \frac{132}{7} \, cm^2 \approx 18.86 \, cm^2$.
Area of the major sector $= \text{Area of the circle} - \text{Area of the minor sector}$
$= \pi r^2 - \frac{132}{7}$
$= \frac{22}{7} \times 6 \times 6 - \frac{132}{7}$
$= \frac{792}{7} - \frac{132}{7} = \frac{660}{7} \, cm^2 \approx 94.29 \, cm^2$.
Thus,the area of the minor sector is $\frac{132}{7} \, cm^2$ and the area of the major sector is $\frac{660}{7} \, cm^2$.

(A) The measure of the angle of the minor sector is $\theta = 90^{\circ}$ (since $OA \perp OB$).
Perimeter of a minor sector = Length of the minor arc $+ 2 \times$ (radius).
$75 = \frac{\pi r \theta}{180} + 2r$
$75 = \frac{22}{7} \times \frac{r \times 90}{180} + 2r$
$75 = r \left( \frac{11}{7} + 2 \right)$
$75 = \frac{25}{7} r$
$r = \frac{75 \times 7}{25} = 21 \, cm$.
Area of minor sector $OACB = \frac{\pi r^2 \theta}{360} = \frac{22}{7} \times \frac{21 \times 21 \times 90}{360} = 346.5 \, cm^2$.
In $\Delta OAB$,$m\angle O = 90^{\circ}$,so Area of $\Delta OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 21 \times 21 = 220.5 \, cm^2$.
Area of minor segment = Area of minor sector $OACB - \text{Area of } \Delta OAB = 346.5 - 220.5 = 126 \, cm^2$.

(A) During $60$ minutes ($1$ hour),the minute hand completes one full rotation,i.e.,it subtends an angle of $360^{\circ}$.
The time duration from $10.10 \, AM$ to $10.30 \, AM$ is $20$ minutes.
Therefore,the angle $\theta$ subtended by the minute hand in $20$ minutes is:
$\theta = \frac{360^{\circ}}{60} \times 20 = 120^{\circ}$.
The radius $r$ of the sector is the length of the minute hand,which is $14 \, cm$.
The area of the sector swept is given by the formula:
$\text{Area} = \frac{\pi r^2 \theta}{360^{\circ}}$
Substituting the values:
$\text{Area} = \frac{22}{7} \times 14 \times 14 \times \frac{120^{\circ}}{360^{\circ}}$
$\text{Area} = \frac{22}{7} \times 196 \times \frac{1}{3}$
$\text{Area} = 22 \times 28 \times \frac{1}{3} = \frac{616}{3} \, cm^2$
$\text{Area} = 205 \frac{1}{3} \, cm^2$.

(C) The circular ground has a radius $R = 35 \, m$. $A$ road of width $w = 3.5 \, m$ runs inside the boundary. The inner radius is $r = R - w = 35 - 3.5 = 31.5 \, m$.
The area to be repaired is the area of the sector of the outer circle minus the area of the corresponding sector of the inner circle,with a central angle $\theta = 72^{\circ}$.
Area of the region $= \frac{\theta}{360^{\circ}} \times \pi \times (R^2 - r^2)$
$= \frac{72}{360} \times \frac{22}{7} \times (35^2 - 31.5^2)$
$= \frac{1}{5} \times \frac{22}{7} \times (35 - 31.5)(35 + 31.5)$
$= \frac{22}{35} \times 3.5 \times 66.5$
$= \frac{22}{35} \times \frac{7}{2} \times 66.5$
$= 11 \times 0.2 \times 66.5 = 2.2 \times 66.5 = 146.3 \, m^{2}$.
The cost of repairing is $146.3 \, m^{2} \times ₹ 80 / m^{2} = ₹ 11,704$.

(D) Suppose the cow is tethered at the vertex $A$ of the square field $ABCD$ by a $3 \, m$ long rope.
Since the angle at each vertex of a square is $90^{\circ}$,the area the cow can graze is a sector of a circle with radius $r = 3 \, m$ and central angle $\theta = 90^{\circ}$.
Area of the sector $= \frac{\pi r^2 \theta}{360^{\circ}}$
$= \frac{3.14 \times (3)^2 \times 90^{\circ}}{360^{\circ}}$
$= \frac{3.14 \times 9 \times 1}{4}$
$= \frac{28.26}{4}$
$= 7.065 \, m^2$
Thus,the area of the region in which the cow can graze is $7.065 \, m^2$.

(A) Given: Radius $r = 6.3 \ cm$,Central angle $\theta = 150^{\circ}$.
Length of the arc $l = \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{150}{360} \times 2 \times \frac{22}{7} \times 6.3 = \frac{5}{12} \times 2 \times 22 \times 0.9 = 16.5 \ cm$.
Area of the sector $A = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{150}{360} \times \frac{22}{7} \times 6.3 \times 6.3 = \frac{5}{12} \times 22 \times 0.9 \times 6.3 = 51.975 \ cm^2$.

(A) Given: Radius $r = 42 \, cm$ and central angle $\theta = 120^{\circ}$.
Length of the arc $l = \frac{\theta}{360^{\circ}} \times 2\pi r = \frac{120^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 42 = \frac{1}{3} \times 2 \times 22 \times 6 = 88 \, cm$.
Area of the sector $A = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{3} \times 22 \times 6 \times 42 = 22 \times 2 \times 42 = 1848 \, cm^2$.

(A) Given: Radius $r = 28 \ cm$,Arc length $l = 22 \ cm$.
$1$. To find the angle $\theta$ subtended at the centre:
The formula for arc length is $l = \frac{\theta}{360^{\circ}} \times 2\pi r$.
$22 = \frac{\theta}{360^{\circ}} \times 2 \times \frac{22}{7} \times 28$.
$22 = \frac{\theta}{360^{\circ}} \times 176$.
$\theta = \frac{22 \times 360^{\circ}}{176} = \frac{360^{\circ}}{8} = 45^{\circ}$.
$2$. To find the area of the sector:
The formula for the area of a sector is $A = \frac{\theta}{360^{\circ}} \times \pi r^2$.
$A = \frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 28 \times 28$.
$A = \frac{1}{8} \times 22 \times 4 \times 28 = 11 \times 28 = 308 \ cm^2$.

(D) The area of a sector of a circle is given by the formula: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$.
Since the two radii are perpendicular to each other,the central angle $\theta = 90^\circ$.
Given,$\text{Area} = 38.5 \, cm^2$ and $\pi = \frac{22}{7}$.
Substituting the values: $38.5 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times r^2$.
$38.5 = \frac{1}{4} \times \frac{22}{7} \times r^2$.
$38.5 = \frac{11}{14} \times r^2$.
$r^2 = \frac{38.5 \times 14}{11}$.
$r^2 = 3.5 \times 14 = 49$.
$r = \sqrt{49} = 7 \, cm$.

(A) Given: Radius $r = 14 \,cm$,Angle $\theta = 90^\circ$.
$1$. Length of arc = $\frac{\theta}{360^\circ} \times 2\pi r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22 \,cm$.
$2$. Area of minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 22 \times 28 = 154 \,cm^2$.
$3$. Area of minor segment = Area of sector - Area of triangle = $154 - \frac{1}{2} \times r^2 \times \sin(90^\circ) = 154 - \frac{1}{2} \times 14 \times 14 \times 1 = 154 - 98 = 56 \,cm^2$.

(N/A) Given: Radius $r = 21 \, cm$,Arc length $l = 33 \, cm$.
$1$. Angle subtended at the centre $(\theta)$: The formula for arc length is $l = \frac{\theta}{360^\circ} \times 2\pi r$. Substituting the values: $33 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 21$. Solving for $\theta$: $33 = \theta \times \frac{44}{60} \implies \theta = \frac{33 \times 60}{44} = 45^\circ$.
$2$. Area of the minor sector: $A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{45}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{8} \times 22 \times 3 \times 21 = 346.5 \, cm^2$.
$3$. Area of the minor segment: $A_{segment} = A_{sector} - A_{triangle} = 346.5 - \frac{1}{2} r^2 \sin(\theta) = 346.5 - \frac{1}{2} \times 21 \times 21 \times \sin(90^\circ)$. Note: The central angle is $90^\circ$ if calculated as $l = \frac{\theta}{360} \times 2\pi r \implies 33 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 21 \implies 33 = \theta \times \frac{132}{360} \implies \theta = 90^\circ$. Area of triangle $= \frac{1}{2} \times 21 \times 21 = 220.5 \, cm^2$. Area of segment $= 346.5 - 220.5 = 126 \, cm^2$.

(A) Given radius $r = 56 \, cm$. Since the two radii are perpendicular,the central angle $\theta = 90^\circ$.
Area of the minor sector = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 56 \times 56 = \frac{1}{4} \times 22 \times 8 \times 56 = 2464 \, cm^2$.
Area of the major sector = Total area of circle - Area of minor sector = $\pi r^2 - 2464 = \frac{22}{7} \times 56 \times 56 - 2464 = 9856 - 2464 = 7392 \, cm^2$.
Area of the minor segment = Area of minor sector - Area of triangle = $2464 - \frac{1}{2} \times r^2 \times \sin(90^\circ) = 2464 - \frac{1}{2} \times 56 \times 56 \times 1 = 2464 - 1568 = 896 \, cm^2$.

(A) Given: Radius $r = 42 \ cm$,Central angle $\theta = 60^{\circ}$.
$1$. Area of minor sector = $\frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{6} \times 22 \times 6 \times 42 = 22 \times 42 = 924 \ cm^2$.
$2$. Area of minor segment = Area of minor sector - Area of $\triangle OAB$.
Area of $\triangle OAB = \frac{1}{2} r^2 \sin(\theta) = \frac{1}{2} \times 42 \times 42 \times \sin(60^{\circ}) = 882 \times \frac{\sqrt{3}}{2} = 441 \times 1.73 = 762.93 \ cm^2$.
Area of minor segment = $924 - 762.93 = 161.07 \ cm^2$.