Areas of Sector and Segment of a Circle Questions in English

(B) The statement is False.
Let the radii of the two circles be $r_1$ and $r_2$,and the lengths of the arcs be $l_1$ and $l_2$. Given that the arc lengths are equal,we have $l_1 = l_2 = l$.
The formula for the length of an arc is $l = r \theta$,where $\theta$ is the angle in radians.
The area of a sector is given by $A = \frac{1}{2} r^2 \theta$.
Since $l = r \theta$,we can write $\theta = \frac{l}{r}$.
Substituting this into the area formula,we get $A = \frac{1}{2} r^2 (\frac{l}{r}) = \frac{1}{2} rl$.
For two different circles with the same arc length $l$,the areas are $A_1 = \frac{1}{2} r_1 l$ and $A_2 = \frac{1}{2} r_2 l$.
Since the circles are different,their radii $r_1$ and $r_2$ are not equal $(r_1 \neq r_2)$.
Therefore,$A_1 \neq A_2$. Thus,the areas are not necessarily equal.

(B) No,it is not necessary that their corresponding arc lengths are equal.
The area of a sector is given by $A = \frac{\theta}{360^\circ} \times \pi r^2$,where $r$ is the radius and $\theta$ is the central angle.
The arc length is given by $L = \frac{\theta}{360^\circ} \times 2\pi r$.
From the area formula,we have $\frac{\theta}{360^\circ} = \frac{A}{\pi r^2}$.
Substituting this into the arc length formula,we get $L = \frac{A}{\pi r^2} \times 2\pi r = \frac{2A}{r}$.
Since the areas $A$ are equal,the arc length $L$ depends inversely on the radius $r$. If the radii of the two circles are different,the arc lengths will be different even if the areas of the sectors are equal.

(B) The statement is False.
The largest circle that can be drawn inside a rectangle of length $a$ and breadth $b$ (where $a > b$) is limited by the smaller dimension of the rectangle,which is the breadth $b$.
The diameter of this largest circle is equal to the breadth of the rectangle,$d = b$.
Therefore,the radius of the circle is $r = \frac{b}{2}$.
The area of the circle is given by the formula $A = \pi r^{2}$.
Substituting the value of $r$,we get $A = \pi \left(\frac{b}{2}\right)^{2} = \frac{\pi b^{2}}{4} \, cm^{2}$.
Thus,the area is $\frac{\pi b^{2}}{4} \, cm^{2}$,not $\pi b^{2} \, cm^{2}$.

(D) The statement is false.
When a square is inscribed in a circle,the diagonal of the square is equal to the diameter of the circle.
Let the side of the square be $a$ and the diameter of the circle be $p$.
According to the property of a square,the diagonal $d = a\sqrt{2}$.
Since the diagonal equals the diameter,we have $a\sqrt{2} = p$,which implies $a = \frac{p}{\sqrt{2}}$.
The area of the square is $a^{2} = (\frac{p}{\sqrt{2}})^{2} = \frac{p^{2}}{2} \, cm^{2}$.
Therefore,the area is $\frac{1}{2} p^{2} \, cm^{2}$,not $p^{2} \, cm^{2}$.

(B) The formula for the area of a sector is given by: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Given,radius $r = 21 \, cm$ and central angle $\theta = 120^{\circ}$.
Substituting the values into the formula:
$\text{Area} = \frac{120}{360} \times \frac{22}{7} \times (21)^{2} \, cm^{2}$.
$\text{Area} = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \, cm^{2}$.
$\text{Area} = \frac{1}{3} \times 22 \times 3 \times 21 \, cm^{2}$.
$\text{Area} = 22 \times 21 \, cm^{2} = 462 \, cm^{2}$.

(A) Given: Radius $r = 36 \, cm$ and Area of the sector $A = 54 \pi \, cm^{2}$.
The formula for the area of a sector is $A = \frac{1}{2} \times l \times r$,where $l$ is the length of the arc.
Substituting the given values into the formula:
$54 \pi = \frac{1}{2} \times l \times 36$
$54 \pi = 18 \times l$
$l = \frac{54 \pi}{18}$
$l = 3 \pi \, cm$.
Thus,the length of the corresponding arc is $3 \pi \, cm$.

(A) Let the radius of the required circle be $r$.
The circumference of a circle is given by the formula $C = 2 \pi r$.
Let the radii of the two given circles be $r_1 = 15 \, cm$ and $r_2 = 18 \, cm$.
According to the problem,the circumference of the new circle is equal to the sum of the circumferences of the two given circles:
$2 \pi r = 2 \pi r_1 + 2 \pi r_2$
Dividing both sides by $2 \pi$,we get:
$r = r_1 + r_2$
Substituting the values:
$r = 15 \, cm + 18 \, cm$
$r = 33 \, cm$
Thus,the radius of the required circle is $33 \, cm$.

(C) Given that,the radius of the circle,$r = 28 \,cm$.
The measure of the central angle,$\theta = 45^{\circ}$.
The formula for the area of a sector of a circle is given by: $\text{Area} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$.
Substituting the given values:
$\text{Area} = \frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (28)^{2}$
$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 28 \times 28$
$\text{Area} = \frac{1}{8} \times 22 \times 4 \times 28$
$\text{Area} = \frac{1}{8} \times 22 \times 112$
$\text{Area} = 22 \times 14 = 308 \,cm^{2}$.
Thus,the area of the sector is $308 \,cm^{2}$.

(A) Let $ABCD$ be a rectangular field of dimensions $20 \, m \times 16 \, m$.
Suppose,a cow is tied at a corner $A$. Let the length of the rope be $r = 14 \, m$.
The area that the cow can graze is a sector of a circle with radius $r = 14 \, m$ and central angle $\theta = 90^{\circ}$ (since the corner of a rectangle is $90^{\circ}$).
Area of the sector $= \frac{\theta}{360^{\circ}} \times \pi r^2$
$= \frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (14)^2$
$= \frac{1}{4} \times \frac{22}{7} \times 196$
$= \frac{1}{4} \times 22 \times 28$
$= 22 \times 7 = 154 \, m^2$.

(N/A) The designed area is the common region between two sectors $BAEC$ and $DAFC$.
Area of sector $BAEC = \frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (8)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times 64$
$= \frac{22 \times 16}{7} = \frac{352}{7} \, cm^{2}$
Area of $\triangle BAC = \frac{1}{2} \times BA \times BC$
$= \frac{1}{2} \times 8 \times 8 = 32 \, cm^{2}$
Area of the designed portion $= 2 \times (\text{Area of segment } AEC)$
$= 2 \times (\text{Area of sector } BAEC - \text{Area of } \triangle BAC)$
$= 2 \times \left(\frac{352}{7} - 32\right) = 2 \left(\frac{352 - 224}{7}\right)$
$= \frac{2 \times 128}{7} = \frac{256}{7} \, cm^{2}$

(B) Given,radius of the circle $(r) = 14 \, cm$ and central angle $(\theta) = 60^{\circ}$.
The area of the minor segment is given by the formula: Area of sector - Area of triangle.
Area of sector $= \frac{\theta}{360^{\circ}} \times \pi r^{2} = \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times 22 \times 2 \times 14 = \frac{308}{3} \approx 102.67 \, cm^{2}$.
Since the central angle is $60^{\circ}$ and the two sides are radii,the triangle formed is an equilateral triangle.
Area of equilateral triangle $= \frac{\sqrt{3}}{4} \times (\text{side})^{2} = \frac{\sqrt{3}}{4} \times 14^{2} = \frac{\sqrt{3}}{4} \times 196 = 49 \sqrt{3} \, cm^{2} \approx 84.87 \, cm^{2}$.
Area of minor segment $= \frac{308}{3} - 49 \sqrt{3} \, cm^{2} \approx 102.67 - 84.87 = 17.8 \, cm^{2}$.
Note: The provided options seem to represent the area of the triangle rather than the segment. Based on the standard calculation for the area of the minor segment,the result is $\frac{308}{3} - 49 \sqrt{3}$. However,if the question implies the area of the triangle formed by the chord,the answer is $49 \sqrt{3}$.

(C) Given,side of square $ABCD = 12 \, cm$.
Since $P, Q, R, S$ are mid-points of sides $AB, BC, CD, DA$ respectively,the radius of each quadrant arc is $r = \frac{12}{2} = 6 \, cm$.
Area of one quadrant = $\frac{1}{4} \pi r^{2} = \frac{1}{4} \times 3.14 \times (6)^{2} = \frac{3.14 \times 36}{4} = 3.14 \times 9 = 28.26 \, cm^{2}$.
Area of four such quadrants = $4 \times 28.26 = 113.04 \, cm^{2}$.
Area of square $ABCD = (side)^{2} = (12)^{2} = 144 \, cm^{2}$.
Area of the shaded region = Area of square $ABCD - $ Area of four quadrants.
Area of the shaded region = $144 - 113.04 = 30.96 \, cm^{2}$.

(D) Given that $ABC$ is an equilateral triangle with side length $10 \, cm$.
Therefore,$\angle A = \angle B = \angle C = 60^{\circ}$.
Since $D, E$ and $F$ are mid-points of sides $BC, CA$ and $AB$ respectively,the radius of each arc is $r = 5 \, cm$.
The shaded region consists of three sectors,each with a central angle of $60^{\circ}$ and radius $r = 5 \, cm$.
Area of one sector $= \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{60^{\circ}}{360^{circ}} \times 3.14 \times (5)^2$.
Area of one sector $= \frac{1}{6} \times 3.14 \times 25 = \frac{78.5}{6} \approx 13.0833 \, cm^2$.
Total area of the shaded region $= 3 \times (\text{Area of one sector}) = 3 \times \frac{78.5}{6} = \frac{78.5}{2} = 39.25 \, cm^2$.

(A) Given that,the radius of each arc $(r)$ $= 14 \, cm$.
The area of a sector with central angle $\theta$ is given by $\frac{\theta}{360^{\circ}} \times \pi r^{2}$.
For the three sectors with central angles $\angle P$,$\angle Q$,and $\angle R$,the sum of their areas is:
Area $= \frac{\angle P}{360^{\circ}} \times \pi r^{2} + \frac{\angle Q}{360^{\circ}} \times \pi r^{2} + \frac{\angle R}{360^{\circ}} \times \pi r^{2}$
Area $= \frac{(\angle P + \angle Q + \angle R)}{360^{\circ}} \times \pi r^{2}$
Since the sum of the interior angles of a triangle is $180^{\circ}$,we have $\angle P + \angle Q + \angle R = 180^{\circ}$.
Area $= \frac{180^{\circ}}{360^{\circ}} \times \pi \times (14)^{2}$
Area $= \frac{1}{2} \times \frac{22}{7} \times 196$
Area $= \frac{1}{2} \times 22 \times 28 = 11 \times 28 = 308 \, cm^{2}$.
Thus,the area of the shaded region is $308 \, cm^{2}$.

(B) Given that,a circular park is surrounded by a road.
Width of the road $= 21\, m$.
Radius of the park $(r_i) = 105\, m$.
Therefore,the radius of the whole circular portion (park + road) is $r_e = 105 + 21 = 126\, m$.
Now,the area of the road $=$ Area of the whole circular portion $-$ Area of the circular park.
Area of road $= \pi r_e^2 - \pi r_i^2 = \pi(r_e^2 - r_i^2)$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
Area of road $= \frac{22}{7} \times (126^2 - 105^2) = \frac{22}{7} \times (126 + 105)(126 - 105)$.
Area of road $= \frac{22}{7} \times 231 \times 21$.
Area of road $= 22 \times 231 \times 3 = 15246\, m^2$.
Hence,the required area of the road is $15246\, m^2$.

(C) Given that the radius of each arc is $r = 21 \, cm$.
The area of a sector with central angle $\theta$ is given by $\frac{\theta}{360^{\circ}} \times \pi r^2$.
The sum of the areas of the four sectors at vertices $A, B, C,$ and $D$ is:
Area $= \frac{\angle A}{360^{\circ}} \times \pi r^2 + \frac{\angle B}{360^{\circ}} \times \pi r^2 + \frac{\angle C}{360^{\circ}} \times \pi r^2 + \frac{\angle D}{360^{\circ}} \times \pi r^2$
Area $= \frac{(\angle A + \angle B + \angle C + \angle D)}{360^{\circ}} \times \pi r^2$
Since the sum of the interior angles of any quadrilateral is $360^{\circ}$,we have:
Area $= \frac{360^{\circ}}{360^{\circ}} \times \pi r^2 = \pi r^2$
Substituting $r = 21 \, cm$ and $\pi = \frac{22}{7}$:
Area $= \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386 \, cm^2$.
Thus,the area of the shaded region is $1386 \, cm^2$.

(D) The length of the arc of the circle is given as $20 \,cm$.
The central angle subtended by the arc is $\theta = 60^{\circ}$.
The formula for the length of an arc is $L = \frac{\theta}{360^{\circ}} \times 2 \pi r$.
Substituting the given values into the formula:
$20 = \frac{60^{\circ}}{360^{\circ}} \times 2 \pi r$
Simplify the fraction:
$20 = \frac{1}{6} \times 2 \pi r$
$20 = \frac{\pi r}{3}$
Solving for $r$:
$r = \frac{20 \times 3}{\pi} = \frac{60}{\pi} \,cm$.
Thus,the radius of the circle is $\frac{60}{\pi} \,cm$.

(D) Given: Radius $r = 20\, cm$,central angle $\theta = 90^{\circ}$.
Area of the major sector $= \frac{360^{\circ} - 90^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{270}{360} \times 3.14 \times (20)^2 = \frac{3}{4} \times 3.14 \times 400 = 3 \times 3.14 \times 100 = 942\, cm^2$.
Area of $\triangle OAB = \frac{1}{2} \times r^2 \times \sin(90^{\circ}) = \frac{1}{2} \times 20 \times 20 \times 1 = 200\, cm^2$.
Area of the major segment = Area of the major sector + Area of $\triangle OAB = 942 + 200 = 1142\, cm^2$.

(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 $= 528 \times 50 = 26400 \, Rs.$

(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 $= (75.36 - 36\sqrt{3}) \, cm^2$.

(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) = $350 - (77 + 77) = 350 - 154 = 196 \, cm^2$.

(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 $42 \, cm^2$.

(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 $18634\, m^2$.

(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 $7\, m$.

(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 $1386 \, cm^2$.

(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: $\text{Area} = \frac{22}{7} \times (56 - 49) \times (56 + 49)$.
$\text{Area} = \frac{22}{7} \times 7 \times 105$.
$\text{Area} = 22 \times 105 = 2310 \, m^2$.

(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 $A = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{150}{360} \times \frac{22}{7} \times 6.3 \times 6.3 = \frac{5}{12} \times 22 \times 0.9 \times 6.3 = 51.975 \ cm^2$.